diff options
| author | priyanka | 2015-06-24 15:03:17 +0530 |
|---|---|---|
| committer | priyanka | 2015-06-24 15:03:17 +0530 |
| commit | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (patch) | |
| tree | ab291cffc65280e58ac82470ba63fbcca7805165 /608 | |
| download | Scilab-TBC-Uploads-b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b.tar.gz Scilab-TBC-Uploads-b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b.tar.bz2 Scilab-TBC-Uploads-b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b.zip | |
initial commit / add all books
Diffstat (limited to '608')
525 files changed, 11136 insertions, 0 deletions
diff --git a/608/CH1/EX1.01/1_01.sce b/608/CH1/EX1.01/1_01.sce new file mode 100755 index 000000000..a069ec0b2 --- /dev/null +++ b/608/CH1/EX1.01/1_01.sce @@ -0,0 +1,11 @@ +//Problem 1.01: If a current of 5 A flows for 2 minutes, find the quantity of electricity transferred.
+
+//initializing the variables:
+I = 5; // in Ampere
+t = 120; // in sec
+
+//calculation:
+Q = I*t
+
+printf("\n\nResult\n\n")
+printf("\nQ: %.0f coulomb(C)\n",Q)
\ No newline at end of file diff --git a/608/CH1/EX1.02/1_02.sce b/608/CH1/EX1.02/1_02.sce new file mode 100755 index 000000000..7bd5e03b5 --- /dev/null +++ b/608/CH1/EX1.02/1_02.sce @@ -0,0 +1,11 @@ +//Problem 1.02: A mass of 5000 g is accelerated at 2 m/s2 by a force. Determine the force needed.
+
+//initializing the variables:
+M = 5; // in Kg
+a = 2; // in m/s2
+
+//calculation:
+F = M*a
+
+printf("\n\nResult\n\n")
+printf("\nForce: %.0f Newton(N)\n",F)
\ No newline at end of file diff --git a/608/CH1/EX1.03/1_03.sce b/608/CH1/EX1.03/1_03.sce new file mode 100755 index 000000000..6d5aa090a --- /dev/null +++ b/608/CH1/EX1.03/1_03.sce @@ -0,0 +1,11 @@ +//Problem 1.03: Find the force acting vertically downwards on a mass of 200 g attached to a wire.
+
+//initializing the variables:
+M = 0.2; // in Kg
+g = 9.81; // in m/s2
+
+//calculation:
+F = M*g
+
+printf("\n\nResult\n\n")
+printf("\nForce: %.3f Newton(N)\n",F)
\ No newline at end of file diff --git a/608/CH1/EX1.04/1_04.sce b/608/CH1/EX1.04/1_04.sce new file mode 100755 index 000000000..197cc64f6 --- /dev/null +++ b/608/CH1/EX1.04/1_04.sce @@ -0,0 +1,13 @@ +//Problem 1.04: A portable machine requires a force of 200 N to move it. How much work is done if the machine is moved 20 m and what average power is utilized if the movement takes 25 s?
+
+//initializing the variables:
+F = 200; // in Newton
+d = 20; // in m
+t = 25; // in sec
+
+//calculation:
+W = F*d
+P = W/t
+
+printf("\n\nResult\n\n")
+printf("\nPower: %.0f watt(W)\n",P)
\ No newline at end of file diff --git a/608/CH1/EX1.05/1_05.sce b/608/CH1/EX1.05/1_05.sce new file mode 100755 index 000000000..585ab8840 --- /dev/null +++ b/608/CH1/EX1.05/1_05.sce @@ -0,0 +1,15 @@ +//Problem 1.05: A mass of 1000 kg is raised through a height of 10m in 20s. What is (a) the work done and (b) the power developed?
+
+//initializing the variables:
+M = 1000; // in Kg
+h = 10; // in m
+t = 20; // in sec
+g = 9.81 // in m/s2
+
+//calculation:
+W = M*g*h
+P = W/t
+
+printf("\n\nResult\n\n")
+printf("\nWork Done: %.0f Joule(J)\n",W)
+printf("\nPower: %.0f watt(W)\n",P)
\ No newline at end of file diff --git a/608/CH1/EX1.06/1_06.sce b/608/CH1/EX1.06/1_06.sce new file mode 100755 index 000000000..1b3d1ae5b --- /dev/null +++ b/608/CH1/EX1.06/1_06.sce @@ -0,0 +1,15 @@ +//Problem 1.06: Find the conductance of a conductor of resistance (a) 10 ohm , (b) 5 kohm and (c) 100 ohm.
+
+//initializing the variables:
+R1 = 10; // in ohm
+R2 = 5000; // in ohm
+R3 = 0.1; // in ohm
+//calculation:
+G1 = 1/R1
+G2 = 1/R2
+G3 = 1/R3
+
+printf("\n\nResult\n\n")
+printf("\nconductance(G1): %.1f seimen(S)n",G1)
+printf("\nconductance(G2): %.4f seimen(S)n",G2)
+printf("\nconductance(G3): %.0f seimen(S)n",G3)
\ No newline at end of file diff --git a/608/CH1/EX1.07/1_07.sce b/608/CH1/EX1.07/1_07.sce new file mode 100755 index 000000000..7fcac6714 --- /dev/null +++ b/608/CH1/EX1.07/1_07.sce @@ -0,0 +1,12 @@ +//Problem 1.07: A source e.m.f. of 5 V supplies a current of 3 A for 10 minutes. How much energy is provided in this time?
+
+//initializing the variables:
+V = 5; // in Volts
+I = 3; // in Ampere
+t = 600; // in sec
+//calculation:
+P = V*I
+E = P*t
+
+printf("\n\nResult\n\n")
+printf("\nEnergy(E): %.0f Joule(J)\n",E)
\ No newline at end of file diff --git a/608/CH1/EX1.08/1_08.sce b/608/CH1/EX1.08/1_08.sce new file mode 100755 index 000000000..a0ed81017 --- /dev/null +++ b/608/CH1/EX1.08/1_08.sce @@ -0,0 +1,14 @@ +//Problem 1.08: An electric heater consumes 1.8 MJ when connected to a 250 V supply for 30 minutes. Find the power rating of the heater and the current taken from the supply.
+
+//initializing the variables:
+E = 18E5; // in Joule
+V = 250; // in Volts
+t = 1800; // in sec
+
+//calculation:
+P = E/t
+I = P/V
+
+printf("\n\nResult\n\n")
+printf("\nPower(P): %.0f Watt(W)",P)
+printf("\nCurrent(I): %.0f Ampere(A)\n",I)
\ No newline at end of file diff --git a/608/CH10/EX10.01/10_01.sce b/608/CH10/EX10.01/10_01.sce new file mode 100755 index 000000000..badfbd465 --- /dev/null +++ b/608/CH10/EX10.01/10_01.sce @@ -0,0 +1,15 @@ +//Problem 10.01: A moving-coil instrument gives a f.s.d. when the current is 40 mA and its resistance is 25 ohms. Calculate the value of the shunt to be connected in parallel with the meter to enable it to be used as an ammeter for measuring currents up to 50 A.
+
+//initializing the variables:
+Ia = 0.040; // in Amperes
+I = 50; // in Amperes
+ra = 25; // in ohms
+
+
+//calculation:
+Is = I - Ia
+V = Ia*ra
+Rs = V/Is
+
+printf("\n\n Result \n\n")
+printf("\n value of the shunt to be connected in parallel = %.3E ohms",Rs)
\ No newline at end of file diff --git a/608/CH10/EX10.02/10_02.sce b/608/CH10/EX10.02/10_02.sce new file mode 100755 index 000000000..76cd8886b --- /dev/null +++ b/608/CH10/EX10.02/10_02.sce @@ -0,0 +1,13 @@ +//Problem 10.02: A moving-coil instrument having a resistance of 10 ohms, gives a f.s.d. when the current is 8 mA. Calculate the value of the multiplier to be connected in series with the instrument so that it can be used as a voltmeter for measuring p.d.s. up to 100 V.
+
+//initializing the variables:
+V = 100; // in volts
+I = 0.008; // in Amperes
+ra = 10; // in ohms
+
+
+//calculation:
+Rm = (V/I) - ra
+
+printf("\n\n Result \n\n")
+printf("\n value of the multiplier to be connected in series = %.3Eohms\n", Rm)
\ No newline at end of file diff --git a/608/CH10/EX10.03/10_03.sce b/608/CH10/EX10.03/10_03.sce new file mode 100755 index 000000000..09ea13730 --- /dev/null +++ b/608/CH10/EX10.03/10_03.sce @@ -0,0 +1,21 @@ +//Problem 10.03: Calculate the power dissipated by the voltmeter and by resistor R in Figure 10.9 when (a) R = 250 ohms (b) R = 2 Mohms. Assume that the voltmeter sensitivity (sometimes called figure of merit) is 10 kohms/V.
+
+//initializing the variables:
+fsd = 200; // in volts
+R1 = 250; // in ohms
+R2 = 2E6; // in ohms
+sensitivity = 10000; // in ohms/V
+
+//calculation:
+Rv = sensitivity*fsd
+Iv = V/Rv
+Pv = V*Iv
+I1 = V/R1
+P1 = V*I1
+I2 = V/R2
+P2 = V*I2
+
+printf("\n\n Result \n\n")
+printf("\n (a)the power dissipated by the voltmeter = %.2E W", Pv)
+printf("\n (b)the power dissipated by resistor 250 ohm = %.0f W", P1)
+printf("\n (c)the power dissipated by resistor 2 Mohm = %.2E W", P2)
\ No newline at end of file diff --git a/608/CH10/EX10.04/10_04.sce b/608/CH10/EX10.04/10_04.sce new file mode 100755 index 000000000..18711398b --- /dev/null +++ b/608/CH10/EX10.04/10_04.sce @@ -0,0 +1,19 @@ +//Problem 10.04: An ammeter has a f.s.d. of 100 mA and a resistance of 50 ohms. The ammeter is used to measure the current in a load of resistance 500 ohms when the supply voltage is 10 V. Calculate (a) the ammeter reading expected (neglecting its resistance), (b) the actual current in the circuit, (c) the power dissipaed in the ammeter, and (d) the power dissipated in the load.
+
+//initializing the variables:
+V = 10; // in volts
+fsd = 0.1; // in Amperes
+ra = 50; // in ohms
+R = 500; // in ohms
+
+//calculation:
+Ie = V/R
+Ia = V/(R + ra)
+Pa = Ia*Ia*ra
+PR = Ia*Ia*R
+
+printf("\n\n Result \n\n")
+printf("\n (a)expected ammeter reading = %.2E A\", Ie)
+printf("\n (b)Actual ammeter reading = %.2E A",Ia)
+printf("\n (c)Power dissipated in the ammeter = %.2E W", Pa)
+printf("\n (d)Power dissipated in the load resistor = %.2E W",PR)
\ No newline at end of file diff --git a/608/CH10/EX10.05/10_05.sce b/608/CH10/EX10.05/10_05.sce new file mode 100755 index 000000000..29219791b --- /dev/null +++ b/608/CH10/EX10.05/10_05.sce @@ -0,0 +1,17 @@ +//Problem 10.05: A voltmeter having a f.s.d. of 100 V and a sensitivity of 1.6 kohms/V is used to measure voltage V1 in the circuit of Figure 10.11. Determine (a) the value of voltage V1 with the voltmeter not connected, and (b) the voltage indicated by the voltmeter when connected between A and B.
+
+//initializing the variables:
+fsd = 100; // in volts
+R1 = 40E3; // in ohms
+R2 = 60E3; // in ohms
+sensitivity = 1600; // in ohms/V
+
+//calculation:
+V1 = (R1/(R1 + R2))*fsd
+Rv = fsd*sensitivity
+Rep = R1*Rv/(R1 + Rv)
+V1n = (Rep/(Rep + R2))*fsd
+
+printf("\n\n Result \n\n")
+printf("\n (a)the value of voltage V1 with the voltmeter6 not connected = %.0f V", V1)
+printf("\n (b)the voltage indicated by the voltmeter when connected between A and B = %.2f V",V1n)
\ No newline at end of file diff --git a/608/CH10/EX10.06/10_06.sce b/608/CH10/EX10.06/10_06.sce new file mode 100755 index 000000000..288f5bf57 --- /dev/null +++ b/608/CH10/EX10.06/10_06.sce @@ -0,0 +1,15 @@ +//Problem 10.06: (a) A current of 20 A flows through a load having a resistance of 2 ohms. Determine the power dissipated in the load. (b) A wattmeter, whose current coil has a resistance of 0.01 ohm is connected as shown in Figure 10.13. Determine the wattmeter reading.
+
+//initializing the variables:
+I = 20; // in amperes
+R = 2; // in ohms
+Rw = 0.01; // in ohms
+
+//calculation:
+PR = I*I*R
+Rt = R + Rw
+Pw = I*I*Rt
+
+printf("\n\n Result \n\n")
+printf("\n (a)the power dissipated in the load = %.0f W", PR)
+printf("\n (b)the wattmeter reading. = %.0f W\n",Pw)
\ No newline at end of file diff --git a/608/CH10/EX10.08/10_08.sce b/608/CH10/EX10.08/10_08.sce new file mode 100755 index 000000000..04d16904d --- /dev/null +++ b/608/CH10/EX10.08/10_08.sce @@ -0,0 +1,18 @@ +//Problem 10.08: (For the c.r.o. square voltage waveform shown in Figure 10.15 determine (a) the periodic time, (b) the frequency and (c) the peak-to-peak voltage. The ‘time/cm’ (or timebase control) switch is on 100 μs/cm and the ‘volts/cm’ (or signal amplitude control) switch is on 20 V/cm.
+//(In Figures 10.15 to 10.18 assume that the squares shown are 1 cm by 1 cm)
+
+//initializing the variables:
+tc = 100E-6; // in s/cm
+Vc = 20; // in V/cm
+w = 5.2; // in cm ( width of one complete cycle )
+h = 3.6; // in cm ( peak-to-peak height of the display )
+
+//calculation:
+T = w*tc
+f = 1/T
+ptpv = h*Vc
+
+printf("\n\n Result \n\n")
+printf("\n (a)the periodic time, T = %.2E sec", T)
+printf("\n (b)Frequency, f = %.0f Hz",f)
+printf("\n (c)the peak-to-peak voltage = %.0f V",ptpv)
\ No newline at end of file diff --git a/608/CH10/EX10.09/10_09.sce b/608/CH10/EX10.09/10_09.sce new file mode 100755 index 000000000..6c7c70aba --- /dev/null +++ b/608/CH10/EX10.09/10_09.sce @@ -0,0 +1,18 @@ +//Problem 10.09: (For the c.r.o. display of a pulse waveform shown in Figure 10.16 the ‘time/cm’ switch is on 50 ms/cm and the ‘volts/cm’ switch is on 0.2 V/cm. Determine (a) the periodic time, (b) the frequency, (c) the magnitude of the pulse voltage.
+//(In Figures 10.15 to 10.18 assume that the squares shown are 1 cm by 1 cm)
+
+//initializing the variables:
+tc = 50E-3; // in s/cm
+Vc = 0.2; // in V/cm
+w = 3.5; // in cm ( width of one complete cycle )
+h = 3.4; // in cm ( peak-to-peak height of the display )
+
+//calculation:
+T = w*tc
+f = 1/T
+ptpv = h*Vc
+
+printf("\n\n Result \n\n")
+printf("\n (a)the periodic time, T = %.2E sec ",T)
+printf("\n (b)Frequency, f = %.2f Hz",f)
+printf("\n (c)the peak-to-peak voltage = %.2f V",ptpv)
\ No newline at end of file diff --git a/608/CH10/EX10.10/10_10.sce b/608/CH10/EX10.10/10_10.sce new file mode 100755 index 000000000..8dc67688a --- /dev/null +++ b/608/CH10/EX10.10/10_10.sce @@ -0,0 +1,21 @@ +//Problem 10.10: A sinusoidal voltage trace displayed by a c.r.o. is shown in Figure 10.17. If the ‘time/cm’ switch is on 500 μs/cm and the ‘volts/cm’ switch is on 5 V/cm, find, for the waveform, (a) the frequency, (b) the peak-to-peak voltage, (c) the amplitude, (d) the r.m.s. value.
+//(In Figures 10.15 to 10.18 assume that the squares shown are 1 cm by 1 cm)
+
+//initializing the variables:
+tc = 500E-6; // in s/cm
+Vc = 5; // in V/cm
+w = 4; // in cm ( width of one complete cycle )
+h = 5; // in cm ( peak-to-peak height of the display )
+
+//calculation:
+T = w*tc
+f = 1/T
+ptpv = h*Vc
+Amp = ptpv/2
+Vrms = Amp/(2^0.5)
+
+printf("\n\n Result \n\n")
+printf("\n (a)Frequency, f = %.0f Hz",f)
+printf("\n (b)the peak-to-peak voltage = %.0f V",ptpv)
+printf("\n (c)Amplitude = %.1f V",Amp)
+printf("\n (d)r.m.s voltage = %.2f V",Vrms)
\ No newline at end of file diff --git a/608/CH10/EX10.11/10_11.sce b/608/CH10/EX10.11/10_11.sce new file mode 100755 index 000000000..a4c2f2d8d --- /dev/null +++ b/608/CH10/EX10.11/10_11.sce @@ -0,0 +1,24 @@ +//Problem 10.11: For the double-beam oscilloscope displays shown in Figure 10.18 determine (a) their frequency, (b) their r.m.s. values, (c) their phase difference. The ‘time/cm’ switch is on 100 μs/cm and the ‘volts/cm’ switch on 2 V/cm.
+//(In Figures 10.15 to 10.18 assume that the squares shown are 1 cm by 1 cm)
+
+//initializing the variables:
+tc = 100E-6; // in s/cm
+Vc = 2; // in V/cm
+w = 5; // in cm ( width of one complete cycle for both waveform )
+h1 = 2; // in cm ( peak-to-peak height of the display )
+h2 = 2.5; // in cm ( peak-to-peak height of the display )
+
+//calculation:
+T = w*tc
+f = 1/T
+ptpv1 = h1*Vc
+Vrms1 = ptpv1/(2^0.5)
+ptpv2 = h2*Vc
+Vrms2 = ptpv2/(2^0.5)
+phi = 0.5*360/w
+
+printf("\n\n Result \n\n")
+printf("\n (a)Frequency, f = %.0f Hz",f)
+printf("\n (b1)r.m.s voltage of 1st waveform = %.2f V",Vrms1)
+printf("\n (b2)r.m.s voltage of 2nd waveform = %.2f V",Vrms2)
+printf("\n (c)Phase difference = %.0f°",phi)
\ No newline at end of file diff --git a/608/CH10/EX10.12/10_12.sce b/608/CH10/EX10.12/10_12.sce new file mode 100755 index 000000000..98c18dd22 --- /dev/null +++ b/608/CH10/EX10.12/10_12.sce @@ -0,0 +1,19 @@ +//Problem 10.12: The ratio of two powers is (a) 3 (b) 20 (c) 400 (d) 1/20. Determine the decibel power ratio in each case.
+
+//initializing the variables:
+rP1 = 3; // ratio of two powers
+rP2 = 20; // ratio of two powers
+rP3 = 400; // ratio of two powers
+rP4 = 1/20; // ratio of two powers
+
+//calculation:
+X1 = 10*log10(3)
+X2 = 10*log10(20)
+X3 = 10*log10(400)
+X4 = 10*log10(1/20)
+
+printf("\n\n Result \n\n")
+printf("\n (a)decibel power ratio for power ratio 3 = %.2f dB ",X1)
+printf("\n (b)decibel power ratio for power ratio 20 = %.1f dB ",X2)
+printf("\n (c)decibel power ratio for power ratio 400 = %.1f dB ",X3)
+printf("\n (d)decibel power ratio for power ratio 1/20 = %.1f dB ",X4)
\ No newline at end of file diff --git a/608/CH10/EX10.13/10_13.sce b/608/CH10/EX10.13/10_13.sce new file mode 100755 index 000000000..85f3b1448 --- /dev/null +++ b/608/CH10/EX10.13/10_13.sce @@ -0,0 +1,11 @@ +//Problem 10.13: The current input to a system is 5 mA and the current output is 20 mA. Find the decibel current ratio assuming the input and load resistances of the system are equal.
+
+//initializing the variables:
+I2 = 0.020; // in ampere
+I1 = 0.005; // in ampere
+
+//calculation:
+X = 20*log10(I2/I1)
+
+printf("\n\n Result \n\n")
+printf("\n decibel current ratio = %.0f dB\n",X)
\ No newline at end of file diff --git a/608/CH10/EX10.14/10_14.sce b/608/CH10/EX10.14/10_14.sce new file mode 100755 index 000000000..504d5b9c1 --- /dev/null +++ b/608/CH10/EX10.14/10_14.sce @@ -0,0 +1,10 @@ +//Problem 10.14: 6% of the power supplied to a cable appears at the output terminals. Determine the power loss in decibels.
+
+//initializing the variables:
+rP = 0.06; // power ratios rP = P2/P1
+
+//calculation:
+X = 10*log10(rP)
+
+printf("\n\n Result \n\n")
+printf("\n decibel Power ratios = %.2f dB\n",X)
\ No newline at end of file diff --git a/608/CH10/EX10.15/10_15.sce b/608/CH10/EX10.15/10_15.sce new file mode 100755 index 000000000..d075a6242 --- /dev/null +++ b/608/CH10/EX10.15/10_15.sce @@ -0,0 +1,12 @@ +//Problem 10.15: An amplifier has a gain of 14 dB. Its input power is 8 mW. Find its output power.
+
+//initializing the variables:
+X = 14; // decibal power ratio in dB
+P1 = 0.008; // in Watt
+
+//calculation:
+rP = 10^(X/10)
+P2 = rP*P1
+
+printf("\n\n Result \n\n")
+printf("\n output power P2 = %.3f W\n",P2)
\ No newline at end of file diff --git a/608/CH10/EX10.16/10_16.sce b/608/CH10/EX10.16/10_16.sce new file mode 100755 index 000000000..d21d36260 --- /dev/null +++ b/608/CH10/EX10.16/10_16.sce @@ -0,0 +1,11 @@ +//Problem 10.16: The output voltage from an amplifier is 4 V. If the voltage gain is 27 dB, calculate the value of the input voltage assuming that the amplifier input resistance and load resistance are equal.
+
+//initializing the variables:
+X = 27; // Voltage gain in decibels
+V2 = 4; // output voltage in Volts
+
+//calculation:
+V1 = V2/(10^(27/20))
+
+printf("\n\n Result \n\n")
+printf("\n input Voltage V1 = %.3f V\n",V1)
\ No newline at end of file diff --git a/608/CH10/EX10.17/10_17.sce b/608/CH10/EX10.17/10_17.sce new file mode 100755 index 000000000..54183347b --- /dev/null +++ b/608/CH10/EX10.17/10_17.sce @@ -0,0 +1,12 @@ +//Problem 10.17: In a Wheatstone bridge ABCD, a galvanometer is connected between A and C, and a battery between B and D. A resistor of unknown value is connected between A and B. When the bridge is balanced, the resistance between B and C is 100 ohms, that between C and D is 10 ohms and that between D and A is 400 ohms. calculate the value of the unknown resistance.
+
+//initializing the variables:
+R2 = 100; // in ohms
+R3 = 400; // in ohms
+R4 = 10; // in ohms
+
+//calculation:
+R1 = R2*R3/R4
+
+printf("\n\n Result \n\n")
+printf("\n unknown resistance, R1 = %.0f Ohms\n",R1)
\ No newline at end of file diff --git a/608/CH10/EX10.18/10_18.sce b/608/CH10/EX10.18/10_18.sce new file mode 100755 index 000000000..be8f613aa --- /dev/null +++ b/608/CH10/EX10.18/10_18.sce @@ -0,0 +1,12 @@ +//Problem 10.18: In a d.c. potentiometer, balance is obtained at a length of 400 mm when using a standard cell of 1.0186 volts. Determine the e.m.f. of a dry cell if balance is obtained with a length of 650 mm
+
+//initializing the variables:
+E1 = 1.0186; // in Volts
+l1 = 0.400; // in m
+l2 = 0.650; // in m
+
+//calculation:
+E2 = (l2/l1)*E1
+
+printf("\n\n Result \n\n")
+printf("\n the e.m.f. of a dry cell = %.3f Volts\n",E2)
\ No newline at end of file diff --git a/608/CH10/EX10.19/10_19.sce b/608/CH10/EX10.19/10_19.sce new file mode 100755 index 000000000..3c0f78657 --- /dev/null +++ b/608/CH10/EX10.19/10_19.sce @@ -0,0 +1,15 @@ +//Problem 10.19: The current flowing through a resistor of 5 kohm(+-)0.4% is measured as 2.5 mA with an accuracy of measurement of (+-)0.5%. Determine the nominal value of the voltage across the resistor and its accuracy.
+
+//initializing the variables:
+I = 0.0025; // in Amperes
+R = 5000; // in ohms
+e1 = 0.4; // in %
+e2 = 0.5; // in %
+
+//calculation:
+V = I*R
+em = e1 + e2
+Ve = em*V/100
+
+printf("\n\n Result \n\n")
+printf("\n voltage V = %.1fV(+-)%.2fV\n",V,Ve)
\ No newline at end of file diff --git a/608/CH10/EX10.20/10_20.sce b/608/CH10/EX10.20/10_20.sce new file mode 100755 index 000000000..962796c86 --- /dev/null +++ b/608/CH10/EX10.20/10_20.sce @@ -0,0 +1,20 @@ +//Problem 10.20: The current I flowing in a resistor R is measured by a 0–10 A ammeter which gives an indication of 6.25 A. The voltage V across the resistor is measured by a 0–50 V voltmeter, which gives an indication of 36.5 V. Determine the resistance of the resistor, and its accuracy of measurement if both instruments have a limit of error of 2% of f.s.d. Neglect any loading effects of the instruments.
+
+//initializing the variables:
+I = 6.25; // in Amperes
+Im = 10; // max in Amperes
+V = 36.5; // in volts
+Vm = 50; // max in volts
+e = 2; // in %
+
+//calculation:
+R = V/I
+Ve = e*Vm/100
+Ve1 = Ve*100/V // in %
+Ie = e*Im/100
+Ie1 = Ie*100/I // in %
+em = Ve1 + Ie1
+Re = em*R/100
+
+printf("\n\n Result \n\n")
+printf("\n Resistance R = %.2f ohms(+-)%.2f ohms\n",R,Re)
\ No newline at end of file diff --git a/608/CH10/EX10.21/10_21.sce b/608/CH10/EX10.21/10_21.sce new file mode 100755 index 000000000..f589dc17d --- /dev/null +++ b/608/CH10/EX10.21/10_21.sce @@ -0,0 +1,17 @@ +//Problem 10.21: The arms of a Wheatstone bridge ABCD have the following resistances: AB: R1 = 1000 ohms (+-) 1.0% ; BC: R2 = 100 ohm (+-) 0.5%; CD: unknown resistance Rx; DA: R3 = 432.5 ohms (+-)0.2%. Determine the value of the unknown resistance and its accuracy of measurement.
+
+//initializing the variables:
+R1 = 1000; // in ohms
+R2 = 100; // in ohms
+R3 = 432.5; // in ohms
+e1 = 1; // in %
+e2 = 0.5; // in %
+e3 = 0.2; // in %
+
+//calculation:
+Rx = R2*R3/R1
+em = e1 + e2 + e3
+Re = em*Rx/100
+
+printf("\n\n Result \n\n")
+printf("\n Resistance R = %.2f ohms(+-)%.2f ohms\n",Rx,Re)
\ No newline at end of file diff --git a/608/CH13/EX13.01/13_01.sce b/608/CH13/EX13.01/13_01.sce new file mode 100755 index 000000000..754cd0c34 --- /dev/null +++ b/608/CH13/EX13.01/13_01.sce @@ -0,0 +1,30 @@ +//Problem 13.01: (a) Find the unknown currents marked in Figure 13.3(a). (b) Determine the value of e.m.f. E in Figure 13.3(b).
+
+//initializing the variables:
+Iab = 50; // in ampere
+Ibc = 20; // in ampere
+Iec = 15; // in ampere
+Idf = 120; // in ampere
+Ifg = 40; // in ampere
+Iab = 50; // in ampere
+I = 2; // in ampere
+V1 = 4; // in volts
+V2 = 3; // in volts
+V3 = 6; // in volts
+R1 = 1; // in ohms
+R2 = 2; // in ohms
+R3 = 2.5; // in ohms
+R4 = 1.5; // in ohms
+
+//calculation:
+I1 = Iab - Ibc
+I2 = Ibc + Iec
+I3 = I1 - Idf
+I4 = Iec - I3
+I5 = Idf - Ifg
+// Applying Kirchhoff’s voltage law and moving clockwise around the loop of Figure 13.3(b) starting at point A:
+E = I*R2 + I*R3 + I*R4 + I*R1 - V2 - V3 + V1
+
+printf("\n\n Result \n\n")
+printf("\n (a) unknown currents I1, I2, I3, I4, I5 are %.0fA, %.0fA, %.0fA, %.0fA, %.0fA respetively",I1, I2, I3, I4, I5)
+printf("\n (b) value of e.m.f. E = %.0f Volts",E)
\ No newline at end of file diff --git a/608/CH13/EX13.06/13_06.sce b/608/CH13/EX13.06/13_06.sce new file mode 100755 index 000000000..008a7d777 --- /dev/null +++ b/608/CH13/EX13.06/13_06.sce @@ -0,0 +1,40 @@ +//Problem 13.06: For the circuit shown in Figure 13.16, find, using the superposition theorem, (a) the current flowing in and the pd across the 18 ohm resistor, (b) the current in the 8 V battery and (c) the current in the 3 V battery.
+
+//initializing the variables:
+E1 = 8; // in volts
+E2 = 3; // in volts
+R1 = 3; // in ohms
+R2 = 2; // in ohms
+R3 = 18; // in ohms
+
+//calculation:
+//Removing source E2 gives the circuit of Figure 13.17(a).
+//The current directions are labelled as shown in Figure 13.17(a), I1 flowing from the positive terminal of E1.
+//From Figure 13.17(b), I1
+r1 = 1/(1/R2 + 1/R3)
+I1 = E1/(R1 + r1)
+//From Figure 13.17(a), I2
+I2 = (R3/(R3 + R2))*I1
+I3 = (R2/(R3 + R2))*I1
+//Removing source E1 gives the circuit of Figure 13.18(a)
+//The current directions are labelled as shown in Figures 13.18(a) and 13.18(b), I4 flowing from the positive terminal of E2
+//From Figure 13.18(c), I4
+r2 = 1/(1/R3 + 1/R1)
+I4 = E2/(r2 + R2)
+//From Figure 13.18(b), I5
+I5 = (R3/(R1 + R3))*I4
+I6 = (R1/(R1 + R3))*I4
+//Resultant current in the R3 resistor
+I18 = I3 - I6
+//P.d. across the R3
+V3 = I18*R3
+//Resultant current in the E1
+Ie1 = I1 + I5
+//Resultant current in the E2
+Ie2 = I2 + I4
+
+printf("\n\n Result \n\n")
+printf("\n Resultant current in the 18 ohm resistor is %.3f A",I18)
+printf("\n P.d. across the 18 ohm resistor is %.3f V",V3)
+printf("\n Resultant current in the E1 is %.3f A",Ie1)
+printf("\n Resultant current in the E2 is %.3f A",Ie2)
\ No newline at end of file diff --git a/608/CH13/EX13.07/13_07.sce b/608/CH13/EX13.07/13_07.sce new file mode 100755 index 000000000..b2720bb64 --- /dev/null +++ b/608/CH13/EX13.07/13_07.sce @@ -0,0 +1,22 @@ +//Problem 13.07: Use Th´evenin’s theorem to find the current flowing in the 10 ohm resistor for the circuit shown in Figure 13.28(a).
+
+//initializing the variables:
+V = 10; // in volts
+R1 = 2; // in ohms
+R2 = 8; // in ohms
+R3 = 5; // in ohms
+R = 10; // in ohms
+
+//calculation:
+//The 10 ohm resistance branch is short-circuited as shown in Figure 13.28(b).
+//Current I1
+I1 = V/(R1 + R2)
+//p.d. across AB, E
+E = R2*I1
+//the resistance ‘looking-in’ at a break made between A and B is given by
+r = R3 + (R1*R2)/(R2 + R1)
+//The equivalent Th´evenin’s circuit is shown in Figure 13.28(d), the current in the 10 ohm resistance is given by:
+I = E/(r + R)
+
+printf("\n\n Result \n\n")
+printf("\n the current in the 10 ohm resistance is given by %.3f A",I)
\ No newline at end of file diff --git a/608/CH13/EX13.08/13_08.sce b/608/CH13/EX13.08/13_08.sce new file mode 100755 index 000000000..7ef395eac --- /dev/null +++ b/608/CH13/EX13.08/13_08.sce @@ -0,0 +1,22 @@ +//Problem 13.08: For the network shown in Figure 13.29(a) determine the current in the 0.8 ohm resistor using Th´evenin’s theorem.
+
+//initializing the variables:
+V = 12; // in volts
+R1 = 5; // in ohms
+R2 = 1; // in ohms
+R3 = 4; // in ohms
+R = 0.8; // in ohms
+
+//calculation:
+//The 0.8 ohm resistance branch is short-circuited as shown in Figure 13.29(b).
+//Current I1
+I1 = V/(R1 + R2 + R3)
+//p.d. across AB, E
+E = R3*I1
+//the resistance ‘looking-in’ at a break made between A and B is given by
+r = R3*(R1 + R2)/(R2 + R1 + R3)
+//The equivalent Th´evenin’s circuit is shown in Figure 13.29(d), the current in the 0.8 ohm resistance is given by:
+I = E/(r + R)
+
+printf("\n\n Result \n\n")
+printf("\n the current in the 0.8 ohm resistance is given by %.1f A",I)
\ No newline at end of file diff --git a/608/CH13/EX13.09/13_09.sce b/608/CH13/EX13.09/13_09.sce new file mode 100755 index 000000000..3a6528418 --- /dev/null +++ b/608/CH13/EX13.09/13_09.sce @@ -0,0 +1,25 @@ +//Problem 13.09: Use Th´evenin’s theorem to determine the current I flowing in the 4 ohm resistor shown in Figure 13.30(a). Find also the power dissipated in the 4 ohm resistor.
+
+//initializing the variables:
+E1 = 4; // in volts
+E2 = 2; // in volts
+R1 = 2; // in ohms
+R2 = 1; // in ohms
+R3 = 4; // in ohms
+
+//calculation:
+//The 4 ohm resistance branch is short-circuited as shown in Figure 13.30(b).
+//Current I1
+I1 = (E1 - E2)/(R1 + R2)
+//p.d. across AB, E
+E = E1 - I1*R1
+//the resistance ‘looking-in’ at a break made between A and B is given by
+r = R2*R1/(R2 + R1)
+//The equivalent Th´evenin’s circuit is shown in Figure 13.30(d), the current in the 4ohm resistance is given by:
+I = E/(r + R3)
+//Power dissipated in R3
+P3 = R3*I^2
+
+printf("\n\n Result \n\n")
+printf("\n the current in the 4 ohm resistance is given by %.3f A",I)
+printf("\n power disipated in 4 ohm resistor is given by %.3f W",P3)
\ No newline at end of file diff --git a/608/CH13/EX13.10/13_10.sce b/608/CH13/EX13.10/13_10.sce new file mode 100755 index 000000000..cf9fc3d20 --- /dev/null +++ b/608/CH13/EX13.10/13_10.sce @@ -0,0 +1,23 @@ +//Problem 13.10: Use Th´evenin’s theorem to determine the current flowing in the 3 ohm resistance of the network shown in Figure 13.31(a). The voltage source has negligible internal resistance.
+
+//initializing the variables:
+V = 24; // in volts
+R1 = 20; // in ohms
+R2 = 5; // in ohms
+R3 = 10; // in ohms
+R4 = 5/3; // in ohms
+R5 = 3; // in ohms
+
+//calculation:
+//The 3 ohm resistance branch is short-circuited as shown in Figure 13.31(b).
+//P.d. across R3
+V3 = (R3/(R3 + R2))*V
+//p.d. across AB, E
+E = V3
+//the resistance ‘looking-in’ at a break made between A and B is given by
+r = R4 + R2*R3/(R2 + R3)
+//The equivalent Th´evenin’s circuit is shown in Figure 13.31(e), the current in the 32 ohm resistance is given by:
+I = E/(r + R5)
+
+printf("\n\n Result \n\n")
+printf("\n the current in the 3 ohm resistance is given by %.0f A",I)
\ No newline at end of file diff --git a/608/CH13/EX13.11/13_11.sce b/608/CH13/EX13.11/13_11.sce new file mode 100755 index 000000000..e57b48d10 --- /dev/null +++ b/608/CH13/EX13.11/13_11.sce @@ -0,0 +1,25 @@ +//Problem 13.11: A Wheatstone Bridge network is shown in Figure 13.32(a). Calculate the current flowing in the 32 ohm resistor, and its direction, using Th´evenin’s theorem. Assume the source of e.m.f. to have negligible resistance.
+
+//initializing the variables:
+E = 54; // in volts
+R1 = 2; // in ohms
+R2 = 14; // in ohms
+R3 = 3; // in ohms
+R4 = 11; // in ohms
+R5 = 32; // in ohms
+
+//calculation:
+//The 32ohm resistance branch is short-circuited as shown in Figure 13.32(b).
+//The p.d. between A and C,
+Vac = (R1/(R1 + R4))*E
+//The p.d. between B and C,
+Vbc = (R2/(R2 + R3))*E
+//Hence the p.d. between A and B
+Vab = Vbc - Vac
+//the resistance ‘looking-in’ at a break made between A and B is given by
+r = R1*R4/(R1 + R4) + R2*R3/(R2 + R3)
+//The equivalent Th´evenin’s circuit is shown in Figure 13.32(f), the current in the 32 ohm resistance is given by:
+I = E/(r + R5)
+
+printf("\n\n Result \n\n")
+printf("\n the current in the 32 ohm resistance is given by %.0f A",I)
\ No newline at end of file diff --git a/608/CH13/EX13.12/13_12.sce b/608/CH13/EX13.12/13_12.sce new file mode 100755 index 000000000..0b4cb7b0e --- /dev/null +++ b/608/CH13/EX13.12/13_12.sce @@ -0,0 +1,20 @@ +//Problem 13.12: Use Norton’s theorem to determine the current flowing in the 10 ohm resistance for the circuit shown in Figure 13.34(a).
+
+//initializing the variables:
+V = 10; // in volts
+R1 = 2; // in ohms
+R2 = 8; // in ohms
+R3 = 5; // in ohms
+R4 = 10; // in ohms
+
+//calculation:
+//The 10ohm resistance branch is short-circuited as shown in Figure 13.34(b).
+//Figure 13.34(c) is equivalent to Figure 13.34(b). Hence
+Isc = V/R1
+//the resistance ‘looking-in’ at a break made between A and B is given by
+r = R1*R2/(R1 + R2)
+//From the Norton equivalent network shown in Figure 13.34(d) the current in the 10 ohm resistance is given by:
+I = (r/(r + R3 + R4))*Isc
+
+printf("\n\n Result \n\n")
+printf("\n the current in the 10 ohm resistance is given by %.3f A",I)
\ No newline at end of file diff --git a/608/CH13/EX13.13/13_13.sce b/608/CH13/EX13.13/13_13.sce new file mode 100755 index 000000000..775953891 --- /dev/null +++ b/608/CH13/EX13.13/13_13.sce @@ -0,0 +1,20 @@ +//Problem 13.13: Use Norton’s theorem to determine the current I flowing in the 4 ohm resistance shown in Figure 13.35(a).
+
+//initializing the variables:
+V1 = 4; // in volts
+V2 = 2; // in volts
+R1 = 2; // in ohms
+R2 = 1; // in ohms
+R3 = 4; // in ohms
+
+//calculation:
+//The 4ohm resistance branch is short-circuited as shown in Figure 13.35(b).
+//Figure 13.35(b)
+Isc = V1/R1 + V2/R2
+//the resistance ‘looking-in’ at a break made between A and B is given by
+r = R1*R2/(R1 + R2)
+//From the Norton equivalent network shown in Figure 13.35(c) the current in the 4ohm resistance is given by:
+I = (r/(r + R3))*Isc
+
+printf("\n\n Result \n\n")
+printf("\n the current in the 4ohm resistance is given by %.3f A",I)
\ No newline at end of file diff --git a/608/CH13/EX13.14/13_14.sce b/608/CH13/EX13.14/13_14.sce new file mode 100755 index 000000000..2272c88de --- /dev/null +++ b/608/CH13/EX13.14/13_14.sce @@ -0,0 +1,21 @@ +//Problem 13.14: Use Norton’s theorem to determine the current flowing in the 3 ohm resistance of the network shown in Figure 13.36(a). The voltage source has negligible internal resistance.
+
+//initializing the variables:
+V = 24; // in volts
+R1 = 20; // in ohms
+R2 = 5; // in ohms
+R3 = 10; // in ohms
+R4 = 5/3; // in ohms
+R5 = 3; // in ohms
+
+//calculation:
+//The 3ohm resistance branch is short-circuited as shown in Figure 13.36(b).
+//Figure 13.36(c) is equivalent to Figure 13.36(b).
+Isc = V/R2
+//the resistance ‘looking-in’ at a break made between A and B is given by
+r = R3*R2/(R3 + R2)
+//From the Norton equivalent network shown in Figure 13.36(f) the current in the 3ohm resistance is given by:
+I = (r/(r + R4 + R5))*Isc
+
+printf("\n\n Result \n\n")
+printf("\n the current in the 3ohm resistance is given by %.0f A",I)
\ No newline at end of file diff --git a/608/CH13/EX13.15/13_15.sce b/608/CH13/EX13.15/13_15.sce new file mode 100755 index 000000000..4e0a951d6 --- /dev/null +++ b/608/CH13/EX13.15/13_15.sce @@ -0,0 +1,21 @@ +//Problem 13.15: Determine the current flowing in the 2ohm resistance in the network shown in Figure 13.37(a).
+
+//initializing the variables:
+I = 15; // in amperes
+R1 = 6; // in ohms
+R2 = 4; // in ohms
+R3 = 8; // in ohms
+R4 = 2; // in ohms
+R5 = 7; // in ohms
+
+//calculation:
+//The 2ohm resistance branch is short-circuited as shown in Figure 13.37(b).
+//Figure 13.37(c) is equivalent to Figure 13.37(b).
+Isc = (R1/(R1 + R2))*I
+//the resistance ‘looking-in’ at a break made between A and B is given by
+r = ((R1 + R2)*(R3 + R5)/(R1 + R2 + R3 + R5))
+//From the Norton equivalent network shown in Figure 13.37(e) the current in the 2ohm resistance is given by:
+I = (r/(r + R4))*Isc
+
+printf("\n\n Result \n\n")
+printf("\n the current in the 2ohm resistance is given by %.2f A",I)
\ No newline at end of file diff --git a/608/CH13/EX13.18/13_18.sce b/608/CH13/EX13.18/13_18.sce new file mode 100755 index 000000000..69fd04087 --- /dev/null +++ b/608/CH13/EX13.18/13_18.sce @@ -0,0 +1,28 @@ +//Problem 13.18: (a) Convert the circuit to the left of terminals AB in Figure 13.45(a) to an equivalent Th´evenin circuit by initially converting to a Norton equivalent circuit. (b) Determine the current flowing in the 1.8 ohm resistor.
+
+//initializing the variables:
+E1 = 12; // in volts
+E2 = 24; // in volts
+R1 = 3; // in ohms
+R2 = 2; // in ohms
+R3 = 1.8; // in ohms
+
+//calculation:
+//For the branch containing the V1 source, converting to a Norton equivalent network gives
+Isc1 = E1/R1
+r1 = R1
+//For the branch containing the V2 source, converting to a Norton equivalent network gives
+Isc2 = E2/R2
+r2 = R2
+//Thus the network of Figure 13.46(a) converts to Figure 13.46(b).
+//total short-circuit current
+Isct = Isc1 + Isc2
+//the resistance is
+z = r1*r2/(r1 + r2)
+//Both of the Norton equivalent networks shown in Figure 13.46(c) may be converted to Th´evenin equivalent circuits. The open-circuit voltage across CD is
+Vcd = Isct*z
+//the current I flowing in a 1.8 ohm resistance connected between A and B is given by:
+I = Vcd/(z + R3)
+
+printf("\n\n Result \n\n")
+printf("\n the current I flowing in a 1.8 ohm resistance connected between A and B is given by %.2f A",I)
\ No newline at end of file diff --git a/608/CH13/EX13.19/13_19.sce b/608/CH13/EX13.19/13_19.sce new file mode 100755 index 000000000..45f649446 --- /dev/null +++ b/608/CH13/EX13.19/13_19.sce @@ -0,0 +1,37 @@ +//Problem 13.19: Determine by successive conversions between Th´evenin and Norton equivalent networks a Th´evenin equivalent circuit for terminals AB of Figure 13.46(a). Hence determine the current flowing in the 200 ohm resistance.
+
+//initializing the variables:
+V1 = 10; // in volts
+V2 = 6; // in volts
+R1 = 2000; // in ohms
+R2 = 3000; // in ohms
+R3 = 600; // in ohms
+R4 = 200; // in ohms
+i = 0.001; // in amperes
+
+//calculation:
+//For the branch containing the V1 source, converting to a Norton equivalent network gives
+Isc1 = V1/R1
+r1 = R1
+//For the branch containing the V2 source, converting to a Norton equivalent network gives
+Isc2 = V2/R2
+r2 = R2
+//Thus the network of Figure 13.46(a) converts to Figure 13.46(b).
+//total short-circuit current
+Isct = Isc1 + Isc2
+//the resistance is
+z = r1*r2/(r1 + r2)
+//Both of the Norton equivalent networks shown in Figure 13.46(c) may be converted to Th´evenin equivalent circuits. The open-circuit voltage across CD is
+Vcd = Isct*z
+//The open-circuit voltage across EF is
+Vef = i*R3
+//the resistance ‘looking-in’ at EF is
+r3 = R3
+//Thus Figure 13.46(c) converts to Figure 13.46(d). Combining the two Th´evenin circuits gives
+E = Vcd - Vef
+r = z + r3
+//the current I flowing in a 200 ohm resistance connected between A and B is given by:
+I = E/(r + R4)
+
+printf("\n\n Result \n\n")
+printf("\n the current I flowing in a 200ohm resistance connected between A and B is given by:is %.2E A",I)
\ No newline at end of file diff --git a/608/CH13/EX13.20/13_20.jpg b/608/CH13/EX13.20/13_20.jpg Binary files differnew file mode 100755 index 000000000..a22bdd4f6 --- /dev/null +++ b/608/CH13/EX13.20/13_20.jpg diff --git a/608/CH13/EX13.20/13_20.sce b/608/CH13/EX13.20/13_20.sce new file mode 100755 index 000000000..c9542b71e --- /dev/null +++ b/608/CH13/EX13.20/13_20.sce @@ -0,0 +1,13 @@ +//Problem 13.20: The circuit diagram of Figure 13.48 shows dry cells of source e.m.f. 6 V, and internal resistance 2.5 ohm. If the load resistance RL is varied from 0 to 5 ohm in 0.5 ohm steps, calculate the power dissipated by the load in each case. Plot a graph of RL (horizontally) against power (vertically) and determine the maximum power dissipated.
+
+//initializing the variables:
+V = 6; // in volts
+r = 2.5; // in ohms
+
+//calculation:
+RL=(0:0.5:5)';
+function [y]=f(RL)
+ y = RL*(V/(r + RL))^2;
+endfunction
+fplot2d(RL, f)
+xtitle("graph of (RL) against power(P)", "RL(ohm)", "P(W)")
\ No newline at end of file diff --git a/608/CH13/EX13.21/13_21.sce b/608/CH13/EX13.21/13_21.sce new file mode 100755 index 000000000..f6d2cc149 --- /dev/null +++ b/608/CH13/EX13.21/13_21.sce @@ -0,0 +1,17 @@ +//Problem 13.21: A d.c. source has an open-circuit voltage of 30 V and an internal resistance of 1.5 ohm. State the value of load resistance that gives maximum power dissipation and determine the value of this power.
+
+//initializing the variables:
+V = 30; // in volts
+r = 1.5; // in ohms
+
+//calculation:
+//current I = E/(r + RL)
+//For maximum power, RL = r
+RL = r
+I = V/(r + RL)
+//Power, P, dissipated in load RL, P
+P = RL*I^2
+
+printf("\n\n Result \n\n")
+printf("\n (a) the value of the load resistor RL is %.1f ohm",RL)
+printf("\n (b) maximum power dissipation = %.0f W",P)
\ No newline at end of file diff --git a/608/CH13/EX13.22/13_22.sce b/608/CH13/EX13.22/13_22.sce new file mode 100755 index 000000000..d54a5d9f2 --- /dev/null +++ b/608/CH13/EX13.22/13_22.sce @@ -0,0 +1,24 @@ +//Problem 13.22: Find the value of the load resistor RL shown in Figure 13.51(a) that gives maximum power dissipation and determine the value of this power.
+
+//initializing the variables:
+V = 15; // in volts
+R1 = 3; // in ohms
+R2 = 12; // in ohms
+
+//calculation:
+//Resistance RL is removed from the circuit as shown in Figure 13.51(b).
+//The p.d. across AB is the same as the p.d. across the 12 ohm resistor.
+E = (R2/(R1 + R2))*V
+//Removing the source of e.m.f. gives the circuit of Figure 13.51(c),
+//from which resistance, r
+r = R1*R2/(R1 + R2)
+//The equivalent Th´evenin’s circuit supplying terminals AB is shown in Figure 13.51(d), from which, current I = E/(r + RL)
+//For maximum power, RL = r
+RL = r
+I = E/(r + RL)
+//Power, P, dissipated in load RL, P
+P = RL*I^2
+
+printf("\n\n Result \n\n")
+printf("\n (a) the value of the load resistor RL is %.1f ohm",RL)
+printf("\n (b) maximum power dissipation = %.0f W",P)
\ No newline at end of file diff --git a/608/CH14/EX14.01/14_01.sce b/608/CH14/EX14.01/14_01.sce new file mode 100755 index 000000000..bf6d19388 --- /dev/null +++ b/608/CH14/EX14.01/14_01.sce @@ -0,0 +1,13 @@ +//Problem 14.01: Determine the periodic time for frequencies of (a) 50 Hz and (b) 20 kHz
+
+//initializing the variables:
+f1 = 50; // in Hz
+f2 = 20000; // in Hz
+
+//calculation:
+T1 = 1/f1
+T2 = 1/f2
+
+printf("\n\n Result \n\n")
+printf("\n (a) Periodic time T = %.2f secs",T1)
+printf("\n (b) Periodic time T = %.2E secs\n",T2)
\ No newline at end of file diff --git a/608/CH14/EX14.02/14_02.sce b/608/CH14/EX14.02/14_02.sce new file mode 100755 index 000000000..73a28a04e --- /dev/null +++ b/608/CH14/EX14.02/14_02.sce @@ -0,0 +1,13 @@ +//Problem 14.02: Determine the frequencies for periodic times of (a) 4 ms, (b) 4 μs
+
+//initializing the variables:
+T1 = 0.004; // in secs
+T2 = 4E-6; // in secs
+
+//calculation:
+f1 = 1/T1
+f2 = 1/T2
+
+printf("\n\n Result \n\n")
+printf("\n (a) Frequency f = %.0f Hz",f1)
+printf("\n (b) Frequency f = %.2E secs",f2)
\ No newline at end of file diff --git a/608/CH14/EX14.03/14_03.sce b/608/CH14/EX14.03/14_03.sce new file mode 100755 index 000000000..4e14a7043 --- /dev/null +++ b/608/CH14/EX14.03/14_03.sce @@ -0,0 +1,10 @@ +//Problem 14.03: An alternating current completes 5 cycles in 8 ms. What is its frequency?
+
+//initializing the variables:
+T = (8E-3)/5; // in secs
+
+//calculation:
+f = 1/T
+
+printf("\n\n Result \n\n")
+printf("\n Frequency f = %.0f Hz\n",f)
\ No newline at end of file diff --git a/608/CH14/EX14.04/14_04.sce b/608/CH14/EX14.04/14_04.sce new file mode 100755 index 000000000..ece0f3186 --- /dev/null +++ b/608/CH14/EX14.04/14_04.sce @@ -0,0 +1,40 @@ +//Problem 14.04: For the periodic waveforms shown in Figure 14.5 determine for each: (i) frequency (ii) average value over half a cycle (iii) rms value (iv) form factor and (v) peak factor
+
+//initializing the variables:
+Ta = 0.02; // Time for 1 complete cycle in secs
+Vamax = 200; // in volts
+Va1 = 25; // in volts
+Va2 = 75; // in volts
+Va3 = 125; // in volts
+Va4 = 175; // in volts
+Tb = 0.016; // Time for 1 complete cycle in secs
+Ibmax = 10; // in Amperes
+
+//calculation:
+//for Triangular waveform (Figure 14.5(a))
+fa = 1/Ta
+Aaw = Ta*Vamax/4
+Vaavg = Aaw*2/Ta
+Varms = (((Va1^2) + (Va2^2) + (Va3^2) + (Va4^2))/4)^0.5 //Note that the greater the number of intervals chosen, the greater the accuracy of the result
+Ffa = Varms/Vaavg
+Pfa = Vamax/Varms
+
+//for Rectangular waveform (Figure 14.5(b))
+fb = 1/Tb
+Abw = Tb*Ibmax/2
+Ibavg = Abw*2/Tb
+Ibrms = 10
+Ffb = Ibrms/Ibavg
+Pfb = Ibmax/Ibrms
+
+printf("\n\n Result \n\n")
+printf("\n (a1)Frequency f = %.0f Hz",fa)
+printf("\n (a2)average value over half a cycle = %.0f V",Vaavg)
+printf("\n (a3)rms value = %.1f V",Varms)
+printf("\n (a4)Form factor = %.2f",Ffa)
+printf("\n (a5)Peak factor = %.2f",Pfa)
+printf("\n (b1)Frequency f = %.1f Hz",fb)
+printf("\n (b2)average value over half a cycle = %.0f A",Ibavg)
+printf("\n (b3)rms value = %.0f A",Ibrms)
+printf("\n (b4)Form factor = %.0f",Ffb)
+printf("\n (b5)Peak factor = %.0f",Pfb)
\ No newline at end of file diff --git a/608/CH14/EX14.06/14_06.sce b/608/CH14/EX14.06/14_06.sce new file mode 100755 index 000000000..ac5abb846 --- /dev/null +++ b/608/CH14/EX14.06/14_06.sce @@ -0,0 +1,11 @@ +//Problem 14.06: Calculate the rms value of a sinusoidal current of maximum value 20 A
+
+//initializing the variables:
+Imax = 20; // in Amperes
+
+//calculation:
+//for a sine wave
+Irms = Imax/(2^0.5)
+
+printf("\n\n Result \n\n")
+printf("\n Rms value = %.2f A\n",Irms)
\ No newline at end of file diff --git a/608/CH14/EX14.07/14_07.sce b/608/CH14/EX14.07/14_07.sce new file mode 100755 index 000000000..aaefcbe7d --- /dev/null +++ b/608/CH14/EX14.07/14_07.sce @@ -0,0 +1,13 @@ +//Problem 14.07: Determine the peak and mean values for a 240 V mains supply.
+
+//initializing the variables:
+Vrms = 240; // in Volts
+
+//calculation:
+//for a sine wave
+Vmax = Vrms*(2^0.5)
+Vmean = 0.637*Vmax
+
+printf("\n\n Result \n\n")
+printf("\n peak value = %.1f V",Vmax)
+printf("\n mean value = %.1f V",Vmean)
\ No newline at end of file diff --git a/608/CH14/EX14.08/14_08.sce b/608/CH14/EX14.08/14_08.sce new file mode 100755 index 000000000..6153070da --- /dev/null +++ b/608/CH14/EX14.08/14_08.sce @@ -0,0 +1,13 @@ +//Problem 14.08: A supply voltage has a mean value of 150 V. Determine its maximum value and its rms value
+
+//initializing the variables:
+Vmean = 150; // in Volts
+
+//calculation:
+//for a sine wave
+Vmax = Vmean/0.637
+Vrms = 0.707*Vmax
+
+printf("\n\n Result \n\n")
+printf("\n peak value = %.1f V",Vmax)
+printf("\n rms value = %.1f V",Vrms)
\ No newline at end of file diff --git a/608/CH14/EX14.09/14_09.sce b/608/CH14/EX14.09/14_09.sce new file mode 100755 index 000000000..19fb0ab28 --- /dev/null +++ b/608/CH14/EX14.09/14_09.sce @@ -0,0 +1,17 @@ +//Problem 14.09: An alternating voltage is given by v = 282.8 sin 314t volts. Find (a) the rms voltage, (b) the frequency and (c) the instantaneous value of voltage when t = 4 ms
+
+//initializing the variables:
+Vmax = 282.8; // in Volts
+w = 314; // in rad/sec
+t = 0.004; // in sec
+
+//calculation:
+//for a sine wave
+Vrms = 0.707*Vmax
+f = w/(2*%pi)
+v = Vmax*sin(w*t)
+
+printf("\n\n Result \n\n")
+printf("\n (a)rms value = %.0f V",Vrms)
+printf("\n (b)frequency f = %.0f Hz",f)
+printf("\n (c)instantaneous value of voltage at 4 ms = %.1f V",v)
\ No newline at end of file diff --git a/608/CH14/EX14.10/14_10.sce b/608/CH14/EX14.10/14_10.sce new file mode 100755 index 000000000..6c54ef3c2 --- /dev/null +++ b/608/CH14/EX14.10/14_10.sce @@ -0,0 +1,24 @@ +//Problem 14.10: An alternating voltage is given by v = 75sin(200*pi*t -0.25) volts. Find (a) the amplitude, (b) the peak-to-peak value, (c) the rms value, (d) the periodic time, (e) the frequency, and (f) the phase angle (in degrees and minutes) relative to 75 sin 200t
+
+//initializing the variables:
+Vmax = 75; // in Volts
+w = 200*%pi; // in rad/sec
+t = 0.004; // in sec
+phi = 0.25; // in radians
+
+//calculation:
+//for a sine wave
+Vptp = 2*Vmax
+Vrms = 0.707*Vmax
+f = w/(2*%pi)
+T = 1/f
+v = Vmax*sin(w*t)
+phid = phi*180/%pi
+
+printf("\n\n Result \n\n")
+printf("\n (a) Amplitude, or peak value = %.0f V",Vmax)
+printf("\n (b) Peak-to-peak value = %.0f V",Vptp)
+printf("\n (c)rms value = %.0f V",Vrms)
+printf("\n (d)periodic time, T = %.2f sec",T)
+printf("\n (e)frequency f = %.0f Hz",f)
+printf("\n (f)phase angle = %.2f°",phid)
\ No newline at end of file diff --git a/608/CH14/EX14.12/14_12.sce b/608/CH14/EX14.12/14_12.sce new file mode 100755 index 000000000..19fc821ae --- /dev/null +++ b/608/CH14/EX14.12/14_12.sce @@ -0,0 +1,26 @@ +//Problem 14.12: The current in an a.c. circuit at any time t seconds is given by: i = 120 sin(100*pit+0.36) amperes. Find: (a) the peak value, the periodic time, the frequency and phase angle relative to 120sin 100pit (b) the value of the current when t = 0 (c) the value of the current when t = 8 ms (d) the time when the current first reaches 60 A, and (e) the time when the current is first a maximum
+
+//initializing the variables:
+Imax = 120; // in Amperes
+w = 100*%pi; // in rad/sec
+phi = 0.36; // in rad
+t1 = 0; // in secs
+t2 = 0.008; // in secs
+i = 60; // in amperes
+
+//calculation:
+//for a sine wave
+f = w/(2*%pi)
+T = 1/f
+phid = phi*180/%pi
+i0 = Imax*sin((w*t1) + phi)
+i8 = Imax*sin((w*t2)+phi)
+ti = (asin(i/Imax) - phi)/w
+tm1 = (asin(Imax/Imax) - phi)/w
+
+printf("\n\n Result \n\n")
+printf("\n (a)Peak value = %.0f A, Periodic time T = %.2f sec, Frequency, f = %.0f Hz Phase angle = %.1f°", Imax, T, f, phid)
+printf("\n (b) When t = 0, i = %.1f A",i0)
+printf("\n (c)When t = 8 ms = %.1f A", i8)
+printf("\n (d)When i is 60 A, then time t = %.2E s",ti)
+printf("\n (e)When the current is a maximum, time, t = %.2E s",tm1)
\ No newline at end of file diff --git a/608/CH15/EX15.01/15_01.sce b/608/CH15/EX15.01/15_01.sce new file mode 100755 index 000000000..6a21fb6be --- /dev/null +++ b/608/CH15/EX15.01/15_01.sce @@ -0,0 +1,15 @@ +//Problem 15.01: (a) Calculate the reactance of a coil of inductance 0.32 H when it is connected to a 50 Hz supply. (b) A coil has a reactance of 124 ohm in a circuit with a supply of frequency 5 kHz. Determine the inductance of the coil.
+
+//initializing the variables:
+L = 0.32; // in Henry
+f1 = 50; // in Hz
+f2 = 5000; // in Hz
+Z = 124; // in ohms
+
+//calculation:
+XL = 2*%pi*f1*L
+L = Z/(2*%pi*f2)
+
+printf("\n\n Result \n\n")
+printf("\n (a)Inductive reactance, XL = %.1f ohms ", XL)
+printf("\n (b)Inductance L = %.2E H",L)
\ No newline at end of file diff --git a/608/CH15/EX15.02/15_02.sce b/608/CH15/EX15.02/15_02.sce new file mode 100755 index 000000000..1d8387de4 --- /dev/null +++ b/608/CH15/EX15.02/15_02.sce @@ -0,0 +1,18 @@ +//Problem 15.02: A coil has an inductance of 40 mH and negligible resistance. Calculate its inductive reactance and the resulting current if connected to (a) a 240 V, 50 Hz supply, and (b) a 100 V, 1 kHz supply.
+
+//initializing the variables:
+L = 0.040; // in Henry
+V1 = 240; // in volts
+V2 = 100; // in volts
+f1 = 50; // in Hz
+f2 = 1000; // in Hz
+
+//calculation:
+XL1 = 2*%pi*f1*L
+I1 = V1/XL1
+XL2 = 2*%pi*f2*L
+I2 = V2/XL2
+
+printf("\n\n Result \n\n")
+printf("\n (a)Inductive reactance, XL = %.2f ohms and current I = %.2f A", XL1, I1)
+printf("\n (b)Inductive reactance, XL = %.1f ohms and current I = %.3f A", XL2, I2)
\ No newline at end of file diff --git a/608/CH15/EX15.03/15_03.sce b/608/CH15/EX15.03/15_03.sce new file mode 100755 index 000000000..3d2f0aaf0 --- /dev/null +++ b/608/CH15/EX15.03/15_03.sce @@ -0,0 +1,14 @@ +//Problem 15.03: Determine the capacitive reactance of a capacitor of 10 μF when connected to a circuit of frequency (a) 50 Hz (b) 20 kHz
+
+//initializing the variables:
+C = 10E-6; // in Farads
+f1 = 50; // in Hz
+f2 = 20000; // in Hz
+
+//calculation:
+Xc1 = 1/(2*%pi*f1*C)
+Xc2 = 1/(2*%pi*f2*C)
+
+printf("\n\n Result \n\n")
+printf("\n (a)Capacitive reactance, Xc = %.1f ohms ", Xc1)
+printf("\n (b)Capacitive reactance, Xc = %.3f ohms ", Xc2)
\ No newline at end of file diff --git a/608/CH15/EX15.04/15_04.sce b/608/CH15/EX15.04/15_04.sce new file mode 100755 index 000000000..c67f74ba3 --- /dev/null +++ b/608/CH15/EX15.04/15_04.sce @@ -0,0 +1,11 @@ +//Problem 15.04: A capacitor has a reactance of 40 ohms when operated on a 50 Hz supply. Determine the value of its capacitance.
+
+//initializing the variables:
+Z = 40; // in ohms
+f = 50; // in Hz
+
+//calculation:
+C = 1/(2*%pi*f1*Z)
+
+printf("\n\n Result \n\n")
+printf("\n Capacitance,C = %.2E F ", C)
\ No newline at end of file diff --git a/608/CH15/EX15.05/15_05.sce b/608/CH15/EX15.05/15_05.sce new file mode 100755 index 000000000..786193555 --- /dev/null +++ b/608/CH15/EX15.05/15_05.sce @@ -0,0 +1,13 @@ +//Problem 15.05: Calculate the current taken by a 23 μF capacitor when connected to a 240 V, 50 Hz supply.
+
+//initializing the variables:
+C = 23E-6; // in Farads
+f = 50; // in Hz
+V = 240; // in volts
+
+//calculation:
+Xc = 1/(2*%pi*f1*C)
+I = V/Xc
+
+printf("\n\n Result \n\n")
+printf("\n current I = %.2f A ",I)
\ No newline at end of file diff --git a/608/CH15/EX15.06/15_06.sce b/608/CH15/EX15.06/15_06.sce new file mode 100755 index 000000000..35c6c45b5 --- /dev/null +++ b/608/CH15/EX15.06/15_06.sce @@ -0,0 +1,13 @@ +//Problem 15.06: In a series R–L circuit the p.d. across the resistance R is 12 V and the p.d. across the inductance L is 5 V. Find the supply voltage and the phase angle between current and voltage.
+
+//initializing the variables:
+Vr = 12; // in volts
+Vl = 5; // in volts
+
+//calculation:
+V = (Vr^2 + Vl^2)^0.5
+phi = atan(Vl/Vr)
+phid = phi*180/%pi
+
+printf("\n\n Result \n\n")
+printf("\n supply voltage V = %.0f V, phase angle between current and voltage is %.2f° ",V, phid)
\ No newline at end of file diff --git a/608/CH15/EX15.07/15_07.sce b/608/CH15/EX15.07/15_07.sce new file mode 100755 index 000000000..b5316aee2 --- /dev/null +++ b/608/CH15/EX15.07/15_07.sce @@ -0,0 +1,19 @@ +//Problem 15.07: A coil has a resistance of 4 ohms and an inductance of 9.55 mH. Calculate (a) the reactance, (b) the impedance, and (c) the current taken from a 240 V, 50 Hz supply. Determine also the phase angle between the supply voltage and current.
+
+//initializing the variables:
+V = 240; // in volts
+R = 4; // in ohms
+L = 0.00955; // in Henry
+f = 50; // in Hz
+
+//calculation:
+XL = 2*%pi*f*L
+Z = (R^2 + XL^2)^0.5
+I = V/Z
+phid = atan(XL/R)*180/%pi
+
+printf("\n\n Result \n\n")
+printf("\n (a)Inductive reactance, XL = %.0f ohms",XL)
+printf("\n (b)Impedance, Z = %.0f ohms",Z)
+printf("\n (c)Current, I = %.0f A",I)
+printf("\n (d)phase angle between the supply voltage and current is %.2f°",phid)
\ No newline at end of file diff --git a/608/CH15/EX15.08/15_08.sce b/608/CH15/EX15.08/15_08.sce new file mode 100755 index 000000000..4f208f339 --- /dev/null +++ b/608/CH15/EX15.08/15_08.sce @@ -0,0 +1,20 @@ +//Problem 15.08: A coil takes a current of 2 A from a 12 V d.c. supply. When connected to a 240 V, 50 Hz supply the current is 20 A. Calculate the resistance, impedance, inductive reactance and inductance of the coil.
+
+//initializing the variables:
+Vdc = 12; // in volts
+Vac = 240; // in volts
+Iac = 20; // in Amperes
+Idc = 2; // in Amperes
+f = 50; // in Hz
+
+//calculation:
+R = Vdc/Idc
+Z = Vac/Iac
+XL = (Z^2 - R^2)^0.5
+L = XL/(2*%pi*f)
+
+printf("\n\n Result \n\n")
+printf("\n (a)Resistance, R = %.0f ohms",R)
+printf("\n (b)Impedance, Z = %.0f ohms",Z)
+printf("\n (c)Inductive reactance, XL = %.2f ohms",XL)
+printf("\n (d)Inductance, L = %.4f H",L)
\ No newline at end of file diff --git a/608/CH15/EX15.09/15_09.sce b/608/CH15/EX15.09/15_09.sce new file mode 100755 index 000000000..d434aea36 --- /dev/null +++ b/608/CH15/EX15.09/15_09.sce @@ -0,0 +1,22 @@ +//Problem 15.09: A coil of inductance 318.3 mH and negligible resistance is connected in series with a 200 ohms resistor to a 240 V, 50 Hz supply. Calculate (a) the inductive reactance of the coil, (b) the impedance of the circuit, (c) the current in the circuit, (d) the p.d. across each component, and (e) the circuit phase angle
+
+//initializing the variables:
+R = 200; // in ohms
+L = 0.3183; // in henry
+Vac = 240; // in volts
+f = 50; // in Hz
+
+//calculation:
+XL = 2*%pi*f*L
+Z = (R^2 + XL^2)^0.5
+I = V/Z
+VL = I*XL
+VR = I*R
+phid = atan(XL/R)*180/%pi
+
+printf("\n\n Result \n\n")
+printf("\n (a)Inductive reactance, XL = %.0f ohms",XL)
+printf("\n (b)Impedance, Z = %.1f ohms",Z)
+printf("\n (c)current, I = %.3f A",I)
+printf("\n (d)p.d. across Inductor, VL = %.1f V and p.d. across resistance, VR = %.1f V",VL, VR)
+printf("\n (e)circuit phase angle is %.2f°",phid)
\ No newline at end of file diff --git a/608/CH15/EX15.10/15_10.sce b/608/CH15/EX15.10/15_10.sce new file mode 100755 index 000000000..897c623ce --- /dev/null +++ b/608/CH15/EX15.10/15_10.sce @@ -0,0 +1,24 @@ +//Problem 15.10: A coil consists of a resistance of 100 ohms and an inductance of 200 mH. If an alternating voltage, v, given by v = 200 sin500t volts is applied across the coil, calculate (a) the circuit impedance, (b) the current flowing, (c) the p.d. across the resistance, (d) the p.d. across the inductance and (e) the phase angle between voltage and current.
+
+//initializing the variables:
+R = 100; // in ohms
+L = 0.2; // in henry
+Vmax = 200; // in volts
+w = 500; // in rad/sec
+
+//calculation:
+Vrms = 0.707*Vmax
+f = w/(2*%pi)
+XL = 2*%pi*f*L
+Z = (R^2 + XL^2)^0.5
+I = Vrms/Z
+VL = I*XL
+VR = I*R
+phid = atan(XL/R)*180/%pi
+
+printf("\n\n Result \n\n")
+printf("\n (a)Impedance, Z = %.1f ohms",Z)
+printf("\n (b)current, I = %.0f A",I)
+printf("\n (c)p.d. across resistance, VR = %.0f V", VR)
+printf("\n (d)p.d. across Inductor, VL = %.0f V",VL)
+printf("\n (e)circuit phase angle is %.0f°",phid)
\ No newline at end of file diff --git a/608/CH15/EX15.11/15_11.sce b/608/CH15/EX15.11/15_11.sce new file mode 100755 index 000000000..40b96b898 --- /dev/null +++ b/608/CH15/EX15.11/15_11.sce @@ -0,0 +1,18 @@ +//Problem 15.11: A pure inductance of 1.273 mH is connected in series with a pure resistance of 30 ohms. If the frequency of the sinusoidal supply is 5 kHz and the p.d. across the 30 ohm resistor is 6 V, determine the value of the supply voltage and the voltage across the 1.273 mH inductance. Draw the phasor diagram.
+
+//initializing the variables:
+R = 30; // in ohms
+L = 1.2273E-3; // in henry
+f = 5000; // in Hz
+VR = 6; // in volts
+
+//calculation:
+I =VR/R
+XL = 2*%pi*f*L
+Z = (R^2 + XL^2)^0.5
+V = I*Z
+VL = I*XL
+
+printf("\n\n Result \n\n")
+printf("\n (a)supply voltage = %.0f Volts",V)
+printf("\n (b)p.d. across Inductor, VL = %.0f V",VL)
\ No newline at end of file diff --git a/608/CH15/EX15.12/15_12.sce b/608/CH15/EX15.12/15_12.sce new file mode 100755 index 000000000..da3e084fb --- /dev/null +++ b/608/CH15/EX15.12/15_12.sce @@ -0,0 +1,29 @@ +//Problem 15.12: A coil of inductance 159.2 mH and resistance 20 ohms is connected in series with a 60 ohms resistor to a 240 V, 50 Hz supply. Determine (a) the impedance of the circuit, (b) the current in the circuit, (c) the circuit phase angle, (d) the p.d. across the 60 ohms resistor and (e) the p.d. across the coil. (f) Draw the circuit phasor diagram showing all voltages.
+
+//initializing the variables:
+R = 60; // in ohms
+Rc = 20; // in ohms
+L = 159.2E-3; // in henry
+f = 50; // in Hz
+V = 240; // in volts
+
+//calculation:
+XL = 2*%pi*f*L
+Rt = R + Rc
+Z = (Rt^2 + XL^2)^0.5
+I = V/Z
+phid = atan(XL/Rt)*180/%pi
+VR = I*R
+Zc = (Rc^2 + XL^2)^0.5
+Vc = I*Zc
+VL = I*XL
+VRc = I*Rc
+
+printf("\n\n Result \n\n")
+printf("\n (a)Impedance, Z = %.2f ohms",Z)
+printf("\n (b)current, I = %.3f A",I)
+printf("\n (c)circuit phase angle is %.0f°",phid)
+printf("\n (d)p.d. across resistance, VR = %.1f V", VR)
+printf("\n (e)p.d. across coil, Vc = %.0f V",Vc)
+printf("\n (f1)p.d. across Inductor, VL = %.1f V",VL)
+printf("\n (f2)p.d. across coil resistance, VRc = %.2f V",VRc)
\ No newline at end of file diff --git a/608/CH15/EX15.13/15_13.sce b/608/CH15/EX15.13/15_13.sce new file mode 100755 index 000000000..bd9d5d6bb --- /dev/null +++ b/608/CH15/EX15.13/15_13.sce @@ -0,0 +1,18 @@ +//Problem 15.13: A resistor of 25 ohms is connected in series with a capacitor of 45 μF. Calculate (a) the impedance, and (b) the current taken from a 240 V, 50 Hz supply. Find also the phase angle between the supply voltage and the current.
+
+//initializing the variables:
+R = 25; // in ohms
+C = 45E-6; // in Farads
+f = 50; // in Hz
+V = 240; // in volts
+
+//calculation:
+Xc = 1/(2*%pi*f*C)
+Z = (R^2 + Xc^2)^0.5
+I = V/Z
+phid = atan(Xc/R)*180/%pi
+
+printf("\n\n Result \n\n")
+printf("\n (a)Impedance, Z = %.2f ohms",Z)
+printf("\n (b)current, I = %.2f A",I)
+printf("\n (c)phase angle between the supply voltage and current is %.2f°",phid)
\ No newline at end of file diff --git a/608/CH15/EX15.14/15_14.sce b/608/CH15/EX15.14/15_14.sce new file mode 100755 index 000000000..fade7204e --- /dev/null +++ b/608/CH15/EX15.14/15_14.sce @@ -0,0 +1,22 @@ +//Problem 15.14: A capacitor C is connected in series with a 40 ohm resistor across a supply of frequency 60 Hz. A current of 3 A flows and the circuit impedance is 50 ohms. Calculate: (a) the value of capacitance, C, (b) the supply voltage, (c) the phase angle between the supply voltage and current, (d) the p.d. across the resistor, and (e) the p.d. across the capacitor. Draw the phasor diagram.
+
+//initializing the variables:
+R = 40; // in ohms
+f = 60; // in Hz
+I = 3; //in amperes
+Z = 50; // in ohms
+
+//calculation:
+Xc = (Z^2 - R^2)^0.5
+C = 1/(2*%pi*f*Xc)
+V = I*Z
+phid = atan(Xc/R)*180/%pi
+VR = I*R
+Vc = I*Xc
+
+printf("\n\n Result \n\n")
+printf("\n (a)capacitance, C = %.2E F",C)
+printf("\n (b)Voltage, V = %.0f Volts",V)
+printf("\n (c)phase angle between the supply voltage and current is %.2f°",phid)
+printf("\n (d)p.d. across resistance, VR = %.0f V", VR)
+printf("\n (e)p.d. across Capacitor, Vc = %.0f V",Vc)
\ No newline at end of file diff --git a/608/CH15/EX15.15/15_15.sce b/608/CH15/EX15.15/15_15.sce new file mode 100755 index 000000000..09ef15de3 --- /dev/null +++ b/608/CH15/EX15.15/15_15.sce @@ -0,0 +1,27 @@ +//Problem 15.15: A coil of resistance 5 ohms and inductance 120 mH in series with a 100 μF capacitor, is connected to a 300 V, 50 Hz supply. Calculate (a) the current flowing, (b) the phase difference between the supply voltage and current, (c) the voltage across the coil and (d) the voltage across the capacitor.
+
+//initializing the variables:
+R = 5; // in ohms
+C = 100E-6; // in Farads
+L = 0.12; // in Henry
+f = 50; // in Hz
+V = 300; // in volts
+
+//calculation:
+XL = 2*%pi*f*L
+Xc = 1/(2*%pi*f*C)
+X = XL - Xc
+//Since XL is greater than Xc, the circuit is inductive.
+Z = (R^2 + (XL-Xc)^2)^0.5
+I = V/Z
+phid = atan((XL-Xc)/R)*180/%pi
+Zcl = (R^2 + XL^2)^0.5
+Vcl = I*Zcl
+phidc = atan(XL/R)*180/%pi
+Vc = I*Xc
+
+printf("\n\n Result \n\n")
+printf("\n (a)Current,I = %.2f A",I)
+printf("\n (b)phase angle between the supply voltage and current is %.2f°",phid)
+printf("\n (c)Voltage across the coil, Vcoil = %.0f Volts",Vcl)
+printf("\n (d)p.d. across Capacitor, Vc = %.0f V",Vc)
\ No newline at end of file diff --git a/608/CH15/EX15.16/15_16.sce b/608/CH15/EX15.16/15_16.sce new file mode 100755 index 000000000..bc89719fa --- /dev/null +++ b/608/CH15/EX15.16/15_16.sce @@ -0,0 +1,30 @@ +//Problem 15.16: The following three impedances are connected in series across a 40 V, 20 kHz supply: (i) a resistance of 8 ohms, (ii) a coil of inductance 130 μH and 5 ohms resistance, and (iii) a 10 ohms resistor in series with a 0.25 μF capacitor. Calculate (a) the circuit current, (b) the circuit phase angle and (c) the voltage drop across each impedance.
+
+//initializing the variables:
+R1 = 8; // in ohms
+C = 0.25E-6; // in Farads
+L = 130E-6; // in Henry
+Rc = 5; // in ohms
+R2 = 10; // in ohms
+f = 20000; // in Hz
+V = 40; // in volts
+
+//calculation:
+XL = 2*%pi*f*L
+Xc = 1/(2*%pi*f*C)
+X = Xc - XL
+R = R1 + R2 + Rc
+//Since Xc is greater than XL, the circuit is capacitive.
+Z = (R^2 + (Xc-XL)^2)^0.5
+I = V/Z
+phid = atan((Xc-XL)/R)*180/%pi
+V1 = I*R1
+V2 = I*((Rc^2 + XL^2)^0.5)
+V3 = I*((R2^2 + Xc^2)^0.5)
+
+printf("\n\n Result \n\n")
+printf("\n (a)Current,I = %.3f A",I)
+printf("\n (b)circuit phase angle is %.2f°",phid)
+printf("\n (c1)Voltage across the Resistance of 8 ohms = %.2f Volts",V1)
+printf("\n (c2)Voltage across the coil, Vcoil = %.2f Volts",V2)
+printf("\n (c3)p.d. across Capacitor resistance circuit = %.2f Volts",V3)
\ No newline at end of file diff --git a/608/CH15/EX15.17/15_17.sce b/608/CH15/EX15.17/15_17.sce new file mode 100755 index 000000000..b3689daed --- /dev/null +++ b/608/CH15/EX15.17/15_17.sce @@ -0,0 +1,25 @@ +//Problem 15.17: Determine the p.d.’s V1 and V2 for the circuit shown in Figure 15.17 if the frequency of the supply is 5 kHz. Draw the phasor diagram and hence determine the supply voltage V and the circuit phase angle.
+
+//initializing the variables:
+R1 = 4; // in ohms
+C = 1.273E-6; // in Farads
+L = 286E-6; // in Henry
+R2 = 8; // in ohms
+f = 5000; // in Hz
+I = 5; // in amperes
+
+//calculation:
+XL = 2*%pi*f*L
+phid1 = atan(XL/R)*180/%pi
+V1 = I*((R1^2 + XL^2)^0.5)
+Xc = 1/(2*%pi*f*C)
+V2 = I*((R2^2 + Xc^2)^0.5)
+phid2 = atan(Xc/R)*180/%pi
+Z = ((R1+R2)^2 + (Xc-XL)^2)^0.5
+V = I*Z
+phid = atan((Xc-XL)/(R1+R2))*180/%pi
+
+printf("\n\n Result \n\n")
+printf("\n Voltage V1 = %.2f V and V2 = %.1f V",V1, V2)
+printf("\n Voltage supply, V = %.0f V",V)
+printf("\n circuit phase angle is %.2f°",phid)
\ No newline at end of file diff --git a/608/CH15/EX15.18/15_18.sce b/608/CH15/EX15.18/15_18.sce new file mode 100755 index 000000000..af750a6d5 --- /dev/null +++ b/608/CH15/EX15.18/15_18.sce @@ -0,0 +1,16 @@ +//Problem 15.18: A coil having a resistance of 10 ohms and an inductanc of 125 mH is connected in series with a 60 μF capacitor across a 120 V supply. At what frequency does resonance occur? Find the current flowing at the resonant frequency.
+
+//initializing the variables:
+R = 10; // in ohms
+C = 60E-6; // in Farads
+L = 125E-3; // in Henry
+V = 120; // in Volts
+
+//calculation:
+fr = 1/(2*%pi*(L*C)^0.5)
+//At resonance, XL = Xc and impedance Z = R
+I = V/R
+
+printf("\n\n Result \n\n")
+printf("\n (a)Resonance frequency,Fr = %.2f Hz",fr)
+printf("\n (b)Current, I = %.0f",I)
\ No newline at end of file diff --git a/608/CH15/EX15.19/15_19.sce b/608/CH15/EX15.19/15_19.sce new file mode 100755 index 000000000..4311da66c --- /dev/null +++ b/608/CH15/EX15.19/15_19.sce @@ -0,0 +1,17 @@ +//Problem 15.19: The current at resonance in a series L–C–R circuit is 100 μA. If the applied voltage is 2 mV at a frequency of 200 kHz, and the circuit inductance is 50 μH, find (a) the circuit resistance, and (b) the circuit capacitance.
+
+//initializing the variables:
+L = 0.05E-3; // in Henry
+fr = 200000; // in Hz
+V = 0.002; // in Volts
+I = 0.1E-3; // in amperes
+
+//calculation:
+// L-C-R
+//At resonance, XL = Xc and impedance Z = R
+R = V/I
+C = 1/(L*(2*%pi*fr)^2)
+
+printf("\n\n Result \n\n")
+printf("\n (a)Resistance, R = %.0f ohms",R)
+printf("\n (b)Capacitance, C = %.2E F",C)
\ No newline at end of file diff --git a/608/CH15/EX15.20/15_20.sce b/608/CH15/EX15.20/15_20.sce new file mode 100755 index 000000000..f07e74d43 --- /dev/null +++ b/608/CH15/EX15.20/15_20.sce @@ -0,0 +1,20 @@ +//Problem 15.20: A coil of inductance 80 mH and negligible resistance is connected in series with a capacitance of 0.25 μF and a resistor of resistance 12.5 ohms across a 100 V, variable frequency supply. Determine (a) the resonant frequency, and (b) the current at resonance. How many times greater than the supply voltage is the voltage across the reactances at resonance?
+
+//initializing the variables:
+L = 80E-3; // in Henry
+C = 0.25E-6; // in Farads
+R = 12.5; // in ohms
+V = 100; // in Volts
+
+//calculation:
+fr = 1/(2*%pi*((L*C)^0.5))
+//At resonance, XL = Xc and impedance Z = R
+I = V/R
+VL = I*(2*%pi*fr*L)
+Vc = I/(2*%pi*fr*C)
+Vm = VL/V
+
+printf("\n\n Result \n\n")
+printf("\n (a)the resonant frequency = %.1f Hz",fr)
+printf("\n (b)Current, I = %.0f A",I)
+printf("\n (b)Voltage magnification at resonance = %.3f V",Vm)
\ No newline at end of file diff --git a/608/CH15/EX15.21/15_21.sce b/608/CH15/EX15.21/15_21.sce new file mode 100755 index 000000000..26283b1fe --- /dev/null +++ b/608/CH15/EX15.21/15_21.sce @@ -0,0 +1,12 @@ +//Problem 15.21: A series circuit comprises a coil of resistance 2 ohms and inductance 60 mH, and a 30 μF capacitor. Determine the Qfactor of the circuit at resonance.
+
+//initializing the variables:
+L = 60E-3; // in Henry
+C = 30E-6; // in Farads
+R = 2; // in ohms
+
+//calculation:
+Q = ((L/C)^0.5)/R
+
+printf("\n\n Result \n\n")
+printf("\n At resonance, Q-factor = %.2f",Q)
\ No newline at end of file diff --git a/608/CH15/EX15.22/15_22.sce b/608/CH15/EX15.22/15_22.sce new file mode 100755 index 000000000..f2143b83b --- /dev/null +++ b/608/CH15/EX15.22/15_22.sce @@ -0,0 +1,21 @@ +//Problem 15.22: A coil of negligible resistance and inductance 100 mH is connected in series with a capacitance of 2 μF and a resistance of 10 across a 50 V, variable frequency supply. Determine (a) the resonant frequency, (b) the current at resonance, (c) the voltages across the coil and the capacitor at resonance, and (d) the Q-factor of the circuit.
+
+//initializing the variables:
+L = 100E-3; // in Henry
+C = 2E-6; // in Farads
+R = 10; // in ohms
+V = 50; // in Volts
+
+//calculation:
+fr = 1/(2*%pi*((L*C)^0.5))
+//At resonance, XL = Xc and impedance Z = R
+I = V/R
+VL = I*(2*%pi*fr*L)
+Vc = I/(2*%pi*fr*C)
+Q = VL/V
+
+printf("\n\n Result \n\n")
+printf("\n (a)the resonant frequency = %.1f Hz",fr)
+printf("\n (b)Current, I = %.0f A",I)
+printf("\n (c)Voltage across coil at resonance is %.0fV and Voltage across capacitance at resonance is %.0fV",VL, Vc)
+printf("\n (d)At resonance, Q-factor = %.2f",Q)
\ No newline at end of file diff --git a/608/CH15/EX15.23/15_23.sce b/608/CH15/EX15.23/15_23.sce new file mode 100755 index 000000000..0ba84a0c2 --- /dev/null +++ b/608/CH15/EX15.23/15_23.sce @@ -0,0 +1,13 @@ +//Problem 15.23: A filter in the form of a series L–R–C circuit is designed to operate at a resonant frequency of 5 kHz. Included within the filter is a 20 mH inductance and 10 ohm resistance. Determine the bandwidth of the filter.
+
+//initializing the variables:
+L = 20E-3; // in Henry
+R = 10; // in ohms
+fr = 5000; // in Hz
+
+//calculation:
+Qr = (2*%pi*fr)*L/R
+bw = fr/Qr
+
+printf("\n\n Result \n\n")
+printf("\n Bandwidth, (f2-f1) = %.1f Hz",bw)
\ No newline at end of file diff --git a/608/CH15/EX15.24/15_24.sce b/608/CH15/EX15.24/15_24.sce new file mode 100755 index 000000000..33dc7f33a --- /dev/null +++ b/608/CH15/EX15.24/15_24.sce @@ -0,0 +1,12 @@ +//Problem 15.24: An instantaneous current, i = 250 sin w t mA flows through a pure resistance of 5 kohm. Find the power dissipated in the resistor.
+
+//initializing the variables:
+R = 5000; // in ohms
+Imax = 0.250; // in Amperes
+
+//calculation:
+Irms = 0.707*Imax
+P = Irms*Irms*R
+
+printf("\n\n Result \n\n")
+printf("\n Power, P = %.1f Watts",P)
\ No newline at end of file diff --git a/608/CH15/EX15.25/15_25.sce b/608/CH15/EX15.25/15_25.sce new file mode 100755 index 000000000..70626c2ef --- /dev/null +++ b/608/CH15/EX15.25/15_25.sce @@ -0,0 +1,16 @@ +//Problem 15.25: A series circuit of resistance 60 ohm and inductance 75 mH is connected to a 110 V, 60 Hz supply. Calculate the power dissipated.
+
+//initializing the variables:
+R = 60; // in ohms
+L = 75E-3; // in Henry
+V = 110; // in Volts
+f = 60; // in Hz
+
+//calculation:
+XL = 2*%pi*f*L
+Z = (R^2 + XL^2)^0.5
+I = V/Z
+P = I*I*R
+
+printf("\n\n Result \n\n")
+printf("\n Power, P = %.0f Watts",P)
\ No newline at end of file diff --git a/608/CH15/EX15.26/15_26.sce b/608/CH15/EX15.26/15_26.sce new file mode 100755 index 000000000..7c6c149fd --- /dev/null +++ b/608/CH15/EX15.26/15_26.sce @@ -0,0 +1,14 @@ +//Problem 15.26: A pure inductance is connected to a 150 V, 50 Hz supply, and the apparent power of the circuit is 300 VA. Find the value of the inductance.
+
+//initializing the variables:
+VI = 300; // in VA
+V = 150; // in Volts
+f = 50; // in Hz
+
+//calculation:
+I = VI/V
+XL = V/I
+L = XL/(2*%pi*f)
+
+printf("\n\n Result \n\n")
+printf("\n Inductance = %.3f H",L)
\ No newline at end of file diff --git a/608/CH15/EX15.27/15_27.sce b/608/CH15/EX15.27/15_27.sce new file mode 100755 index 000000000..083544e0e --- /dev/null +++ b/608/CH15/EX15.27/15_27.sce @@ -0,0 +1,12 @@ +//Problem 15.27: A transformer has a rated output of 200 kVA at a power factor of 0.8. Determine the rated power output and the corresponding reactive power.
+
+//initializing the variables:
+VI = 200000; // in VA
+pf = 0.8; // power factor
+
+//calculation:
+P = VI*pf
+Q = VI*sin(acos(pf))
+
+printf("\n\n Result \n\n")
+printf("\n rated power output is %.0fW and the corresponding reactive power is %.0f var",P,Q)
\ No newline at end of file diff --git a/608/CH15/EX15.28/15_28.sce b/608/CH15/EX15.28/15_28.sce new file mode 100755 index 000000000..0b1f95c1d --- /dev/null +++ b/608/CH15/EX15.28/15_28.sce @@ -0,0 +1,21 @@ +//Problem 15.28: The power taken by an inductive circuit when connected to a 120 V, 50 Hz supply is 400 W and the current is 8 A. Calculate (a) the resistance, (b) the impedance, (c) the reactance, (d) the power factor, and (e) the phase angle between voltage and current.
+
+//initializing the variables:
+V = 120; // in Volts
+f = 50; // in Hz
+P = 400; // in Watt
+I = 8; // in Amperes
+
+//calculation:
+R = P/(I*I)
+Z = V/I
+XL = (Z^2 - R^2)^0.5
+pf = P/(V*I)
+phi = acos(pf)*180/%pi
+
+printf("\n\n Result \n\n")
+printf("\n (a)resistance = %.2f ohm ",R)
+printf("\n (b)Impedance Z = %.0f Ohm ",Z)
+printf("\n (c)reactance = %.2f ohm ",XL)
+printf("\n (d)Power factor = %.4f",pf)
+printf("\n (e)phase angle = %.2f°",phi)
\ No newline at end of file diff --git a/608/CH15/EX15.29/15_29.sce b/608/CH15/EX15.29/15_29.sce new file mode 100755 index 000000000..cae8816d3 --- /dev/null +++ b/608/CH15/EX15.29/15_29.sce @@ -0,0 +1,22 @@ +//Problem 15.29: A circuit consisting of a resistor in series with a capacitor takes 100 watts at a power factor of 0.5 from a 100 V, 60 Hz supply. Find (a) the current flowing, (b) the phase angle, (c) the resistance, (d) the impedance, and (e) the capacitance.
+
+//initializing the variables:
+V = 100; // in Volts
+f = 60; // in Hz
+P = 100; // in Watt
+pf = 0.5; // power factor
+
+//calculation:
+I = P/(pf*V)
+phi = acos(pf)*180/%pi
+R = P/(I*I)
+Z = V/I
+Xc = (Z^2 - R^2)^0.5
+C = 1/(2*%pi*f*Xc)
+
+printf("\n\n Result \n\n")
+printf("\n (a)Current I = %.0f A ",I)
+printf("\n (b)phase angle = %.0f°",phi)
+printf("\n (c)resistance = %.0f ohm ",R)
+printf("\n (d)Impedance Z = %.0f Ohm ",Z)
+printf("\n (e)capacitance = %.2E F ",C)
\ No newline at end of file diff --git a/608/CH16/EX16.01/16_01.sce b/608/CH16/EX16.01/16_01.sce new file mode 100755 index 000000000..a2a148717 --- /dev/null +++ b/608/CH16/EX16.01/16_01.sce @@ -0,0 +1,24 @@ +//Problem 16.01: A 20 ohm resistor is connected in parallel with an inductance of 2.387 mH across a 60 V, 1 kHz supply. Calculate (a) the current in each branch, (b) the supply current, (c) the circuit phase angle, (d) the circuit impedance, and (e) the power consumed.
+
+//initializing the variables:
+R = 20; // in Ohms
+L = 2.387E-3; // in Henry
+V = 60; // in Volts
+f = 1000; // in Hz
+
+//calculation:
+IR = V/R
+XL = 2*%pi*f*L
+IL = V/XL
+I = (IR^2 + IL^2)^0.5
+phi = atan(IL/IR)
+phid = phi*180/%pi
+Z = V/I
+P = V*I*cos(phi)
+
+printf("\n\n Result \n\n")
+printf("\n (a)Current through resistor is %.0f A and current through Inductor is %.0f A",IR, IL)
+printf("\n (b)current, I = %.0f A ",I)
+printf("\n (c)phase angle = %.2f°",phid)
+printf("\n (d)Impedance Z = %.0f Ohm ",Z)
+printf("\n (e)Power consumed = %.0f Watt ",P)
\ No newline at end of file diff --git a/608/CH16/EX16.02/16_02.sce b/608/CH16/EX16.02/16_02.sce new file mode 100755 index 000000000..11f69afb1 --- /dev/null +++ b/608/CH16/EX16.02/16_02.sce @@ -0,0 +1,26 @@ +//Problem 16.02: A 30 μF capacitor is connected in parallel with an 80 ohms resistor across a 240 V, 50 Hz supply. Calculate (a) the current in each branch, (b) the supply current, (c) the circuit phase angle, (d) the circuit impedance, (e) the power dissipated, and (f) the apparent power.
+
+//initializing the variables:
+R = 80; // in Ohms
+C = 30E-6; // in Farads
+V = 240; // in Volts
+f = 50; // in Hz
+
+//calculation:
+IR = V/R
+Xc = 1/(2*%pi*f*C)
+Ic = V/Xc
+I = (IR^2 + Ic^2)^0.5
+phi = atan(Ic/IR)
+phid = phi*180/%pi
+Z = V/I
+P = V*I*cos(phi)
+S = V*I
+
+printf("\n\n Result \n\n")
+printf("\n (a)Current through resistor is %.0f A and current through capacitor is %.3f A",IR, Ic)
+printf("\n (b)current, I = %.3f A ",I)
+printf("\n (c)phase angle = %.2f°",phid)
+printf("\n (d)Impedance Z = %.2f Ohm ",Z)
+printf("\n (e)Power consumed = %.0f Watt ",P)
+printf("\n (f)apparent Power = %.1f VA ",S)
\ No newline at end of file diff --git a/608/CH16/EX16.03/16_03.sce b/608/CH16/EX16.03/16_03.sce new file mode 100755 index 000000000..b97cb57ac --- /dev/null +++ b/608/CH16/EX16.03/16_03.sce @@ -0,0 +1,19 @@ +//Problem 16.03: A capacitor C is connected in parallel with a resistor R across a 120 V, 200 Hz supply. The supply current is 2 A at a power factor of 0.6 leading. Determine the values of C and R.
+
+//initializing the variables:
+pf = 0.6; // power factor
+V = 120; // in Volts
+f = 200; // in Hz
+I = 2; // in Amperes
+
+//calculation:
+phi = acos(pf)
+phid = phi*180/%pi
+IR = I*cos(phi)
+Ic = I*sin(phi)
+R = V/IR
+C = Ic/(2*%pi*f*V)
+
+printf("\n\n Result \n\n")
+printf("\n (a)Resistance R = %.0f Ohm ",R)
+printf("\n (b)Capacitance,C = %.2E F ",C)
\ No newline at end of file diff --git a/608/CH16/EX16.04/16_04.sce b/608/CH16/EX16.04/16_04.sce new file mode 100755 index 000000000..f78bf0237 --- /dev/null +++ b/608/CH16/EX16.04/16_04.sce @@ -0,0 +1,27 @@ +//Problem 16.04: A pure inductance of 120 mH is connected in parallel with a 25 μF capacitor and the network is connected to a 100 V, 50 Hz supply. Determine (a) the branch currents, (b) the supply current and its phase angle, (c) the circuit impedance, and (d) the power consumed.
+
+//initializing the variables:
+C = 25E-6; // in Farads
+L = 120E-3; // in Henry
+V = 100; // in Volts
+f = 50; // in Hz
+
+//calculation:
+XL = 2*%pi*f*L
+IL = V/XL
+Xc = 1/(2*%pi*f*C)
+Ic = V/Xc
+//IL and Ic are anti-phase. Hence supply current,
+I = IL - Ic
+//the current lags the supply voltage V by 90°
+phi = %pi/2
+phid = phi*180/%pi
+Z = V/I
+P = V*I*cos(phi)
+
+printf("\n\n Result \n\n")
+printf("\n (a)Current through Inductor is %.3f A and current through capacitor is %.3f A",IL, Ic)
+printf("\n (b)current, I = %.3f A ",I)
+printf("\n (c)phase angle = %.0f°",phid)
+printf("\n (d)Impedance Z = %.2f Ohm ",Z)
+printf("\n (e)Power consumed = %.0f Watt ",P)
\ No newline at end of file diff --git a/608/CH16/EX16.05/16_05.sce b/608/CH16/EX16.05/16_05.sce new file mode 100755 index 000000000..71f8f3d98 --- /dev/null +++ b/608/CH16/EX16.05/16_05.sce @@ -0,0 +1,27 @@ +//Problem 16.05: Repeat Problem 16.04 for the condition when the frequency is changed to 150 Hz.
+
+//initializing the variables:
+C = 25E-6; // in Farads
+L = 120E-3; // in Henry
+V = 100; // in Volts
+f = 150; // in Hz
+
+//calculation:
+XL = 2*%pi*f*L
+IL = V/XL
+Xc = 1/(2*%pi*f*C)
+Ic = V/Xc
+//IL and Ic are anti-phase. Hence supply current,
+I = Ic - IL
+//the current leads the supply voltage V by 90°
+phi = %pi/2
+phid = phi*180/%pi
+Z = V/I
+P = V*I*cos(phi)
+
+printf("\n\n Result \n\n")
+printf("\n (a)Current through Inductor is %.3f A and current through capacitor is %.3f A",IL, Ic)
+printf("\n (b)current, I = %.3f A ",I)
+printf("\n (c)phase angle = %.0f°",phid)
+printf("\n (d)Impedance Z = %.2f Ohm ",Z)
+printf("\n (e)Power consumed = %.0f Watt ",P)
\ No newline at end of file diff --git a/608/CH16/EX16.06/16_06.sce b/608/CH16/EX16.06/16_06.sce new file mode 100755 index 000000000..35bbb8004 --- /dev/null +++ b/608/CH16/EX16.06/16_06.sce @@ -0,0 +1,37 @@ +//Problem 16.06: A coil of inductance 159.2 mH and resistance 40 ohm is connected in parallel with a 30 μF capacitor across a 240 V, 50 Hz supply. Calculate (a) the current in the coil and its phase angle, (b) the current in the capacitor and its phase angle, (c) the supply current and its phase angle,(d) the circuit impedance, (e) the power consumed, (f) the apparent power, and (g) the reactive power. Draw the phasor diagram.
+
+//initializing the variables:
+C = 30E-6; // in Farads
+R = 40; // in Ohms
+L = 159.2E-3; // in Henry
+V = 240; // in Volts
+f = 50; // in Hz
+
+//calculation:
+XL = 2*%pi*f*L
+Z1 = (R^2 + XL^2)^0.5
+ILR = V/Z1
+phi1 = atan(XL/R)
+phi1d = phi1*180/%pi
+Xc = 1/(2*%pi*f*C)
+Ic = V/Xc
+phi2 = %pi/2
+phi2d = phi2*180/%pi
+Ih = ILR*cos(phi1) + Ic*cos(phi2)
+Iv = -1*ILR*sin(phi1) + Ic*sin(phi2)
+I = (Ih^2 + Iv^2)^0.5
+phi = atan(abs(Iv)/Ih)
+Z = V/I
+P = V*I*cos(phi)
+phid = phi*180/%pi
+S = V*I
+Q = V*I*sin(phi)
+
+printf("\n\n Result \n\n")
+printf("\n (a)Current through coil is %.3f A and lagged by phase angle is %.2f°",ILR,phi1d)
+printf("\n (b)Current through capacitor is %.3f A and lead by phase angle is %.0f°",Ic,phi2d)
+printf("\n (c)supply Current is %.3f A and lagged by phase angle is %.2f°",I,phid)
+printf("\n (d)Impedance Z = %.2f Ohm ",Z)
+printf("\n (e)Power consumed = %.0f Watt ",P)
+printf("\n (f)apparent Power = %.1f VA ",S)
+printf("\n (g)reactive Power = %.1f var ",Q)
\ No newline at end of file diff --git a/608/CH16/EX16.07/16_07.sce b/608/CH16/EX16.07/16_07.sce new file mode 100755 index 000000000..0fe4f6499 --- /dev/null +++ b/608/CH16/EX16.07/16_07.sce @@ -0,0 +1,33 @@ +//Problem 16.07: A coil of inductance 0.12 H and resistance 3 kohm is connected in parallel with a 0.02 μF capacitor and is supplied at 40 V at a frequency of 5 kHz. Determine (a) the current in the coil, and (b) the current in the capacitor. (c) Draw to scale the phasor diagram and measure the supply current and its phase angle; check the answer by calculation. Determine (d) the circuit impedance and (e) the power consumed.
+
+//initializing the variables:
+C = 0.02E-6; // in Farads
+R = 3000; // in Ohms
+L = 120E-3; // in Henry
+V = 40; // in Volts
+f = 5000; // in Hz
+
+//calculation:
+XL = 2*%pi*f*L
+Z1 = (R^2 + XL^2)^0.5
+ILR = V/Z1
+phi1 = atan(XL/R)
+phi1d = phi1*180/%pi
+Xc = 1/(2*%pi*f*C)
+Ic = V/Xc
+phi2 = %pi/2
+phi2d = phi2*180/%pi
+Ih = ILR*cos(phi1) + Ic*cos(phi2)
+Iv = -1*ILR*sin(phi1) + Ic*sin(phi2)
+I = (Ih^2 + Iv^2)^0.5
+phi = atan((Iv)/Ih)
+phid = phi*180/%pi
+Z = V/I
+P = V*I*cos(phi)
+
+printf("\n\n Result \n\n")
+printf("\n (a)Current through coil is %.5f A and lagged by phase angle is %.2f°",ILR,phi1d)
+printf("\n (b)Current through capacitor is %.5f A and lead by phase angle is %.0f°",Ic,phi2d)
+printf("\n (c)supply Current is %.5f A and lagged by phase angle is %.2f°",I,phid)
+printf("\n (d)Impedance Z = %.2f Ohm ",Z)
+printf("\n (e)Power consumed = %.4f Watt ",P)
\ No newline at end of file diff --git a/608/CH16/EX16.08/16_08.sce b/608/CH16/EX16.08/16_08.sce new file mode 100755 index 000000000..1205eac31 --- /dev/null +++ b/608/CH16/EX16.08/16_08.sce @@ -0,0 +1,16 @@ +//Problem 16.08: A pure inductance of 150 mH is connected in parallel with a 40 μF capacitor across a 50 V, variable frequency supply. Determine (a) the resonant frequency of the circuit and (b) the current circulating in the capacitor and inductance at resonance.
+
+//initializing the variables:
+C = 40E-6; // in Farads
+R = 0; // in Ohms
+L = 150E-3; // in Henry
+V = 50; // in Volts
+
+//calculation:
+fr = ((1/(L*C) - R*R/(L*L))^0.5)/(2*%pi)
+Xc = 1/(2*%pi*fr*C)
+Icirc = V/Xc
+
+printf("\n\n Result \n\n")
+printf("\n (a)Parallel resonant frequency, fr = %.2f Hz ",fr)
+printf("\n (b)Current circulating in L and C at resonance = %.3f A ",Icirc)
\ No newline at end of file diff --git a/608/CH16/EX16.09/16_09.sce b/608/CH16/EX16.09/16_09.sce new file mode 100755 index 000000000..707d3a9f8 --- /dev/null +++ b/608/CH16/EX16.09/16_09.sce @@ -0,0 +1,19 @@ +//Problem 16.09: A coil of inductance 0.20 H and resistance 60 ohm is connected in parallel with a 20 μF capacitor across a 20 V, variable frequency supply. Calculate (a) the resonant frequency, (b) the dynamic resistance, (c) the current at resonance and (d) the circuit Q-factor at resonance.
+
+//initializing the variables:
+C = 20E-6; // in Farads
+R = 60; // in Ohms
+L = 200E-3; // in Henry
+V = 20; // in Volts
+
+//calculation:
+fr = ((1/(L*C) - R*R/(L*L))^0.5)/(2*%pi)
+Rd = L/(R*C)
+Ir = V/Rd
+Q = 2*%pi*fr*L/R
+
+printf("\n\n Result \n\n")
+printf("\n (a)Parallel resonant frequency, fr = %.2f Hz ",fr)
+printf("\n (b)the dynamic resistance,RD = %.1f ohm ",Rd)
+printf("\n (c)Current at resonance = %.2f A ",Ir)
+printf("\n (d)Q-factor = %.2f",Q)
\ No newline at end of file diff --git a/608/CH16/EX16.10/16_10.sce b/608/CH16/EX16.10/16_10.sce new file mode 100755 index 000000000..cc9562ba1 --- /dev/null +++ b/608/CH16/EX16.10/16_10.sce @@ -0,0 +1,19 @@ +//Problem 16.10: A coil of inductance 100 mH and resistance 800 ohm is connected in parallel with a variable capacitor across a 12 V, 5 kHz supply. Determine for the condition when the supply current is a minimum: (a) the capacitance of the capacitor, (b) the dynamic resistance, (c) the supply current, and (d) the Q-factor.
+
+//initializing the variables:
+fr = 5000; // in ohm
+R = 800; // in Ohms
+L = 100E-3; // in Henry
+V = 12; // in Volts
+
+//calculation:
+C = 1/(L*{(2*%pi*fr)^2 + R*R/(L*L)})
+Rd = L/(R*C)
+Ir = V/Rd
+Q = 2*%pi*fr*L/R
+
+printf("\n\n Result \n\n")
+printf("\n (a)capacitance, C = %.3E F ",C)
+printf("\n (b)the dynamic resistance,RD = %.0f ohm ",Rd)
+printf("\n (c)Current at resonance = %.3E A ",Ir)
+printf("\n (d)Q-factor = %.2f ",Q)
\ No newline at end of file diff --git a/608/CH16/EX16.11/16_11.sce b/608/CH16/EX16.11/16_11.sce new file mode 100755 index 000000000..71c4c4613 --- /dev/null +++ b/608/CH16/EX16.11/16_11.sce @@ -0,0 +1,17 @@ +//Problem 16.11: A single-phase motor takes 50 A at a power factor of 0.6 lagging from a 240 V, 50 Hz supply. Determine (a) the current taken by a capacitor connected in parallel with the motor to correct the power factor to unity, and (b) the value of the supply current after power factor correction.
+
+//initializing the variables:
+f = 50; // in ohm
+V = 240; // in Volts
+pf = 0.6// power factor
+Im = 50; // in amperes
+
+//calculation:
+phi = acos(pf)
+phid = phi*180/%pi
+Ic = Im*sin(phi)
+I = Im*cos(phi)
+
+printf("\n\n Result \n\n")
+printf("\n (a)the capacitor current Ic must be %.0f A for the power factor to be unity. ",Ic)
+printf("\n (b)Supply current I = %.0f A ",I)
\ No newline at end of file diff --git a/608/CH16/EX16.12/16_12.sce b/608/CH16/EX16.12/16_12.sce new file mode 100755 index 000000000..3f5bbe10c --- /dev/null +++ b/608/CH16/EX16.12/16_12.sce @@ -0,0 +1,34 @@ +//Problem 16.12: A motor has an output of 4.8 kW, an efficiency of 80% and a power factor of 0.625 lagging when operated from a 240 V, 50 Hz supply. It is required to improve the power factor to 0.95 lagging by connecting a capacitor in parallel with the motor. Determine (a) the current taken by the motor, (b) the supply current after power factor correction, (c) the current taken by the capacitor, (d) the capacitance of the capacitor, and (e) the kvar rating of the capacitor.
+
+//initializing the variables:
+Pout = 4800; // in Watt
+eff = 0.8;// effficiency
+f = 50; // in ohm
+V = 240; // in Volts
+pf1 = 0.625// power factor
+pf2 = 0.95// power factor
+
+//calculation:
+Pin = Pout/eff
+Im = Pin/(V*pf1)
+phi1 = acos(pf1)
+phi1d = phi1*180/%pi
+//When a capacitor C is connected in parallel with the motor a current Ic flows which leads V by 90°.
+phi2 = acos(pf2)
+phi2d = phi2*180/%pi
+Imh = Im*cos(phi1)
+//Ih = I*cos(phi2)
+Ih = Imh
+I = Ih/cos(phi2)
+Imv = Im*sin(phi1)
+Iv = I*sin(phi2)
+Ic = Imv - Iv
+C = Ic/(2*%pi*f*V)
+kvar = V*Ic/1000
+
+printf("\n\n Result \n\n")
+printf("\n (a)current taken by the motor, Im = %.0f A",Im)
+printf("\n (b)supply current after p.f. correction, I = %.2f A ",I)
+printf("\n (c)magnitude of the capacitor current Ic = %.0f A",Ic)
+printf("\n (d)capacitance, C = %.0f μF ",(C/1E-6))
+printf("\n (d)kvar rating of the capacitor = %.2f kvar ",kvar)
\ No newline at end of file diff --git a/608/CH16/EX16.13/16_13.sce b/608/CH16/EX16.13/16_13.sce new file mode 100755 index 000000000..60106a64e --- /dev/null +++ b/608/CH16/EX16.13/16_13.sce @@ -0,0 +1,39 @@ +//Problem 16.13: A 250 V, 50 Hz single-phase supply feeds the following loads (i) incandescent lamps taking a current of 10 A at unity power factor, (ii) fluorescent lamps taking 8 A at a power factor of 0.7 lagging, (iii) a 3 kVA motor operating at full load and at a power factor of 0.8 lagging and (iv) a static capacitor. Determine, for the lamps and motor, (a) the total current, (b) the overall power factor and (c) the total power. (d) Find the value of the static capacitor to improve the overall power factor to 0.975 lagging.
+
+//initializing the variables:
+S = 3000; // in VA
+f = 50; // in ohm
+V = 250; // in Volts
+Iil = 10; // in Amperes
+Ifl = 8; // in Amperes
+pfil = 1// power factor
+pffl = 0.7// power factor
+pfm = 0.8// power factor
+pf0 = 0.975// power factor
+
+//calculation:
+phiil = acos(pfil)
+phiild = phiil*180/%pi
+phifl = acos(pffl)
+phifld = phifl*180/%pi
+phim = acos(pfm)
+phimd = phim*180/%pi
+phi0 = acos(pf0)
+phi0d = phi0*180/%pi
+Im = S/V
+Ih = Iil*cos(phiil) + Ifl*cos(phifl) + Im*cos(phim)
+Iv = Iil*sin(phiil) - Ifl*sin(phifl) - Im*sin(phim)
+Il = (Ih^2 + Iv^2)^0.5
+phi = atan(abs(Iv)/Ih)
+phid = phi*180/%pi
+pf = cos(phi)
+P = V*Il*pf
+I = Il*cos(phi)/cos(phi0)
+Ic = Il*sin(phi) - I*sin(phi0)
+C = Ic/(2*%pi*f*V)
+
+printf("\n\n Result \n\n")
+printf("\n (a)total current, Il = %.2f A",Il)
+printf("\n (b)Power factor = %.3f",pf)
+printf("\n (c)Total power, P = %.3f Watt",P)
+printf("\n (d)capacitance, C = %.2E F ",C)
\ No newline at end of file diff --git a/608/CH17/EX17.01/17_01.JPG b/608/CH17/EX17.01/17_01.JPG Binary files differnew file mode 100755 index 000000000..d6fc590f9 --- /dev/null +++ b/608/CH17/EX17.01/17_01.JPG diff --git a/608/CH17/EX17.01/17_01.sce b/608/CH17/EX17.01/17_01.sce new file mode 100755 index 000000000..1b2d5ec1d --- /dev/null +++ b/608/CH17/EX17.01/17_01.sce @@ -0,0 +1,21 @@ +//Problem 17.01: A 15 μF uncharged capacitor is connected in series with a 47 kohm resistor across a 120 V, d.c. supply. Use the tangential graphical method to draw the capacitor voltage/time characteristic of the circuit. From the characteristic, determine the capacitor voltage at a time equal to one time constant after being connected to the supply, and also two seconds after being connected to the supply. Also, find the time for the capacitor voltage to reach one half of its steady state value.
+
+//initializing the variables:
+C = 15E-6; // in Farads
+R = 47000; // in ohms
+V = 120; // in Volts
+
+//calculation:
+tou = R*C
+t1 = tou
+Vctou = V*(1-%e^(-1*t1/tou))
+Vct = V/2
+t = 0:0.1:10
+Vc = V*(1-%e^(-1*t/tou));
+plot(t,Vc)
+xtitle("capacitor voltage/time characteristic", "t", "Vc")
+t = -1*tou*log(1 - Vct/V)
+
+printf("\ = \n\n Result \n\n")
+printf("\n (a)the capacitor voltage at a time equal to one time constant = %.2f V",Vctou)
+printf("\n (b)the time for the capacitor voltage to reach one half of its steady state value = %.1f sec",t)
\ No newline at end of file diff --git a/608/CH17/EX17.02/17_02.JPG b/608/CH17/EX17.02/17_02.JPG Binary files differnew file mode 100755 index 000000000..136b4d003 --- /dev/null +++ b/608/CH17/EX17.02/17_02.JPG diff --git a/608/CH17/EX17.02/17_02.sce b/608/CH17/EX17.02/17_02.sce new file mode 100755 index 000000000..9e3710384 --- /dev/null +++ b/608/CH17/EX17.02/17_02.sce @@ -0,0 +1,29 @@ +//Problem 17.02: A 4 μF capacitor is charged to 24 V and then discharged through a 220 kohms resistor. Use the ‘initial slope and three point’ method to draw: (a) the capacitor voltage/time characteristic, (b) the resistor voltage/time characteristic and (c) the current/time characteristic, for the transients which occur. From the characteristics determine the value of capacitor voltage, resistor voltage and current one and a half seconds after discharge has started.
+
+//initializing the variables:
+C = 4E-6; // in Farads
+R = 220000; // in ohms
+V = 24; // in Volts
+t1 = 1.5; // in secs
+
+//calculation:
+tou = R*C
+t = 0:0.1:10
+Vc = V*(1-%e^(-1*t/tou));
+plot2d(t,Vc)
+xtitle("capacitor voltage/time characteristic", "t", "Vc")
+xset('window',1)
+VR = V*(1-%e^(-1*t/tou));
+plot2d(t,VR)
+xtitle("resistor voltage/time characteristic", "t", "VR")
+xset('window',2)
+I = V/R
+i = I*%e^(-1*t/tou)
+plot2d(t,i)
+xtitle("current/time characteristic", "t", "i")
+Vct1 = V*%e^(-1*t1/tou)
+VRt1 = V*%e^(-1*t1/tou)
+it1 = I*%e^(-1*t1/tou)
+
+printf("\ = \n\n Result \n\n")
+printf("\n the value of capacitor voltage is %.2f V, resistor voltage is %.2f V, current is %.1E A at one and a half seconds after discharge has started.",Vct1, VRt1,it1)
\ No newline at end of file diff --git a/608/CH17/EX17.03/17_03.sce b/608/CH17/EX17.03/17_03.sce new file mode 100755 index 000000000..ba030cf86 --- /dev/null +++ b/608/CH17/EX17.03/17_03.sce @@ -0,0 +1,23 @@ +//Problem 17.03: A 20 μF capacitor is connected in series with a 50 kohm resistor and the circuit is connected to a 20 V, d.c. supply. Determine (a) the initial value of the current flowing, (b) the time constant of the circuit, (c) the value of the current one second after connection, (d) the value of the capacitor voltage two seconds after connection, and (e) the time after connection when the resistor voltage is 15 V
+
+//initializing the variables:
+C = 20E-6; // in Farads
+R = 50000; // in ohms
+V = 20; // in Volts
+t1 = 1; // in secs
+t2 = 2; // in secs
+VRt = 15; // in Volts
+
+//calculation:
+tou = R*C
+I = V/R
+Vct1 = V*(1-%e^(-1*t2/tou))
+t3 = -1*tou*log(VRt/V)
+it1 = I*%e^(-1*t1/tou)
+
+printf("\ = \n\n Result \n\n")
+printf("\n (a)initial value of the current flowing is %.4f A",I)
+printf("\n (b)time constant of the circuit %.0f Sec",tou)
+printf("\n (c)the value of the current one second after connection, %.3f mA",(it1/1E-3))
+printf("\n (d)the value of the capacitor voltage two seconds after connection %.1f V",Vct1)
+printf("\n (e)the time after connection when the resistor voltage is 15 V is %.3f sec",t3)
\ No newline at end of file diff --git a/608/CH17/EX17.04/17_04.sce b/608/CH17/EX17.04/17_04.sce new file mode 100755 index 000000000..9a6eb8b1c --- /dev/null +++ b/608/CH17/EX17.04/17_04.sce @@ -0,0 +1,15 @@ +//Problem 17.04: A circuit consists of a resistor connected in series with a 0.5 μF capacitor and has a time constant of 12 ms. Determine (a) the value of the resistor, and (b) the capacitor voltage 7 ms after connecting the circuit to a 10 V supply
+
+//initializing the variables:
+C = 0.5E-6; // in Farads
+V = 10; // in Volts
+tou = 0.012; // in secs
+t1 = 0.007; // in secs
+
+//calculation:
+R = tou/C
+Vc = V*(1-%e^(-1*t1/tou))
+
+printf("\ = \n\n Result \n\n")
+printf("\n (a)value of the resistor is %.0f ohm",R)
+printf("\n (b)capacitor voltage is %.2f V",Vc)
\ No newline at end of file diff --git a/608/CH17/EX17.05/17_05.sce b/608/CH17/EX17.05/17_05.sce new file mode 100755 index 000000000..6acf05968 --- /dev/null +++ b/608/CH17/EX17.05/17_05.sce @@ -0,0 +1,22 @@ +//Problem 17.05: A capacitor is charged to 100 V and then discharged through a 50 kohm resistor. If the time constant of the circuit is 0.8 s, determine: (a) the value of the capacitor, (b) the time for the capacitor voltage to fall to 20 V, (c) the current flowing when the capacitor has been discharging for 0.5 s, and (d) the voltage drop across the resistor when the capacitor has been discharging for one second.
+
+//initializing the variables:
+R = 50000; // in ohms
+V = 100; // in Volts
+Vc1 = 20; // in Volts
+tou = 0.8; // in secs
+t1 = 0.5; // in secs
+t2 = 1; // in secs
+
+//calculation:
+C = tou/R
+t = -1*tou*log(Vc1/V)
+I = V/R
+it1 = I*%e^(-1*t1/tou)
+Vc = V*%e^(-1*t2/tou)
+
+printf("\ = \n\n Result \n\n")
+printf("\n (a)the value of the capacitor is %.2E F",C)
+printf("\n (b)the time for the capacitor voltage to fall to 20 V is %.2f sec",t)
+printf("\n (c)the current flowing when the capacitor has been discharging for 0.5 s is %.5f A",it1)
+printf("\n (d)the voltage drop across the resistor when the capacitor has been discharging for one second is %.1f V",Vc)
\ No newline at end of file diff --git a/608/CH17/EX17.06/17_06.sce b/608/CH17/EX17.06/17_06.sce new file mode 100755 index 000000000..7d096c851 --- /dev/null +++ b/608/CH17/EX17.06/17_06.sce @@ -0,0 +1,17 @@ +//Problem 17.06: A 0.1 μF capacitor is charged to 200 V before being connected across a 4 kohm resistor. Determine (a) the initial discharge current, (b) the time constant of the circuit, and (c) the minimum time required for the voltage across the capacitor to fall to less than 2 V
+
+//initializing the variables:
+C = 0.1E-6; // in Farads
+R = 4000; // in ohms
+V = 200; // in Volts
+Vc1 = 2; // in Volts
+
+//calculation:
+tou = R*C
+I = V/R
+t = -1*tou*log(Vc1/V)
+
+printf("\ = \n\n Result \n\n")
+printf("\n (a) initial discharge current is %.2f A",I)
+printf("\n (b)Time constant tou is %.4f sec",tou)
+printf("\n (c)minimum time required for the voltage across the capacitor to fall to less than 2 V is %.3f sec",t)
\ No newline at end of file diff --git a/608/CH17/EX17.07/17_07.JPG b/608/CH17/EX17.07/17_07.JPG Binary files differnew file mode 100755 index 000000000..be8fd0bf4 --- /dev/null +++ b/608/CH17/EX17.07/17_07.JPG diff --git a/608/CH17/EX17.07/17_07.sce b/608/CH17/EX17.07/17_07.sce new file mode 100755 index 000000000..875988363 --- /dev/null +++ b/608/CH17/EX17.07/17_07.sce @@ -0,0 +1,22 @@ +//Problem 17.07: A relay has an inductance of 100 mH and a resistance of 20 ohm. It is connected to a 60 V, d.c. supply. Use the ‘initial slope and three point’ method to draw the current/time characteristic and hence determine the value of current flowing at a time equal to two time constants and the time for the current to grow to 1.5 A.
+
+//initializing the variables:
+L = 0.1; // in Henry
+R = 20; // in ohms
+V = 60; // in Volts
+i2 = 1.5; // in Amperes
+
+//calculation:
+tou = L/R
+t1 = 2*tou
+t = 0:0.0001:0.025
+I = V/R
+i = I*(1 - %e^(-1*t/tou))
+plot2d(t,i)
+xtitle("current/time characteristic", "t", "i")
+i1 = I*(1 - %e^(-1*t1/tou))
+t2 = -1*tou*log(1 - i2/I)
+
+printf("\ = \n\n Result \n\n")
+printf("\n (a) the value of current flowing at a time equal to two time constants is %.3f A",i1)
+printf("\n (b)the time for the current to grow to 1.5 A is %.7f sec",t2)
\ No newline at end of file diff --git a/608/CH17/EX17.08/17_08.sce b/608/CH17/EX17.08/17_08.sce new file mode 100755 index 000000000..b0fe0c456 --- /dev/null +++ b/608/CH17/EX17.08/17_08.sce @@ -0,0 +1,21 @@ +//Problem 17.08: A coil of inductance 0.04 H and resistance 10 ohm is connected to a 120 V, d.c. supply. Determine (a) the final value of current, (b) the time constant of the circuit, (c) the value of current after a time equal to the time constant from the instant the supply voltage is connected, (d) the expected time for the current to rise to within 1% of its final value.
+
+//initializing the variables:
+L = 0.04; // in Henry
+R = 10; // in ohms
+V = 120; // in Volts
+i2 = 0.01*I; // in amperes
+
+//calculation:
+tou = L/R
+t1 = tou
+I = V/R
+i1 = I*(1 - %e^(-1*t1/tou))
+i2 = 0.01*I
+t2 = -1*tou*log(i2/I)
+
+printf("\ = \n\n Result \n\n")
+printf("\n (a) the final value of current is %.0f A",I)
+printf("\n (b)time constant of the circuit is %.3f sec",tou)
+printf("\n (c) value of current after a time equal to the time constant is %.2f A",i1)
+printf("\n (d)the expected time for the current to rise to within 0.01 times of its final value is %.2f sec",t2)
\ No newline at end of file diff --git a/608/CH17/EX17.09/17_09.sce b/608/CH17/EX17.09/17_09.sce new file mode 100755 index 000000000..eca15d991 --- /dev/null +++ b/608/CH17/EX17.09/17_09.sce @@ -0,0 +1,23 @@ +//Problem 17.09: The winding of an electromagnet has an inductance of 3 H and a resistance of 15 ohm. When it is connected to a 120 V, d.c. supply, calculate: (a) the steady state value of current flowing in the winding, (b) the time constant of the circuit, (c) the value of the induced e.m.f. after 0.1 s, (d) the time for the current to rise to 85% of its final value, and (e) the value of the current after 0.3 s.
+
+//initializing the variables:
+L = 3; // in Henry
+R = 15; // in ohms
+V = 120; // in Volts
+t1 = 0.1; // in secs
+t3 = 0.3; // in secs
+
+//calculation:
+tou = L/R
+I = V/R
+i2 = 0.85*I
+VL = V*%e^(-1*t1/tou)
+t2 = -1*tou*log(1 - (i2/I))
+i3 = I*(1 - %e^(-1*t3/tou))
+
+printf("\ = \n\n Result \n\n")
+printf("\n (a) steady state value of current is %.0f A",I)
+printf("\n (b)time constant of the circuit is %.3f sec",tou)
+printf("\n (c)value of the induced e.m.f. after 0.1 s is %.2f V",VL)
+printf("\n (d) time for the current to rise to 0.85 times of its final values is %.3f sec",t2)
+printf("\n (e)value of the current after 0.3 s is %.3f A",i3)
\ No newline at end of file diff --git a/608/CH17/EX17.10/17_10.JPG b/608/CH17/EX17.10/17_10.JPG Binary files differnew file mode 100755 index 000000000..28da724af --- /dev/null +++ b/608/CH17/EX17.10/17_10.JPG diff --git a/608/CH17/EX17.10/17_10.sce b/608/CH17/EX17.10/17_10.sce new file mode 100755 index 000000000..2336b6422 --- /dev/null +++ b/608/CH17/EX17.10/17_10.sce @@ -0,0 +1,23 @@ +//Problem 17.10: The field winding of a 110 V, d.c. motor has a resistance of 15 ohms and a time constant of 2 s. Determine the inductance and use the tangential method to draw the current/time characteristic when the supply is removed and replaced by a shorting link. From the characteristic determine (a) the current flowing in the winding 3 s after being shorted-out and (b) the time for the current to decay to 5 A.
+
+//initializing the variables:
+R = 15; // in ohms
+V = 110; // in Volts
+tou = 2; // in secs
+t1 = 3; // in secs
+i2 =5; // in amperes
+
+//calculation:
+L = tou*R
+t = 0:0.1:10
+I = V/R
+i = I*(%e^(-1*t/tou))
+plot2d(t,i)
+xtitle("current/time characteristic", "t", "i")
+i1 = I*(%e^(-1*t1/tou))
+t2 = -1*tou*log((i2/I))
+
+printf("\ = \n\n Result \n\n")
+printf("\n inductance is %.0f H",L)
+printf("\n (a)the current flowing in the winding 3 s after being shorted-out is %.2f A",i1)
+printf("\n (b)the time for the current to decay to 5 A is %.3f sec",t2)
\ No newline at end of file diff --git a/608/CH17/EX17.11/17_11.sce b/608/CH17/EX17.11/17_11.sce new file mode 100755 index 000000000..408204a89 --- /dev/null +++ b/608/CH17/EX17.11/17_11.sce @@ -0,0 +1,21 @@ +//Problem 17.11 A coil having an inductance of 6 H and a resistance of R is connected in series with a resistor of 10 to a 120 V, d.c. supply. The time constant of the circuit is 300 ms. When steady-state conditions have been reached, the supply is replaced instantaneously by a short-circuit. Determine: (a) the resistance of the coil,(b) the current flowing in the circuit one second after the shorting link has been placed in the circuit, and (c) the time taken for the current to fall to 10% of its initial value.
+
+//initializing the variables:
+L = 6; // in Henry
+r = 10; // in ohms
+V = 120; // in Volts
+tou = 0.3; // in secs
+t1 = 1; // in secs
+
+//calculation:
+R = (L/tou) - r
+Rt = R + r
+I = V/Rt
+i2 = 0.1*I
+i1 = I*(%e^(-1*t1/tou))
+t2 = -1*tou*log((i2/I))
+
+printf("\ = \n\n Result \n\n")
+printf("\n (a) resistance of the coil is %.0f ohm",R)
+printf("\n (b) current flowing in the circuit one second after the shorting link has been placed is %.3f A",i1)
+printf("\n (c)the time for the current to decay to 0.1 times of initial value is %.3f sec",t2)
\ No newline at end of file diff --git a/608/CH17/EX17.12/17_12.sce b/608/CH17/EX17.12/17_12.sce new file mode 100755 index 000000000..0bb477e1f --- /dev/null +++ b/608/CH17/EX17.12/17_12.sce @@ -0,0 +1,22 @@ +//Problem 17.12: An inductor has a negligible resistance and an inductance of 200 mH and is connected in series with a 1 kohm resistor to a 24 V, d.c. supply. Determine the time constant of the circuit and the steady-state value of the current flowing in the circuit. Find (a) the current flowing in the circuit at a time equal to one time constant, (b) the voltage drop across the inductor at a time equal to two time constants and (c) the voltage drop across the resistor after a time equal to three time constants.
+
+//initializing the variables:
+L = 0.2; // in Henry
+R = 1000; // in ohms
+V = 24; // in Volts
+
+//calculation:
+tou = L/R
+t1 = 1*tou // in secs
+t2 = 2*tou // in secs
+t3 = 3*tou // in secs
+I = V/R
+i1 = I*(1 - %e^(-1*t1/tou))
+VL = V*(%e^(-1*t2/tou))
+VR = V*(1 - %e^(-1*t3/tou))
+
+printf("\ = \n\n Result \n\n")
+printf("\n time constant of the circuit is %.4f sec, and the steady-state value of the current is %.3f A",tou, I)
+printf("\n (a) urrent flowing in the circuit at a time equal to one time constant is %.5f A",i1)
+printf("\n (b) voltage drop across the inductor at a time equal to two time constants is %.3f V",VL)
+printf("\n (c)the voltage drop across the resistor after a time equal to three time constants is %.2f V",VR)
\ No newline at end of file diff --git a/608/CH18/EX18.01/18_01.sce b/608/CH18/EX18.01/18_01.sce new file mode 100755 index 000000000..40e5ece49 --- /dev/null +++ b/608/CH18/EX18.01/18_01.sce @@ -0,0 +1,12 @@ +//Problem 18.01: A differential amplifier has an open-loop voltage gain of 120. The input signals are 2.45 V and 2.35 V. Calculate the output voltage of the amplifier
+
+//initializing the variables:
+Vi2 = 2.45; // in Volts
+Vi1 = 2.35; // in Volts
+A0 = 120; // open-loop voltage gain
+
+//calculation:
+Vo = A0*(Vi2 - Vi1)
+
+printf("\n\n Result \n\n")
+printf("\n the output voltage is %.0f V",Vo)
\ No newline at end of file diff --git a/608/CH18/EX18.02/18_02.sce b/608/CH18/EX18.02/18_02.sce new file mode 100755 index 000000000..8064ecc4b --- /dev/null +++ b/608/CH18/EX18.02/18_02.sce @@ -0,0 +1,11 @@ +//Problem 18.02: Determine the common-mode gain of an op amp that has a differential voltage gain of 150E3 and a CMRR of 90 dB.
+
+//initializing the variables:
+Vg = 150E3; // differential voltage gain
+CMRR = 90; // in dB
+
+//calculation:
+CMG = Vg/(10^(CMRR/20))
+
+printf("\n\n Result \n\n")
+printf("\n common-mode gain is %.2f",CMG)
\ No newline at end of file diff --git a/608/CH18/EX18.03/18_03.sce b/608/CH18/EX18.03/18_03.sce new file mode 100755 index 000000000..9aadaa6a5 --- /dev/null +++ b/608/CH18/EX18.03/18_03.sce @@ -0,0 +1,13 @@ +//Problem 18.03: A differential amplifier has an open-loop voltage gain of 120 and a common input signal of 3.0 V to both terminals. An output signal of 24 mV results. Calculate the common-mode gain and the CMRR.
+
+//initializing the variables:
+Vg = 120; // differential voltage gain
+Vi = 3; // in Volts
+Vo = 0.024; // in Volts
+
+//calculation:
+CMG = Vo/Vi
+CMRR = 20*log10(Vg/CMG)
+
+printf("\n\n Result \n\n")
+printf("\n common-mode gain is %.3f and CMRR is %.2f dB",CMG, CMRR)
\ No newline at end of file diff --git a/608/CH18/EX18.04/18_04.sce b/608/CH18/EX18.04/18_04.sce new file mode 100755 index 000000000..d0adb3b57 --- /dev/null +++ b/608/CH18/EX18.04/18_04.sce @@ -0,0 +1,14 @@ +//Problem 18.04: In the inverting amplifier of Figure 18.5, Ri = 1 kohm and Rf = 2 kohm. Determine the output voltage when the input voltage is: (a)+0.4 V (b) -1.2 V
+
+//initializing the variables:
+Rf = 2000; // in ohms
+Ri = 1000; // in ohms
+Vi1 = 0.4; // in Volts
+Vi2 = -1.2; // in Volts
+
+//calculation:
+Vo1 = -1*Rf*Vi1/Ri
+Vo2 = -1*Rf*Vi2/Ri
+
+printf("\n\n Result \n\n")
+printf("\n output voltage when the input voltage is 0.4V is %.1f V and when the input voltage is -1.2V is %.1f V",Vo1, Vo2)
\ No newline at end of file diff --git a/608/CH18/EX18.05/18_05.sce b/608/CH18/EX18.05/18_05.sce new file mode 100755 index 000000000..1df26bb38 --- /dev/null +++ b/608/CH18/EX18.05/18_05.sce @@ -0,0 +1,15 @@ +//Problem 18.05: The op amp shown in Figure 18.6 has an input bias current of 100 nA at 20°C. Calculate (a) the voltage gain, and (b) the output offset voltage due to the input bias current. (c) How can the effect of input bias current be minimised?
+
+//initializing the variables:
+Ii = 100E-9; // in Amperes
+T = 20; // in °C
+Rf = 1E6; // in ohms
+Ri = 10000; // in ohms
+
+//calculation:
+A = -1*Rf/Ri
+Vos = Ii*Ri*Rf/(Ri+Rf)
+
+printf("\n\n Result \n\n")
+printf("\n (a)the voltage gain is %.0f",A)
+printf("\n (b)output offset voltage is %.5f V",Vos)
\ No newline at end of file diff --git a/608/CH18/EX18.06/18_06.sce b/608/CH18/EX18.06/18_06.sce new file mode 100755 index 000000000..c09c5c0b0 --- /dev/null +++ b/608/CH18/EX18.06/18_06.sce @@ -0,0 +1,14 @@ +//Problem 18.06: Design an inverting amplifier to have a voltage gain of 40 dB, a closed-loop bandwidth of 5 kHz and an input resistance of 10 kohm.
+
+//initializing the variables:
+Vg = 40; // in dB
+bf = 5000; // in Hz
+Ri = 10000; // in ohms
+
+//calculation:
+A = 10^(Vg/20)
+Rf = A*Ri
+f = A*bf
+
+printf("\n\n Result \n\n")
+printf("\n the voltage gain is %.0f, Rf = %.0f ohm and frequency = %.0f Hz",A, Rf, f)
\ No newline at end of file diff --git a/608/CH18/EX18.07/18_07.sce b/608/CH18/EX18.07/18_07.sce new file mode 100755 index 000000000..0fe86458a --- /dev/null +++ b/608/CH18/EX18.07/18_07.sce @@ -0,0 +1,14 @@ +//Problem 18.07: For the op amp shown in Figure 18.8, R1 = 4.7 kohm and R2 = 10 kohm. If the input voltage is- 0.4 V, determine (a) the voltage gain (b) the output voltage
+
+//initializing the variables:
+Vi = -0.4; // in Volts
+R1 = 4700; // in ohms
+R2 = 10000; // in ohms
+
+//calculation:
+A = 1 + (R2/R1)
+Vo = A*Vi
+
+printf("\n\n Result \n\n")
+printf("\n(a) the voltage gain is %.2f",A)
+printf("\n(b) output voltageis %.2f V",Vo)
\ No newline at end of file diff --git a/608/CH18/EX18.08/18_08.sce b/608/CH18/EX18.08/18_08.sce new file mode 100755 index 000000000..526809b08 --- /dev/null +++ b/608/CH18/EX18.08/18_08.sce @@ -0,0 +1,16 @@ +//Problem 18.08: For the summing op amp shown in Figure 18.11, determine the output voltage, Vo
+
+//initializing the variables:
+V1 = 0.5; // in Volts
+V2 = 0.8; // in Volts
+V3 = 1.2; // in Volts
+R1 = 10000; // in ohms
+R2 = 20000; // in ohms
+R3 = 30000; // in ohms
+Rf = 50000; // in ohms
+
+//calculation:
+Vo = -1*Rf*(V1/R1 + V2/R2 + V3/R3)
+
+printf("\n\n Result \n\n")
+printf("\n output voltageis %.1f V",Vo)
\ No newline at end of file diff --git a/608/CH18/EX18.10/18_10.sce b/608/CH18/EX18.10/18_10.sce new file mode 100755 index 000000000..709985d98 --- /dev/null +++ b/608/CH18/EX18.10/18_10.sce @@ -0,0 +1,13 @@ +//Problem 18.10: A steady voltage of -0.75V is applied to an op amp integrator having component values of R = 200 kohm and C = 2.5 μF. Assuming that the initial capacitor charge is zero, determine the value of the output voltage 100 ms after application of the input.
+
+//initializing the variables:
+Vs = -0.75; // in Volts
+R = 200000; // in ohms
+C = 2.5E-6; // in Farads
+t = 0.1; // in secs
+
+//calculation:
+Vo = (-1/(C*R))*integrate('-0.75', 't', 0, 0.1)
+
+printf("\n\n Result \n\n")
+printf("\n output voltage is %.2f V",Vo)
\ No newline at end of file diff --git a/608/CH18/EX18.11/18_11.sce b/608/CH18/EX18.11/18_11.sce new file mode 100755 index 000000000..e07676410 --- /dev/null +++ b/608/CH18/EX18.11/18_11.sce @@ -0,0 +1,31 @@ +//Problem 18.11: In the differential amplifier shown in Figure 18.16, R1 = 10 kohm, R2 = 10 kohm, R3 = 100 kohm and Rf = 100 kohm. Determine the output voltage Vo if:
+//(a) V1 = 5 mV and V2 = 0
+//(b) V1 = 0 and V2 = 5mV
+//(c) V1 = 50 mV and V2 = 25mV
+//(d) V1 = 25 mV and V2 = 50mV
+
+//initializing the variables:
+V1a = 0.005; // in Volts
+V2a = 0; // in Volts
+V1b = 0; // in Volts
+V2b = 0.005; // in Volts
+V1c = 0.05; // in Volts
+V2c = 0.025; // in Volts
+V1d = 0.025; // in Volts
+V2d = 0.05; // in Volts
+R1 = 10000; // in ohms
+R2 = 10000; // in ohms
+R3 = 100000; // in ohms
+Rf = 100000; // in ohms
+
+//calculation:
+Vo1 = -1*Rf*V1a/R1
+Vo2 = (R3/(R2+R3))*(1 + (Rf/R1))*V2b
+Vo3 = -1*Rf*(V1c-V2c)/R1
+Vo4 = (R3/(R2+R3))*(1 + (Rf/R1))*(V2d-V1d)
+
+printf("\n\n Result \n\n")
+printf("\n (a)output voltage is %.3f V",Vo1)
+printf("\n (b)output voltage is %.3f V",Vo2)
+printf("\n (c)output voltage is %.3f V",Vo3)
+printf("\n (d)output voltage is %.3f V",Vo4)
\ No newline at end of file diff --git a/608/CH19/EX19.01/19_01.sce b/608/CH19/EX19.01/19_01.sce new file mode 100755 index 000000000..9dbde4867 --- /dev/null +++ b/608/CH19/EX19.01/19_01.sce @@ -0,0 +1,15 @@ +//Problem 19.01: Three loads, each of resistance 30 ohm
, are connected in star to a 415 V, 3-phase supply. Determine (a) the system phase voltage, (b) the phase current and (c) the line current.
+
+//initializing the variables:
+Vl = 415; // in Volts
+Rp = 30; // in ohms
+
+//calculation:
+Vp = Vl/(3^0.5)
+Ip = Vp/Rp
+Il = Ip
+
+printf("\n\n Result \n\n")
+printf("\n (a)the system phase voltage is %.1f V",Vp)
+printf("\n (b)phase current is %.0f A",Ip)
+printf("\n (c)line current is %.0f A",Il)
\ No newline at end of file diff --git a/608/CH19/EX19.02/19_02.sce b/608/CH19/EX19.02/19_02.sce new file mode 100755 index 000000000..0d65800d1 --- /dev/null +++ b/608/CH19/EX19.02/19_02.sce @@ -0,0 +1,17 @@ +//Problem 19.02: A star-connected load consists of three identical coils each of resistance 30
ohm and inductance 127.3 mH. If the line current is 5.08 A, calculate the line voltage if the supply frequency is 50 Hz.
+
+//initializing the variables:
+R = 30; // in ohms
+L = 0.1273; // in Henry
+Ip = 5.08; // in Amperes
+f = 50; // in Hz
+
+//calculation:
+XL = 2*%pi*f*L
+Zp = (R*R + XL*XL)^0.5
+Il = Ip
+Vp = Ip*Zp
+Vl = Vp*(3^0.5)
+
+printf("\n\n Result \n\n")
+printf("\n (a)line voltage is %.0f V",Vl)
\ No newline at end of file diff --git a/608/CH19/EX19.04/19_04.sce b/608/CH19/EX19.04/19_04.sce new file mode 100755 index 000000000..08e14076f --- /dev/null +++ b/608/CH19/EX19.04/19_04.sce @@ -0,0 +1,28 @@ +//Problem 19.04: A 415 V, 3-phase, 4 wire, star-connected system supplies three resistive loads as shown in Figure 19.7. Determine (a) the current in each line and (b) the current in the neutral conductor.
+
+//initializing the variables:
+V = 415; // in Volts
+PR = 24000; // in Watt
+Py = 18000; // in Watt
+Pb = 12000; // in Watt
+VR = 240; // in Volts
+Vy = 240; // in Volts
+Vb = 240; // in Volts
+
+//calculation:
+//For a star-connected system VL = Vp*(3^0.5)
+Vp = Vl/(3^0.5)
+phir = 90*%pi/180
+phiy = 330*%pi/180
+phib = 210*%pi/180
+// I = P/V for a resistive load
+IR = PR/VR
+Iy = Py/Vy
+Ib = Pb/Vb
+Inh = IR*cos(phir) + Ib*cos(phib) + Iy*cos(phiy)
+Inv = IR*sin(phir) + Ib*sin(phib) + Iy*sin(phiy)
+In = (Inh^2 + Inv^2)^0.5
+
+printf("\n\n Result \n\n")
+printf("\n (a)cuurnt in R line is %.0f A, cuurnt in Y line is %.0f A and cuurnt in B line is %.0f A",IR,Iy,Ib)
+printf("\n (b)cuurnt in neutral line is %.1f A",In)
\ No newline at end of file diff --git a/608/CH19/EX19.05/19_05.sce b/608/CH19/EX19.05/19_05.sce new file mode 100755 index 000000000..0a6a2d31e --- /dev/null +++ b/608/CH19/EX19.05/19_05.sce @@ -0,0 +1,20 @@ +//Problem 19.05: Three identical coils each of resistance 30 ohm
and inductance 127.3 mH are connected in delta to a 440 V, 50 Hz, 3-phase supply. Determine (a) the phase current, and (b) the line current.
+
+//initializing the variables:
+R = 30; // in ohms
+L = 0.1273; // in Henry
+VL = 440; // in Volts
+f = 50; // in Hz
+
+//calculation:
+XL = 2*%pi*f*L
+Zp = (R*R + XL*XL)^0.5
+Vp = VL
+//Phase current
+Ip = Vp/Zp
+//For a delta connection,
+IL = Ip*(3^0.5)
+
+printf("\n\n Result \n\n")
+printf("\n (a)the phase current %.1f A",Ip)
+printf("\n (b)line current %.2f A",IL)
\ No newline at end of file diff --git a/608/CH19/EX19.06/19_06.sce b/608/CH19/EX19.06/19_06.sce new file mode 100755 index 000000000..0c80d075c --- /dev/null +++ b/608/CH19/EX19.06/19_06.sce @@ -0,0 +1,17 @@ +//Problem 19.06: Three identical capacitors are connected in delta to a 415 V, 50 Hz, 3-phase supply. If the line current is 15 A, determine the capacitance of each of the capacitors.
+
+//initializing the variables:
+IL = 15; // in Amperes
+VL = 415; // in Volts
+f = 50; // in Hz
+
+//calculation:
+//For a delta connection
+Ip = IL/(3^0.5) //phase current
+Vp = VL
+//Capacitive reactance per phase
+Xc = Vp/Ip
+C = 1/(2*%pi*f*Xc)
+
+printf("\n\n Result \n\n")
+printf("\n capacitance is %.3E F",C)
\ No newline at end of file diff --git a/608/CH19/EX19.07/19_07.sce b/608/CH19/EX19.07/19_07.sce new file mode 100755 index 000000000..6d4ee76a4 --- /dev/null +++ b/608/CH19/EX19.07/19_07.sce @@ -0,0 +1,28 @@ +//Problem 19.07: Three coils each having resistance 3
ohm and inductive reactance 4 ohm
are connected (i) in star and (ii) in delta to a 415 V, 3-phase supply. Calculate for each connection (a) the line and phase voltages and (b) the phase and line currents.
+
+//initializing the variables:
+R = 3; // in ohms
+XL = 4; // in ohms
+VL = 415; // in Volts
+
+//calculation:
+//For a star connection:
+//IL = Ip
+//VL = Vp*(3^0.5)
+VLs = VL
+Vps = VLs/(3^0.5)
+//Impedance per phase,
+Zp = (R*R + XL*XL)^0.5
+Ips = Vps/Zp
+ILs = Ips
+//For a delta connection:
+//VL = Vp
+//IL = Ip*(3^0.5)
+VLd = VL
+Vpd = VLd
+Ipd = Vpd/Zp
+ILd = Ipd*(3^0.5)
+
+printf("\n\n Result \n\n")
+printf("\n (a)the line voltage for star connection is %.0f V and the phase voltage for star connection is %.0f V and the line voltage for delta connection is %.0f V and the phase voltage for delta connection is %.0f V",VLs,Vps,VLd,Vpd)
+printf("\n (b)the line current for star connection is %.0f A and the phase current for star connection is %.0f A and the line current for delta connection is %.0f A and the phase current for delta connection is %.0f A",ILs,Ips,ILd,Ipd)
\ No newline at end of file diff --git a/608/CH19/EX19.08/19_08.sce b/608/CH19/EX19.08/19_08.sce new file mode 100755 index 000000000..bfbff96d8 --- /dev/null +++ b/608/CH19/EX19.08/19_08.sce @@ -0,0 +1,20 @@ +//Problem 19.08: Three 12
ohms resistors are connected in star to a 415 V, 3-phase supply. Determine the total power dissipated by the resistors.
+
+//initializing the variables:
+Rp = 12; // in ohms
+VL = 415; // in Volts
+
+//calculation:
+//Power dissipated, P = VL*IL*(3^0.5)*cos(phi) or P = 3*Ip*Ip*Rp)
+Vp = VL/(3^0.5) // since the resistors are star-connected
+//Phase current, Ip
+Zp = Rp
+Ip = Vp/Zp
+//For a star connection
+IL = Ip
+// For a purely resistive load, the power factor cos(phi) = 1
+pf = 1
+P = VL*IL*(3^0.5)*pf
+
+printf("\n\n Result \n\n")
+printf("\n (a)total power dissipated by the resistors is %.0f W",P)
\ No newline at end of file diff --git a/608/CH19/EX19.09/19_09.sce b/608/CH19/EX19.09/19_09.sce new file mode 100755 index 000000000..67e7c16e6 --- /dev/null +++ b/608/CH19/EX19.09/19_09.sce @@ -0,0 +1,13 @@ +//Problem 19.09: The input power to a 3-phase a.c. motor is measured as 5 kW. If the voltage and current to the motor are 400 V and 8.6 A respectively, determine the power factor of the system.
+
+//initializing the variables:
+P = 5000; // in Watts
+IL = 8.6; // in amperes
+VL = 400; // in Volts
+
+//calculation:
+//Power dissipated, P = VL*IL*(3^0.5)*cos(phi) or P = 3*Ip*Ip*Rp)
+pf = P/(VL*IL*(3^0.5))
+
+printf("\n\n Result \n\n")
+printf("\n power factor is %.3f",pf)
\ No newline at end of file diff --git a/608/CH19/EX19.10/19_10.sce b/608/CH19/EX19.10/19_10.sce new file mode 100755 index 000000000..5d53a7dec --- /dev/null +++ b/608/CH19/EX19.10/19_10.sce @@ -0,0 +1,36 @@ +//Problem 19.10: Three identical coils, each of resistance 10 ohm
and inductance 42 mH are connected (a) in star and (b) in delta to a 415 V, 50 Hz, 3-phase supply. Determine the total power dissipated in each case.
+
+//initializing the variables:
+R = 10; // in ohms
+L = 0.042; // in Henry
+VL = 415; // in Volts
+f = 50; // in Hz
+
+//calculation:
+//For a star connection:
+//IL = Ip
+//VL = Vp*(3^0.5)
+XL = 2*%pi*f*L
+Zp = (R*R + XL*XL)^0.5
+VLs = VL
+Vps = VLs/(3^0.5)
+//Impedance per phase,
+Ips = Vps/Zp
+ILs = Ips
+//Power dissipated, P = VL*IL*(3^0.5)*cos(phi) or P = 3*Ip*Ip*Rp)
+pfs = R/Zp
+Ps = VLs*ILs*(3^0.5)*pfs
+
+//For a delta connection:
+//VL = Vp
+//IL = Ip*(3^0.5)
+VLd = VL
+Vpd = VLd
+Ipd = Vpd/Zp
+ILd = Ipd*(3^0.5)
+//Power dissipated, P = VL*IL*(3^0.5)*cos(phi) or P = 3*Ip*Ip*Rp)
+pfd = R/Zp
+Pd = VLd*ILd*(3^0.5)*pfd
+
+printf("\n\n Result \n\n")
+printf("\n total power dissipated in star is %.1E W and in delta is %.2E W",Ps, Pd)
\ No newline at end of file diff --git a/608/CH19/EX19.11/19_11.sce b/608/CH19/EX19.11/19_11.sce new file mode 100755 index 000000000..1f7682661 --- /dev/null +++ b/608/CH19/EX19.11/19_11.sce @@ -0,0 +1,22 @@ +//Problem 19.11: A 415 V, 3-phase a.c. motor has a power output of 12.75 kW and operates at a power factor of 0.77 lagging and with an efficiency of 85%. If the motor is delta-connected, determine (a) the power input, (b) the line current and (c) the phase current.
+
+//initializing the variables:
+Po = 12750; // in Watts
+pf = 0.77; // power factor
+eff = 0.85;
+VL = 415; // in Volts
+
+//calculation:
+//eff = power_out/power_in
+Pi = Po/eff
+//Power P = VL*IL*(3^0.5)*cos(phi) or P = 3*Ip*Ip*Rp)
+IL = Pi/(VL*(3^0.5)*pf) // line current
+
+//For a delta connection:
+//IL = Ip*(3^0.5)
+Ip = IL/(3^0.5)
+
+printf("\n\n Result \n\n")
+printf("\n (a)power input is %.2E W",Pi)
+printf("\n (b)line current is %.2f A",IL)
+printf("\n (c)phase current is %.2f A",Ip)
\ No newline at end of file diff --git a/608/CH19/EX19.13/19_13.sce b/608/CH19/EX19.13/19_13.sce new file mode 100755 index 000000000..dc2b38b31 --- /dev/null +++ b/608/CH19/EX19.13/19_13.sce @@ -0,0 +1,24 @@ +//Problem 19.13: A 400 V, 3-phase star connected alternator supplies a delta-connected load, each phase of which has a resistance of 30 ohm
and inductive reactance 40
ohm. Calculate (a) the current supplied by the alternator and (b) the output power and the kVA of the alternator, neglecting losses in the line between the alternator and load.
+
+//initializing the variables:
+R = 30; // in ohms
+XL = 40; // in ohms
+VL = 400; // in Volts
+
+//calculation:
+Zp = (R*R + XL*XL)^0.5
+//a delta-connected load
+Vp = VL
+//Phase current
+Ip = Vp/Zp
+IL = Ip*(3^0.5)
+//Alternator output power is equal to the power dissipated by the load.
+//Power P = VL*IL*(3^0.5)*cos(phi) or P = 3*Ip*Ip*Rp)
+pf = R/Zp
+P = VL*IL*(3^0.5)*pf
+//Alternator output kVA,
+S = VL*IL*(3^0.5)
+
+printf("\n\n Result \n\n")
+printf("\n (a)the current supplied by the alternator is %.2f A",IL)
+printf("\n (b)output power is %.2E W and kVA of the alternator is %.2E kVA",P, S)
\ No newline at end of file diff --git a/608/CH19/EX19.14/19_14.sce b/608/CH19/EX19.14/19_14.sce new file mode 100755 index 000000000..f72824daa --- /dev/null +++ b/608/CH19/EX19.14/19_14.sce @@ -0,0 +1,27 @@ +//Problem 19.14: Each phase of a delta-connected load comprises a resistance of 30 ohm
and an 80 μF capacitor in series. The load is connected to a 400 V, 50 Hz, 3-phase supply. Calculate (a) the phase current, (b) the line current, (c) the total power dissipated and (d) the kVA rating of the load. Draw the complete phasor diagram for the load.
+
+//initializing the variables:
+R = 30; // in ohms
+C = 80E-6; // in Farads
+f = 50; // in Hz
+VL = 400; // in Volts
+
+//calculation:
+//Capacitive reactance
+Xc = 1/(2*%pi*f*C)
+Zp = (R*R + Xc*Xc)^0.5
+pf = R/Zp
+//a delta-connected load
+Vp = VL
+//Phase current
+Ip = Vp/Zp
+IL = Ip*(3^0.5)
+//Power P = VL*IL*(3^0.5)*cos(phi) or P = 3*Ip*Ip*Rp)
+P = VL*IL*(3^0.5)*pf
+//Alternator output kVA,
+S = VL*IL*(3^0.5)
+
+printf("\n\n Result \n\n")
+printf("\n (a)the phase current is %.3f A",Ip)
+printf("\n (b)the line current is %.2f A",IL)
+printf("\n (c) power is %.2E W and \n(d)kVA of the alternator is %.2E kVA",P, S)
\ No newline at end of file diff --git a/608/CH19/EX19.15/19_15.sce b/608/CH19/EX19.15/19_15.sce new file mode 100755 index 000000000..52c2c97be --- /dev/null +++ b/608/CH19/EX19.15/19_15.sce @@ -0,0 +1,16 @@ +//Problem 19.15: Two wattmeters are connected to measure the input power to a balanced 3-phase load by the two-wattmeter method. If the instrument readings are 8 kW and 4 kW, determine (a) the total power input and (b) the load power factor.
+
+//initializing the variables:
+Pi1 = 8000; // in Watts
+Pi2 = 4000; // in Watts
+
+//calculation:
+//Total input power
+Pi = Pi1 + Pi2
+phi = atan((Pi1 - Pi2)*(3^0.5)/(Pi1 + Pi2))
+//Power factor
+pf = cos(phi)
+
+printf("\n\n Result \n\n")
+printf("\n (a)power input is %.1E W",Pi)
+printf("\n (b)power factor is %.3f",pf)
\ No newline at end of file diff --git a/608/CH19/EX19.16/19_16.sce b/608/CH19/EX19.16/19_16.sce new file mode 100755 index 000000000..54e019ad8 --- /dev/null +++ b/608/CH19/EX19.16/19_16.sce @@ -0,0 +1,21 @@ +//Problem 19.16: Two wattmeters connected to a 3-phase motor indicate the total power input to be 12 kW. The power factor is 0.6. Determine the readings of each wattmeter.
+
+//initializing the variables:
+Pi = 12000; // in Watts
+pf = 0.6; // power factor
+
+//calculation:
+//If the two wattmeters indicate Pi1 and Pi2 respectively
+// Pit = Pi1 + Pi2
+Pit = Pi
+// Pid = Pi1 - Pi2
+//power factor = 0.6 = cos(phi)
+phi = acos(pf)
+Pid = Pit*tan(phi)/(3^0.5)
+//Hence wattmeter 1 reads
+Pi1 = (Pid + Pit)/2
+//wattmeter 2 reads
+Pi2 = Pit - Pi1
+
+printf("\n\n Result \n\n")
+printf("\n reading in each wattameter are %.2E W and %.2E W",Pi1,Pi2)
\ No newline at end of file diff --git a/608/CH19/EX19.17/19_17.sce b/608/CH19/EX19.17/19_17.sce new file mode 100755 index 000000000..5539e2044 --- /dev/null +++ b/608/CH19/EX19.17/19_17.sce @@ -0,0 +1,16 @@ +//Problem 19.17: Two wattmeters indicate 10 kW and 3 kW respectively when connected to measure the input power to a 3-phase balanced load, the reverse switch being operated on the meter indicating the 3 kW reading. Determine (a) the input power and (b) the load power factor.
+
+//initializing the variables:
+Pi1 = 10000; // in Watts
+Pi2 = -3000; // in Watts
+
+//calculation:
+//Total input power
+Pi = Pi1 + Pi2
+phi = atan((Pi1 - Pi2)*(3^0.5)/(Pi1 + Pi2))
+//Power factor
+pf = cos(phi)
+
+printf("\n\n Result \n\n")
+printf("\n (a)power input is %.2E W",Pi)
+printf("\n (b)power factor is %.3f",pf)
\ No newline at end of file diff --git a/608/CH19/EX19.18/19_18.sce b/608/CH19/EX19.18/19_18.sce new file mode 100755 index 000000000..382591ce6 --- /dev/null +++ b/608/CH19/EX19.18/19_18.sce @@ -0,0 +1,51 @@ +//Problem 19.18: Three similar coils, each having a resistance of 8 ohm and an inductive reactance of 8 ohm are connected (a) in star and (b) in delta, across a 415 V, 3-phase supply. Calculate for each connection the readings on each of two wattmeters connected to measure the power by the two-wattmeter method.
+
+//initializing the variables:
+R = 8; // in ohms
+XL = 8; // in ohms
+VL = 415; // in Volts
+
+//calculation:
+//For a star connection:
+//IL = Ip
+//VL = Vp*(3^0.5)
+VLs = VL
+Vps = VLs/(3^0.5)
+//Impedance per phase,
+Zp = (R*R + XL*XL)^0.5
+Ips = Vps/Zp
+ILs = Ips
+//Power dissipated, P = VL*IL*(3^0.5)*cos(phi) or P = 3*Ip*Ip*Rp)
+pf = R/Zp
+Ps = VLs*ILs*(3^0.5)*pf
+//If wattmeter readings are P1 and P2 then P1 + P2 = Pst
+Pst = Ps
+// Pid = Pi1 - Pi2
+phi = acos(pf)
+Psd = Pst*tan(phi)/(3^0.5)
+//Hence wattmeter 1 reads
+Ps1 = (Psd + Pst)/2
+//wattmeter 2 reads
+Ps2 = Pst - Ps1
+
+//For a delta connection:
+//VL = Vp
+//IL = Ip*(3^0.5)
+VLd = VL
+Vpd = VLd
+Ipd = Vpd/Zp
+ILd = Ipd*(3^0.5)
+//Power dissipated, P = VL*IL*(3^0.5)*cos(phi) or P = 3*Ip*Ip*Rp)
+Pd = VLd*ILd*(3^0.5)*pf
+//If wattmeter readings are P1 and P2 then P1 + P2 = Pdt
+Pdt = Pd
+// Pid = Pi1 - Pi2
+Pdd = Pdt*tan(phi)/(3^0.5)
+//Hence wattmeter 1 reads
+Pd1 = (Pdd + Pdt)/2
+//wattmeter 2 reads
+Pd2 = Pdt - Pd1
+
+printf("\n\n Result \n\n")
+printf("\n (a)When the coils are star-connected the wattmeter readings are %.3E W and %.3E W",Ps1,Ps2)
+printf("\n (b)When the coils are delta-connected the wattmeter readings are are %.3E W and %.3E W",Pd1,Pd2)
\ No newline at end of file diff --git a/608/CH2/EX2.01/2_01.sce b/608/CH2/EX2.01/2_01.sce new file mode 100755 index 000000000..ee51783f1 --- /dev/null +++ b/608/CH2/EX2.01/2_01.sce @@ -0,0 +1,11 @@ +//Problem 2.01: What current must flow if 0.24 coulombs is to be transferred in 15 ms?
+
+//initializing the variables:
+Q = 0.24; // in Coulomb
+t = 0.015; // in sec
+
+//calculation:
+I = Q/t
+
+printf("\n\nResult\n\n")
+printf("\nCurrent(I): %.0f Ampere(A)\n",I)
\ No newline at end of file diff --git a/608/CH2/EX2.02/2_02.sce b/608/CH2/EX2.02/2_02.sce new file mode 100755 index 000000000..bcdb5b4e1 --- /dev/null +++ b/608/CH2/EX2.02/2_02.sce @@ -0,0 +1,11 @@ +//Problem 2.02: If a current of 10 A flows for four minutes, find the quantity of electricity transferred.
+
+//initializing the variables:
+I = 10; // in Ampere
+t = 240; // in sec
+
+//calculation:
+Q = I*t
+
+printf("\n\nResult\n\n")
+printf("\nCharge(Q): %.0f Coulomb(C)\n",Q)
\ No newline at end of file diff --git a/608/CH2/EX2.03/2_03.sce b/608/CH2/EX2.03/2_03.sce new file mode 100755 index 000000000..91c2f5255 --- /dev/null +++ b/608/CH2/EX2.03/2_03.sce @@ -0,0 +1,11 @@ +//Problem 2.03: The current flowing through a resistor is 0.8 A when a p.d. of 20 V is applied. Determine the value of the resistance.
+
+//initializing the variables:
+I = 0.8; // in Ampere
+V = 20; // in Volts
+
+//calculation:
+R = V/I
+
+printf("\n\nResult\n\n")
+printf("\nResistance(R): %.0f Ohms\n",R)
\ No newline at end of file diff --git a/608/CH2/EX2.04/2_04.sce b/608/CH2/EX2.04/2_04.sce new file mode 100755 index 000000000..b20cc31ce --- /dev/null +++ b/608/CH2/EX2.04/2_04.sce @@ -0,0 +1,11 @@ +//Problem 2.04: Determine the p.d. which must be applied to a 2 kohm resistor in order that a current of 10 mA may flow.
+
+//initializing the variables:
+I = 0.010; // in Ampere
+R = 2000; // in ohms
+
+//calculation:
+V = I*R
+
+printf("\n\nResult\n\n")
+printf("\np.d.(V): %.0f Volts(V)\n",V)
\ No newline at end of file diff --git a/608/CH2/EX2.05/2_05.sce b/608/CH2/EX2.05/2_05.sce new file mode 100755 index 000000000..482ea9946 --- /dev/null +++ b/608/CH2/EX2.05/2_05.sce @@ -0,0 +1,11 @@ +//Problem 2.05: A coil has a current of 50 mA flowing through it when the applied voltage is 12 V. What is the resistance of the coil?
+
+//initializing the variables:
+I = 0.050; // in Ampere
+V = 12; // in Volts
+
+//calculation:
+R = V/I
+
+printf("\n\nResult\n\n")
+printf("\nResistance(R): %.0f Ohms\n",R)
\ No newline at end of file diff --git a/608/CH2/EX2.06/2_06.sce b/608/CH2/EX2.06/2_06.sce new file mode 100755 index 000000000..54213d54c --- /dev/null +++ b/608/CH2/EX2.06/2_06.sce @@ -0,0 +1,16 @@ +//Problem 2.06: A 100 V battery is connected across a resistor and causes a current of 5 mA to flow. Determine the resistance of the resistor. If the voltage is now reduced to 25 V, what will be the new value of the current flowing?
+
+//initializing the variables:
+I = 0.005; // in Ampere
+V1 = 100; // in Volts
+V2 = 25; // in Volts
+
+//calculation:
+//resistance
+R = V1/I
+//Current when voltage is reduced to 25 V,
+I = V2/R
+
+printf("\n\nResult\n\n")
+printf("\nResistance(R): %.0f Ohms",R)
+printf("\n Current when voltage is reduced to 25 V is %.2E A",I)
\ No newline at end of file diff --git a/608/CH2/EX2.07/2_07.sce b/608/CH2/EX2.07/2_07.sce new file mode 100755 index 000000000..82a36b2a0 --- /dev/null +++ b/608/CH2/EX2.07/2_07.sce @@ -0,0 +1,14 @@ +//Problem 2.07: What is the resistance of a coil which draws a current of (a) 50 mA and (b) 200 μA from a 120 V supply?
+
+//initializing the variables:
+I1 = 0.050; // in Ampere
+I2 = 200E-6; // in Ampere
+V = 120; // in Volts
+
+//calculation:
+R1 = V/I1
+R2 = V/I2
+
+printf("\n\nResult\n\n")
+printf("\nResistance(R1): %.0f Ohms",R1)
+printf("\nResistance(R2): %.0f Ohms\n",R2)
\ No newline at end of file diff --git a/608/CH2/EX2.08/2_08.sce b/608/CH2/EX2.08/2_08.sce new file mode 100755 index 000000000..1484ef851 --- /dev/null +++ b/608/CH2/EX2.08/2_08.sce @@ -0,0 +1,13 @@ +//Problem 2.08: A 100 W electric light bulb is connected to a 250 V supply. Determine (a) the current flowing in the bulb, and (b) the resistance of the bulb.
+
+//initializing the variables:
+P = 100; // in Watt
+V = 250; // in Volts
+
+//calculation:
+I = P/V
+R = V/I
+
+printf("\n\nResult\n\n")
+printf("\nCurrent(I): %.1f Ampere(A)",I)
+printf("\nResistance(R): %.0f Ohms\n",R)
\ No newline at end of file diff --git a/608/CH2/EX2.09/2_09.sce b/608/CH2/EX2.09/2_09.sce new file mode 100755 index 000000000..edceb587e --- /dev/null +++ b/608/CH2/EX2.09/2_09.sce @@ -0,0 +1,11 @@ +//Problem 2.09: Calculate the power dissipated when a current of 4 mA flows through a resistance of 5 k
+
+//initializing the variables:
+I = 0.004; // in ampere
+R = 5000; // in ohms
+
+//calculation:
+P = I*I*R
+
+printf("\n\nResult\n\n")
+printf("\nPower(P): %.2f Watt(W)\n",P)
\ No newline at end of file diff --git a/608/CH2/EX2.10/2_10.sce b/608/CH2/EX2.10/2_10.sce new file mode 100755 index 000000000..b26bb40dd --- /dev/null +++ b/608/CH2/EX2.10/2_10.sce @@ -0,0 +1,13 @@ +//Problem 2.10: An electric kettle has a resistance of 30. What current will flow when it is connected to a 240 V supply? Find also the power rating of the kettle.
+
+//initializing the variables:
+V = 240; // in Volts
+R = 30; // in ohms
+
+//calculation:
+I = V/R
+P = V*I
+
+printf("\n\nResult\n\n")
+printf("\nCurrent(I): %.0f Ampere(A)",I)
+printf("\nPower(P): %.0f Watt(W)\n",P)
\ No newline at end of file diff --git a/608/CH2/EX2.11/2_11.sce b/608/CH2/EX2.11/2_11.sce new file mode 100755 index 000000000..9a5eabe4a --- /dev/null +++ b/608/CH2/EX2.11/2_11.sce @@ -0,0 +1,13 @@ +//Problem 2.11: A current of 5 A flows in the winding of an electric motor, the resistance of the winding being 100. Determine (a) the p.d. across the winding, and (b) the power dissipated by the coil.
+
+//initializing the variables:
+I = 5; // in ampere
+R = 100; // in ohms
+
+//calculation:
+V = I*R
+P = I*R*I
+
+printf("\n\nResult\n\n")
+printf("\np.d(V): %.0f Volts(V)",V)
+printf("\nPower(P): %.0f Watt(W)\n",P)
\ No newline at end of file diff --git a/608/CH2/EX2.12/2_12.sce b/608/CH2/EX2.12/2_12.sce new file mode 100755 index 000000000..cdd3a7a43 --- /dev/null +++ b/608/CH2/EX2.12/2_12.sce @@ -0,0 +1,15 @@ +//Problem 2.12: The current/voltage relationship for two resistors A and B is as shown in Figure 2.5. Determine the value of the resistance of each resistor.
+
+//initializing the variables:
+I1 = 0.020; // in ampere
+V1 = 20; // in Volts
+I2 = 0.005; // in ampere
+V2 = 16; // in Volts
+
+//calculation:
+R1 = V1/I1
+R2 = V2/I2
+
+printf("\n\nResult\n\n")
+printf("\nResistance(R1): %.0f Ohms",R1)
+printf("\nResistance(R2): %.0f Ohms\n",R2)
\ No newline at end of file diff --git a/608/CH2/EX2.13/2_13.sce b/608/CH2/EX2.13/2_13.sce new file mode 100755 index 000000000..a0d96c6da --- /dev/null +++ b/608/CH2/EX2.13/2_13.sce @@ -0,0 +1,13 @@ +//Problem 2.13: The hot resistance of a 240 V filament lamp is 960. Find the current taken by the lamp and its power rating.
+
+//initializing the variables:
+V = 240; // in Volts
+R = 960; // in ohms
+
+//calculation:
+I = V/R
+P = I*V
+
+printf("\n\nResult\n\n")
+printf("\nCurrent(I): %.2f Ampere(A)",I)
+printf("\nPower(P): %.0f Watt(W)\n",P)
\ No newline at end of file diff --git a/608/CH2/EX2.14/2_14.sce b/608/CH2/EX2.14/2_14.sce new file mode 100755 index 000000000..f2c22469e --- /dev/null +++ b/608/CH2/EX2.14/2_14.sce @@ -0,0 +1,16 @@ +//Problem 2.14: A 12 V battery is connected across a load having a resistance of 40ohms. Determine the current flowing in the load, the power consumed and the energy dissipated in 2 minutes.
+
+//initializing the variables:
+V = 12; // in Volts
+R = 40; // in ohms
+t = 120; // in sec
+
+//calculation:
+I = V/R
+P = I*V
+E = P*t
+
+printf("\n\nResult\n\n")
+printf("\nCurrent(I): %.1f Ampere(A)",I)
+printf("\nPower(P): %.1f Watt(W)",P)
+printf("\nEnergy(E): %.0f Joule(J)\n",E)
\ No newline at end of file diff --git a/608/CH2/EX2.15/2_15.sce b/608/CH2/EX2.15/2_15.sce new file mode 100755 index 000000000..5489f0ba1 --- /dev/null +++ b/608/CH2/EX2.15/2_15.sce @@ -0,0 +1,12 @@ +//Problem 2.15: A source of e.m.f. of 15 V supplies a current of 2 A for six minutes. How much energy is provided in this time?
+
+//initializing the variables:
+V = 15; // in Volts
+I = 2; // in ampere
+t = 360; // in sec
+
+//calculation:
+E = V*I*t
+
+printf("\n\nResult\n\n")
+printf("\nEnergy(E): %.0f Joule(J)\n",E)
\ No newline at end of file diff --git a/608/CH2/EX2.16/2_16.sce b/608/CH2/EX2.16/2_16.sce new file mode 100755 index 000000000..64388ba2e --- /dev/null +++ b/608/CH2/EX2.16/2_16.sce @@ -0,0 +1,15 @@ +//Problem 2.16: Electrical equipment in an office takes a current of 13 A from a 240 V supply. Estimate the cost per week of electricity if the equipment is used for 30 hours each week and 1 kWh of energy costs 7p?
+
+//initializing the variables:
+V = 240; // in Volts
+I = 13; // in ampere
+t = 30; // in hours
+p = 7; // in paise per kWh
+
+//calculation:
+P = V*I
+E = P*t/1000 // in kWh
+C = E*p
+
+printf("\n\nResult\n\n")
+printf("\nCost per week: %.1f Paise(p)\n",C)
\ No newline at end of file diff --git a/608/CH2/EX2.17/2_17.sce b/608/CH2/EX2.17/2_17.sce new file mode 100755 index 000000000..72d9d7b99 --- /dev/null +++ b/608/CH2/EX2.17/2_17.sce @@ -0,0 +1,14 @@ +//Problem 2.17: An electric heater consumes 3.6 MJ when connected to a 250 V supply for 40 minutes. Find the power rating of the heater and the current taken from the supply.
+
+//initializing the variables:
+V = 250; // in Volts
+E = 3.6E6; // energy in J
+t = 2400; // in sec
+
+//calculation:
+P = E/t
+I = P/V
+
+printf("\n\nResult\n\n")
+printf("\nPower(P): %.0f Watt(W)",P)
+printf("\nCurrent(I): %.0f Ampere(A)\n",I)
\ No newline at end of file diff --git a/608/CH2/EX2.18/2_18.sce b/608/CH2/EX2.18/2_18.sce new file mode 100755 index 000000000..ea0cc9eb6 --- /dev/null +++ b/608/CH2/EX2.18/2_18.sce @@ -0,0 +1,16 @@ +//Problem 2.18: Determine the power dissipated by the element of an electric fire of resistance 20ohms when a current of 10 A flows through it. If the fire is on for 6 hours determine the energy used and the cost if 1 unit of electricity costs 7p.
+
+//initializing the variables:
+R = 20; // in ohms
+I = 10; // in ampere
+t = 6; // in hours
+p = 7; // in paise per kWh
+
+//calculation:
+P = I*I*R
+E = P*t/1000 // in kWh
+C = E*p
+
+printf("\n\nResult\n\n")
+printf("\nPower(P): %.0f Watt(W)",P)
+printf("\nCost per week: %.0f Paise(p)\n",C)
\ No newline at end of file diff --git a/608/CH2/EX2.19/2_19.sce b/608/CH2/EX2.19/2_19.sce new file mode 100755 index 000000000..19201ad65 --- /dev/null +++ b/608/CH2/EX2.19/2_19.sce @@ -0,0 +1,18 @@ +//Problem 2.19: A business uses two 3 kW fires for an average of 20 hours each per week, and six 150 W lights for 30 hours each per week. If the cost of electricity is 7p per unit, determine the weekly cost of electricity to the business.
+
+//initializing the variables:
+P1 = 3; // in kW
+P2 = 150; // in Watt
+n1 = 2; // no. of P1 Equips
+n2 = 6; // no. of P2 Equips
+t1 = 20; // in hours each per week
+t2 = 30; // in hours each per week
+p = 7; // in paise per kWh
+
+//calculation:
+E1 = P1*t1*n1 // in kWh by two P1 eqips
+E2 = P2*t2*n2/1000 // in kWh by six P2 eqips
+Et = E1 + E2
+C = Et * 7
+printf("\n\nResult\n\n")
+printf("\nCost per week: %.0f Paise(p)\n",C)
\ No newline at end of file diff --git a/608/CH20/EX20.01/20_01.sce b/608/CH20/EX20.01/20_01.sce new file mode 100755 index 000000000..66175a6ea --- /dev/null +++ b/608/CH20/EX20.01/20_01.sce @@ -0,0 +1,13 @@ +//Problem 20.01: A transformer has 500 primary turns and 3000 secondary turns. If the primary voltage is 240 V, determine the secondary voltage, assuming an ideal transformer.
+
+//initializing the variables:
+N1 = 500; // primary turns
+N2 = 3000; // secondary turns
+V1 = 240; // in Volts
+
+//calculation:
+//For an ideal transformer, voltage ratio = turns ratio
+V2 = V1*N2/N1
+
+printf("\n\n Result \n\n")
+printf("\n secondary voltage %.2E V",V2)
\ No newline at end of file diff --git a/608/CH20/EX20.02/20_02.sce b/608/CH20/EX20.02/20_02.sce new file mode 100755 index 000000000..5927a2338 --- /dev/null +++ b/608/CH20/EX20.02/20_02.sce @@ -0,0 +1,12 @@ +//Problem 20.02: An ideal transformer with a turns ratio of 2:7 is fed from a 240 V supply. Determine its output voltage.
+
+//initializing the variables:
+tr = 2/7; // turns ratio
+V1 = 240; // in Volts
+
+//calculation:
+//A turns ratio of 2:7 means that the transformer has 2 turns on the primary for every 7 turns on the secondary
+V2 = V1/tr
+
+printf("\n\n Result \n\n")
+printf("\n output voltage %.0f V",V2)
\ No newline at end of file diff --git a/608/CH20/EX20.03/20_03.sce b/608/CH20/EX20.03/20_03.sce new file mode 100755 index 000000000..b604c3daa --- /dev/null +++ b/608/CH20/EX20.03/20_03.sce @@ -0,0 +1,15 @@ +//Problem 20.03: An ideal transformer has a turns ratio of 8:1 and the primary current is 3 A when it is supplied at 240 V. Calculate the secondary voltage and current.
+
+//initializing the variables:
+tr = 8/1; // turns ratio
+I1 = 3; // in Amperes
+V1 = 240; // in Volts
+
+//calculation:
+//A turns ratio of 8:1 means that the transformer has 28 turns on the primary for every 1turns on the secondary
+V2 = V1/tr
+//secondary current
+I2 = I1*tr
+
+printf("\n\n Result \n\n")
+printf("\n secondary voltage is %.0f V and secondary current is %.0f A",V2, I2)
\ No newline at end of file diff --git a/608/CH20/EX20.04/20_04.sce b/608/CH20/EX20.04/20_04.sce new file mode 100755 index 000000000..cedbbae7e --- /dev/null +++ b/608/CH20/EX20.04/20_04.sce @@ -0,0 +1,17 @@ +//Problem 20.04: An ideal transformer, connected to a 240 V mains, supplies a 12 V, 150 W lamp. Calculate the transformer turns ratio and the current taken from the supply.
+
+//initializing the variables:
+V1 = 240; // in Volts
+V2 = 12; // in Volts
+P = 150; // in Watts
+
+//calculation:
+I2 = P/V2
+//A turns ratio = Vp/Vs
+tr = V1/V2 // turn ratio
+// V1/V2 = I2/I1
+//current taken from the supply
+I1 = I2*V2/V1
+
+printf("\n\n Result \n\n")
+printf("\n turn ratio is %.0f and current taken from the supply is %.3f A",tr, I1)
\ No newline at end of file diff --git a/608/CH20/EX20.05/20_05.sce b/608/CH20/EX20.05/20_05.sce new file mode 100755 index 000000000..af54d8dff --- /dev/null +++ b/608/CH20/EX20.05/20_05.sce @@ -0,0 +1,21 @@ +//Problem 20.05: A 5 kVA single-phase transformer has a turns ratio of 10:1 and is fed from a 2.5 kV supply. Neglecting losses, determine (a) the full-load secondary current, (b) the minimum load resistance which can be connected across the secondary winding to give full load kVA, (c) the primary current at full load kVA.
+
+//initializing the variables:
+S = 5000; // in VA
+tr = 10; // turn ratio
+V1 = 2500; // in Volts
+
+//calculation:
+//A turns ratio of 8:1 means that the transformer has 28 turns on the primary for every 1turns on the secondary
+V2 = V1/tr
+//transformer rating in volt-amperes = Vs*Is
+I2 = S/V2
+//Minimum value of load resistance
+RL = V2/I2
+// tr = I2/I1
+I1 = I2/tr
+
+printf("\n\n Result \n\n")
+printf("\n (a)full-load secondary current is %.0f A",I2)
+printf("\n (b)minimum load resistance is %.1f ohm",RL)
+printf("\n (c) primary current is %.0f A",I1)
\ No newline at end of file diff --git a/608/CH20/EX20.06/20_06.sce b/608/CH20/EX20.06/20_06.sce new file mode 100755 index 000000000..7135a3ee3 --- /dev/null +++ b/608/CH20/EX20.06/20_06.sce @@ -0,0 +1,19 @@ +//Problem 20.06: A 2400 V/400 V single-phase transformer takes a no load current of 0.5 A and the core loss is 400 W. Determine the values of the magnetizing and core loss components of the no load current. Draw to scale the no-load phasor diagram for the transformer.
+
+//initializing the variables:
+V1 = 2400; // in Volts
+V2 = 400; // in Volts
+I0 = 0.5; // in Amperes
+Pc = 400; // in Watts
+
+//calculation:
+//Core loss (i.e. iron loss) P = V1*I0*cos(phi0)
+pf = Pc/(V1*I0)
+phi0 = acos(pf)
+//Magnetizing component
+Im = I0*sin(phi0)
+//Core loss component
+Ic = I0*cos(phi0)
+
+printf("\n\n Result \n\n")
+printf("\n (a)magnetizing component is %.3f A and Core loss component is %.3f A",Im, Ic)
\ No newline at end of file diff --git a/608/CH20/EX20.07/20_07.sce b/608/CH20/EX20.07/20_07.sce new file mode 100755 index 000000000..80200d877 --- /dev/null +++ b/608/CH20/EX20.07/20_07.sce @@ -0,0 +1,21 @@ +//Problem 20.07: A transformer takes a current of 0.8 A when its primary is connected to a 240 volt, 50 Hz supply, the secondary being on open circuit. If the power absorbed is 72 watts, determine (a) the iron loss current, (b) the power factor on no-load, and (c) the magnetizing current.
+
+//initializing the variables:
+V = 240; // in Volts
+I0 = 0.8; // in Amperes
+P = 72; // in Watts
+f = 50; // in Hz
+
+//calculation:
+//Power absorbed = total core loss, P = V*I0*cos(phi0)
+//Ic = I0*cos(phi0)
+Ic = P/V
+pf = Ic/I0
+//From the right-angled triangle in Figure 20.2(b) and using
+//Pythagoras’ theorem,
+Im = (I0*I0 - Ic*Ic)^0.5
+
+printf("\n\n Result \n\n")
+printf("\n (a) Core loss component is %.2f A", Ic)
+printf("\n (b) power factor is %.3f", pf)
+printf("\n (c)magnetizing component is %.2f A",Im)
\ No newline at end of file diff --git a/608/CH20/EX20.08/20_08.sce b/608/CH20/EX20.08/20_08.sce new file mode 100755 index 000000000..22b05fa68 --- /dev/null +++ b/608/CH20/EX20.08/20_08.sce @@ -0,0 +1,25 @@ +//Problem 20.08: A 100 kVA, 4000 V/200 V, 50 Hz single-phase transformer has 100 secondary turns. Determine (a) the primary and secondary current, (b) the number of primary turns, and (c) the maximum value of the flux.
+
+//initializing the variables:
+S = 100000; // in VA
+V1 = 4000; // in Volts
+V2 = 200; // in Volts
+N2 = 100; // sec turns
+f = 50; // in Hz
+
+//calculation:
+//Transformer rating = V1*I1 = V2*I2
+//primary current
+I1 = S/V1
+//secondary current
+I2 = S/V2
+//primary turns
+N1 = N2*V1/V2
+//maximum flux
+//assuming E2 = V2
+Phim = V2/(4.44*f*N2)
+
+printf("\n\n Result \n\n")
+printf("\n (a)primary current is %.0f A and secondary current is %.0f A", I1, I2)
+printf("\n (b)number of primary turns is %.0f", N1)
+printf("\n (c)maximum value of the flux is %.2E Wb",Phim)
\ No newline at end of file diff --git a/608/CH20/EX20.09/20_09.sce b/608/CH20/EX20.09/20_09.sce new file mode 100755 index 000000000..d60056c0d --- /dev/null +++ b/608/CH20/EX20.09/20_09.sce @@ -0,0 +1,22 @@ +//Problem 20.09: A single-phase, 50 Hz transformer has 25 primary turns and 300 secondary turns. The cross-sectional area of the core is 300 cm2. When the primary winding is connected to a 250 V supply, determine (a) the maximum value of the flux density in the core, and (b) the voltage induced in the secondary winding.
+
+//initializing the variables:
+V1 = 250; // in Volts
+A = 0.03; // in m2
+N2 = 300; // sec turns
+N1 = 25; // prim turns
+f = 50; // in Hz
+
+//calculation:
+//e.m.f. E1 = 4.44*f*Phim*N1
+//maximum flux density,
+Phim = V1/(4.44*f*N1)
+//Phim = Bm*A, where Bm = maximum core flux density and A = cross-sectional area of the core
+//maximum core flux density
+Bm = Phim/A
+//voltage induced in the secondary winding,
+V2 = V1*N2/N1
+
+printf("\n\n Result \n\n")
+printf("\n (a)maximum core flux density %.2f T", Bm)
+printf("\n (b)voltage induced in the secondary winding is %.0f V", V2)
\ No newline at end of file diff --git a/608/CH20/EX20.10/20_10.sce b/608/CH20/EX20.10/20_10.sce new file mode 100755 index 000000000..80c3d9f57 --- /dev/null +++ b/608/CH20/EX20.10/20_10.sce @@ -0,0 +1,21 @@ +//Problem 20.10: A single-phase 500 V/100 V, 50 Hz transformer has a maximum core flux density of 1.5 T and an effective core crosssectional area of 50 cm2. Determine the number of primary and secondary turns.
+
+//initializing the variables:
+V1 = 500; // in Volts
+V2 = 100; // in Volts
+Bm = 1.5; // in Tesla
+A = 0.005; // in m2
+f = 50; // in Hz
+
+//calculation:
+//Phim = Bm*A, where Bm = maximum core flux density and A = cross-sectional area of the core
+//maximum core flux density
+Phim = Bm*A
+//e.m.f. E1 = 4.44*f*Phim*N1
+//primary turns,
+N1 = V1/(4.44*f*Phim)
+//secondary turns,
+N2 = V2*N1/V1
+
+printf("\n\n Result \n\n")
+printf("\n no. of primary and secondary turns are %.0f turns, and %.0f turns respectively", N1, N2)
\ No newline at end of file diff --git a/608/CH20/EX20.11/20_11.sce b/608/CH20/EX20.11/20_11.sce new file mode 100755 index 000000000..20163a3ba --- /dev/null +++ b/608/CH20/EX20.11/20_11.sce @@ -0,0 +1,25 @@ +//Problem 20.11: A 4500 V/225 V, 50 Hz single-phase transformer is to have an approximate e.m.f. per turn of 15 V and operate with a maximum flux of 1.4 T. Calculate (a) the number of primary and secondary turns and (b) the cross-sectional area of the core.
+
+//initializing the variables:
+emfpt = 15; // in Volts
+V1 = 4500; // in Volts
+V2 = 225; // in Volts
+Bm = 1.4; // in Tesla
+f = 50; // in Hz
+
+//calculation:
+//E.m.f. per turn, V1/N1 = V2/N2 = emfpt
+//primary turns,
+N1 = V1/emfpt
+//secondary turns,
+N2 = V2/emfpt
+//e.m.f. E1 = 4.44*f*Phim*N1
+//maximum flux density,
+Phim = V1/(4.44*f*N1)
+//Phim = Bm*A, where Bm = maximum core flux density and A = cross-sectional area of the core
+//cross-sectional area
+A = Phim/Bm
+
+printf("\n\n Result \n\n")
+printf("\n (a)no. of primary and secondary turns are %.0f turns, and %.0f turns respectively", N1, N2)
+printf("\n (b)cross-sectional area is %.2E m2", A)
\ No newline at end of file diff --git a/608/CH20/EX20.12/20_12.sce b/608/CH20/EX20.12/20_12.sce new file mode 100755 index 000000000..96fa84f82 --- /dev/null +++ b/608/CH20/EX20.12/20_12.sce @@ -0,0 +1,25 @@ +//Problem 20.12: A single-phase transformer has 2000 turns on the primary and 800 turns on the secondary. Its no-load current is 5 A at a power factor of 0.20 lagging. Assuming the volt drop in the windings is negligible, determine the primary current and power factor when the secondary current is 100 A at a power factor of 0.85 lagging.
+
+//initializing the variables:
+N1 = 2000; // prim turns
+N2 = 800; // sec turns
+I0 = 5; // in Amperes
+pf0 = 0.20; // power factor
+I2 = 100; // in Amperes
+pf2 = 0.85; // power factor
+
+//calculation:
+//Let I01 be the component of the primary current which provides the restoring mmf. Then I01*N1 = I2*N2
+I01 = I2*N2/N1
+//If the power factor of the secondary is 0.85
+phi2 = acos(pf2)
+//If the power factor on no-load is 0.20,
+phi0 = acos(pf0)
+I1h = I0*cos(phi0) + I01*cos(phi2)
+I1v = I0*sin(phi0) + I01*sin(phi2)
+//Hence the magnitude of I1
+I1 = (I1h*I1h + I1v*I1v)^0.5
+pf1 = cos(atan(I1v/I1h))
+
+printf("\n\n Result \n\n")
+printf("\n Primary currnt is %.2f A, and Power factor is %.2f", I1, pf1)
\ No newline at end of file diff --git a/608/CH20/EX20.13/20_13.sce b/608/CH20/EX20.13/20_13.sce new file mode 100755 index 000000000..ff9e94745 --- /dev/null +++ b/608/CH20/EX20.13/20_13.sce @@ -0,0 +1,29 @@ +//Problem 20.13: A transformer has 600 primary turns and 150 secondary turns. The primary and secondary resistances are 0.25 ohm and 0.01 ohm respectively and the corresponding leakage reactances are 1.0 ohm and 0.04 ohm respectively. Determine (a) the equivalent resistance referred to the primary winding, (b) the equivalent reactance referred to the primary winding, (c) the equivalent impedance referred to the primary winding, and (d) the phase angle of the impedance.
+
+//initializing the variables:
+N1 = 600; // prim turns
+N2 = 150; // sec turns
+R1 = 0.25; // in ohms
+R2 = 0.01; // in ohms
+X1 = 1.0; // in ohms
+X2 = 0.04; // in ohms
+
+//calculation:
+tr = N1/N2 // turn ratio
+vr = tr // voltage ratio = turn raio, vr = V1/V2
+//equivalent resistance Re
+Re = R1 + R2*(vr^2)
+//equivalent reactance, Xe
+Xe = X1 + X2*(vr^2)
+//equivalent impedance, Ze
+Ze = (Re*Re + Xe*Xe)^0.5
+//cos(phie) = Re/Ze
+pfe = Re/Ze
+phie = acos(pfe)
+phied = phie*180/%pi // in °(deg)
+
+printf("\n\n Result \n\n")
+printf("\n (a)the equivalent resistance referred to the primary winding is %.2f ohm", Re)
+printf("\n (b)the equivalent reactance referred to the primary winding is %.2f ohm", Xe)
+printf("\n (c)the equivalent impedance referred to the primary winding is %.2f ohm", Ze)
+printf("\n (d)phase angle is %.2f°", phied)
\ No newline at end of file diff --git a/608/CH20/EX20.14/20_14.sce b/608/CH20/EX20.14/20_14.sce new file mode 100755 index 000000000..aad51cb54 --- /dev/null +++ b/608/CH20/EX20.14/20_14.sce @@ -0,0 +1,14 @@ +//Problem 20.14: A 5 kVA, 200 V/400 V, single-phase transformer has a secondary terminal voltage of 387.6 volts when loaded. Determine the regulation of the transformer.
+
+//initializing the variables:
+V1 = 200; // in Volts
+V2 = 400; // in Volts
+V2L = 387.6; // in Volts
+S = 5000; // in VA
+
+//calculation:
+//regulation =(No-load secondary voltage - terminal voltage on load)*100/no-load secondary voltage in %
+reg = (V2 - V2L)*100/V2
+
+printf("\n\n Result \n\n")
+printf("\n the regulation of the transformer is %.1f percent ", reg)
\ No newline at end of file diff --git a/608/CH20/EX20.15/20_15.sce b/608/CH20/EX20.15/20_15.sce new file mode 100755 index 000000000..206a833f4 --- /dev/null +++ b/608/CH20/EX20.15/20_15.sce @@ -0,0 +1,12 @@ +//Problem 20.15: The open circuit voltage of a transformer is 240 V. A tap changing device is set to operate when the percentage regulation drops below 2.5%. Determine the load voltage at which the mechanism operates.
+
+//initializing the variables:
+VnL = 240; // in Volts
+reg = 2.5; // in percent
+
+//calculation:
+//regulation =(No-load secondary voltage - terminal voltage on load)*100/no-load secondary voltage in %
+VL = VnL - reg*VnL/100
+
+printf("\n\n Result \n\n")
+printf("\n the load voltage at which the mechanism operates is %.0f V ", VL)
\ No newline at end of file diff --git a/608/CH20/EX20.16/20_16.sce b/608/CH20/EX20.16/20_16.sce new file mode 100755 index 000000000..9cca90770 --- /dev/null +++ b/608/CH20/EX20.16/20_16.sce @@ -0,0 +1,22 @@ +//Problem 20.16: A 200 kVA rated transformer has a full-load copper loss of 1.5 kW and an iron loss of 1 kW. Determine the transformer efficiency at full load and 0.85 power factor.
+
+//initializing the variables:
+S = 200000; // in VA
+Pc = 1500; // in Watt
+Pi = 1000; // in Watt
+pf = 0.85; // power factor
+
+//calculation:
+//Efficiency = output power/input power = (input power—losses)/input power
+//Efficiency = 1 - losses/input power
+//Full-load output power = V*I*pf
+Po = S*pf
+//Total losses
+Pl = Pc + Pi
+//Input power = output power + losses
+PI = Po + Pl
+//efficiency
+eff = 1-(Pl/PI)
+
+printf("\n\n Result \n\n")
+printf("\n the transformer efficiency at full load is %.4f", eff)
\ No newline at end of file diff --git a/608/CH20/EX20.17/20_17.sce b/608/CH20/EX20.17/20_17.sce new file mode 100755 index 000000000..872c0a8fb --- /dev/null +++ b/608/CH20/EX20.17/20_17.sce @@ -0,0 +1,26 @@ +//Problem 20.17: Determine the efficiency of the transformer in Problem 20.16 at half full-load and 0.85 power factor.
+
+//initializing the variables:
+S = 200000; // in VA
+Pc = 1500; // in Watt
+Pi = 1000; // in Watt
+pf = 0.85; // power factor
+
+//calculation:
+//Efficiency = output power/input power = (input power—losses)/input power
+//Efficiency = 1 - losses/input power
+//Half full-load power output = V*I*pf/2
+Po = S*pf/2
+//Copper loss (or I*I*R loss) is proportional to current squared
+//Hence the copper loss at half full-load is
+Pch = Pc/(2*2)
+//Iron loss = 1000 W (constant)
+//Total losses
+Pl = Pch + Pi
+//Input power at half full-load = output power at half full-load + losses
+PI = Po + Pl
+//efficiency
+eff = (1-(Pl/PI))*100
+
+printf("\n\n Result \n\n")
+printf("\n the transformer efficiency at half full load is %.2f percent", eff)
\ No newline at end of file diff --git a/608/CH20/EX20.18/20_18.sce b/608/CH20/EX20.18/20_18.sce new file mode 100755 index 000000000..bd5a69e6a --- /dev/null +++ b/608/CH20/EX20.18/20_18.sce @@ -0,0 +1,45 @@ +//Problem 20.18: A 400 kVA transformer has a primary winding resistance of 0.5 ohm and a secondary winding resistance of 0.001 ohm . The iron loss is 2.5 kW and the primary and secondary voltages are 5 kV and 320 V respectively. If the power factor of the load is 0.85, determine the efficiency of the transformer (a) on full load, and (b) on half load.
+
+//initializing the variables:
+S = 400000; // in VA
+R1 = 0.5; // in Ohm
+R2 = 0.001; // in Ohm
+V1 = 5000; // in Volts
+V2 = 320; // in Volts
+Pi = 2500; // in Watt
+pf = 0.85; // power factor
+
+//calculation:
+//Rating = 400 kVA = V1*I1 = V2*I2
+//Hence primary current
+I1 = S/V1
+//secondary current
+I2 = S/V2
+//Total copper loss = I1*I1*R1 + I2*I2*R2,
+Pcf = I1*I1*R1 + I2*I2*R2
+//On full load, total loss = copper loss + iron loss
+Plf = Pcf + Pi
+// full-load power output = V2*I2*pf
+Pof = S*pf
+//Input power at full-load = output power at full-load + losses
+PIf = Pof + Plf
+//Efficiency = output power/input power = (input power—losses)/input power
+//Efficiency = 1 - losses/input power
+efff = (1-(Plf/PIf))*100
+
+//Half full-load power output = V*I*pf/2
+Poh = S*pf/2
+//Copper loss (or I*I*R loss) is proportional to current squared
+//Hence the copper loss at half full-load is
+Pch = Pcf/(2*2)
+//Iron loss = 2500 W (constant)
+//Total losses
+Plh = Pch + Pi
+//Input power at half full-load = output power at half full-load + losses
+PIh = Poh + Plh
+//efficiency
+effh = (1-(Plh/PIh))*100
+
+printf("\n\n Result \n\n")
+printf("\n (a)the transformer efficiency at full load is %.2f percent", efff)
+printf("\n (b)the transformer efficiency at half full load is %.2f percent", effh)
\ No newline at end of file diff --git a/608/CH20/EX20.19/20_19.sce b/608/CH20/EX20.19/20_19.sce new file mode 100755 index 000000000..d99231ada --- /dev/null +++ b/608/CH20/EX20.19/20_19.sce @@ -0,0 +1,29 @@ +//Problem 20.19: A 500 kVA transformer has a full load copper loss of 4 kW and an iron loss of 2.5 kW. Determine (a) the output kVA at which the efficiency of the transformer is a maximum, and (b) the maximum efficiency, assuming the power factor of the load is 0.75.
+
+//initializing the variables:
+S = 500000; // in VA
+Pcf = 4000; // in Watt
+Pi = 2500; // in Watt
+pf = 0.75; // power factor
+
+//calculation:
+//Let x be the fraction of full load kVA at which the efficiency is a maximum.
+//The corresponding total copper loss = (4 kW)*(x^2)
+//At maximum efficiency, copper loss = iron loss Hence
+x = (Pi/Pcf)^0.5
+//Hence the output kVA at maximum efficiency
+So = x*S
+//Total loss at maximum efficiency
+Pl = 2*Pi
+//Output power
+Po = So*pf
+//Input power = output power + losses
+PI = Po + Pl
+//Efficiency = output power/input power = (input power—losses)/input power
+//Efficiency = 1 - losses/input power
+//Maximum efficiency
+effm = (1 - Pl/PI)*100
+
+printf("\n\n Result \n\n")
+printf("\n the output kVA at maximum efficiency is %.2E VA", So)
+printf("\n max. efficiency is %.2f pecent", effm)
\ No newline at end of file diff --git a/608/CH20/EX20.20/20_20.sce b/608/CH20/EX20.20/20_20.sce new file mode 100755 index 000000000..c441abf4d --- /dev/null +++ b/608/CH20/EX20.20/20_20.sce @@ -0,0 +1,12 @@ +//Problem 20.20: A transformer having a turns ratio of 4:1 supplies a load of resistance 100 ohm . Determine the equivalent input resistance of the transformer.
+
+//initializing the variables:
+tr = 4; // turn ratio
+RL = 100; // in Ohms
+
+//calculation:
+//the equivalent input resistance,
+Ri = RL*(tr^2)
+
+printf("\n\n Result \n\n")
+printf("\n the equivalent input resistance is %.0f ohm", Ri)
\ No newline at end of file diff --git a/608/CH20/EX20.21/20_21.sce b/608/CH20/EX20.21/20_21.sce new file mode 100755 index 000000000..67d67f53c --- /dev/null +++ b/608/CH20/EX20.21/20_21.sce @@ -0,0 +1,14 @@ +//Problem 20.21: The output stage of an amplifier has an output resistance of 112 ohm . Calculate the optimum turns ratio of a transformer which would match a load resistance of 7 ohm to the output resistance of the amplifier.
+
+//initializing the variables:
+R1 = 112; // in Ohms
+RL = 7; // in Ohms
+
+//calculation:
+//The equivalent input resistance, R1 of the transformer needs to be 112 ohm for maximum power transfer.
+//R1 = RL*(tr^2)
+// tr = N1/N2 turn ratio
+tr = (R1/RL)^0.5
+
+printf("\n\n Result \n\n")
+printf("\n the optimum turns ratio is %.0f ", tr)
\ No newline at end of file diff --git a/608/CH20/EX20.22/20_22.sce b/608/CH20/EX20.22/20_22.sce new file mode 100755 index 000000000..c30b07558 --- /dev/null +++ b/608/CH20/EX20.22/20_22.sce @@ -0,0 +1,13 @@ +//Problem 20.22: Determine the optimum value of load resistance for maximum power transfer if the load is connected to an amplifier of output resistance 150 ohm through a transformer with a turns ratio of 5:1..
+
+//initializing the variables:
+tr = 5; // turn ratio
+R1 = 150; // in Ohms
+
+//calculation:
+//The equivalent input resistance, R1 of the transformer needs to be 150 ohm for maximum power transfer.
+//R1 = RL*(tr^2)
+RL = R1/(tr^2)
+
+printf("\n\n Result \n\n")
+printf("\n the optimum value of load resistance is %.0f ohm", RL)
\ No newline at end of file diff --git a/608/CH20/EX20.23/20_23.sce b/608/CH20/EX20.23/20_23.sce new file mode 100755 index 000000000..c4c8bf577 --- /dev/null +++ b/608/CH20/EX20.23/20_23.sce @@ -0,0 +1,27 @@ +//Problem 20.23: A single-phase, 220 V/1760 V ideal transformer is supplied from a 220 V source through a cable of resistance 2 ohm . If the load across the secondary winding is 1.28 k ohm determine (a) the primary current flowing and (b) the power dissipated in the load resistor.
+
+//initializing the variables:
+V1 = 220; // in Volts
+V2 = 1760; // in Volts
+V = 220; // in Volts
+RL = 1280; // in Ohms
+R = 2; // in Ohms
+
+//calculation:
+//Turns ratio, tr = N1/N2 = V1/V2
+tr = V1/V2
+//Equivalent input resistance of the transformer,
+//R1 = RL*(tr^2)
+R1 = RL*(tr^2)
+//Total input resistance
+Rin = R + R1
+// Primary current
+I1 = V1/Rin
+//For an ideal transformer V1/V2 = I2/I1,
+I2 = I1*tr
+//Power dissipated in load resistor RL
+P = I2*I2*RL
+
+printf("\n\n Result \n\n")
+printf("\n (a) primary current flowing is %.0f A", I1)
+printf("\n (b) power dissipated in the load resistor is %.0f W", P)
\ No newline at end of file diff --git a/608/CH20/EX20.24/20_24.sce b/608/CH20/EX20.24/20_24.sce new file mode 100755 index 000000000..bdd04556e --- /dev/null +++ b/608/CH20/EX20.24/20_24.sce @@ -0,0 +1,24 @@ +//Problem 20.24: An a.c. source of 24 V and internal resistance 15 kohm is matched to a load by a 25:1 ideal transformer. Determine (a) the value of the load resistance and (b) the power dissipated in the load.
+
+//initializing the variables:
+tr = 25; // teurn ratio
+V = 24; // in Volts
+R1 = 15000; // in Ohms
+Rin = 15000; // in ohms
+
+//calculation:
+//Turns ratio, tr = N1/N2 = V1/V2
+//For maximum power transfer R1 needs to be equal to 15 kohm
+RL = R1/(tr^2)
+//The total input resistance when the source is connected to the matching transformer is
+Rt = Rin + R1
+//Primary current,
+I1 = V/Rt
+//N1/N2 = I2/I1
+I2 = I1*tr
+//Power dissipated in load resistor RL
+P = I2*I2*RL
+
+printf("\n\n Result \n\n")
+printf("\n (a) the load resistance is %.0f ohm", RL)
+printf("\n (b) power dissipated in the load resistor is %.2E W", P)
\ No newline at end of file diff --git a/608/CH20/EX20.25/20_25.sce b/608/CH20/EX20.25/20_25.sce new file mode 100755 index 000000000..d5c35763b --- /dev/null +++ b/608/CH20/EX20.25/20_25.sce @@ -0,0 +1,19 @@ +//Problem 20.25: A single-phase auto transformer has a voltage ratio 320 V:250 V and supplies a load of 20 kVA at 250 V. Assuming an ideal transformer, determine the current in each section of the winding.
+
+//initializing the variables:
+V1 = 320; // in Volts
+V2 = 250; // in Volts
+S = 20000; // in VA
+
+//calculation:
+//Rating = 20 kVA = V1*I1 = V2*I2
+//Hence primary current, I1
+I1 = S/V1
+//secondary current, I2
+I2 = S/V2
+//Hence current in common part of the winding
+I = I2 - I1
+
+printf("\n\n Result \n\n")
+printf("\n current in common part of the winding is %.1f A", I)
+printf("\n primary current and secondary current are %.1f A and %.0f A respectively",I1, I2)
\ No newline at end of file diff --git a/608/CH20/EX20.26/20_26.sce b/608/CH20/EX20.26/20_26.sce new file mode 100755 index 000000000..da2bf9723 --- /dev/null +++ b/608/CH20/EX20.26/20_26.sce @@ -0,0 +1,25 @@ +//Problem 20.26: Determine the saving in the volume of copper used in an auto transformer compared with a double-wound transformer for (a) a 200 V:150 V transformer, and (b) a 500 V:100 V transformer.
+
+//initializing the variables:
+V1a = 200; // in Volts
+V2a = 150; // in Volts
+V1b = 500; // in Volts
+V2b = 100; // in Volts
+
+//calculation:
+//For a 200 V:150 V transformer, xa
+xa = V2a/V1a
+//volume of copper in auto transformer
+vca = (1 - xa)*100 // of copper in a double-wound transformer
+//the saving is
+vsa = 100 - vca
+//For a 500 V:100 V transformer, xb
+xb = V2b/V1b
+//volume of copper in auto transformer
+vcb = (1 - xb)*100 // of copper in a double-wound transformer
+//the saving is
+vsb = 100 - vcb
+
+printf("\n\n Result \n\n")
+printf("\n (a)For a 200 V:150 V transformer, the saving is %.0f percent", vsa)
+printf("\n (b)For a 500 V:100 V transformer, the saving is %.0f percent", vsb)
\ No newline at end of file diff --git a/608/CH20/EX20.27/20_27.sce b/608/CH20/EX20.27/20_27.sce new file mode 100755 index 000000000..8e8c4dd8f --- /dev/null +++ b/608/CH20/EX20.27/20_27.sce @@ -0,0 +1,29 @@ +//Problem 20.27: A three-phase transformer has 500 primary turns and 50 secondary turns. If the supply voltage is 2.4 kV find the secondary line voltage on no-load when the windings are connected a) star-delta, (b) delta-star.
+
+//initializing the variables:
+N1 = 500; // prim turns
+N2 = 50; // sec turns
+VL = 2400; // in Volts
+
+//calculation:
+//For a star-connection, VL = Vp*(3^0.5)
+VL1s = VL
+//Primary phase voltage
+Vp1s = VL1s/(3^0.5)
+//For a delta-connection, VL = Vp
+//N1/N2 = V1/V2, from which,
+//secondary phase voltage, Vp2s
+Vp2s = Vp1s*N2/N1
+VL2d = Vp2s
+
+//For a delta-connection, VL = Vp
+VL1d = VL
+//primary phase voltage Vp1d
+Vp1d = VL1d
+//Secondary phase voltage, Vp2d
+Vp2d = Vp1d*N2/N1
+//For a star-connection, VL = Vp*(3^0.5)
+VL2s = Vp2d*(3^0.5)
+
+printf("\n\n Result \n\n")
+printf("\n the secondary line voltage for star and delta connection are %.0f V and %.1f V respectively", VL2s, VL2d)
\ No newline at end of file diff --git a/608/CH20/EX20.28/20_28.sce b/608/CH20/EX20.28/20_28.sce new file mode 100755 index 000000000..0bffdb1eb --- /dev/null +++ b/608/CH20/EX20.28/20_28.sce @@ -0,0 +1,25 @@ +//Problem 20.28: A current transformer has a single turn on the primary winding and a secondary winding of 60 turns. The secondary winding is connected to an ammeter with a resistance of 0.15 ohm . The resistance of the secondary winding is 0.25 ohm . If the current in the primary winding is 300 A, determine (a) the reading on the ammeter, (b) the potential difference across the ammeter and (c) the total load (in VA) on the secondary.
+
+//initializing the variables:
+N1 = 1; // prim turns
+N2 = 60; // sec turns
+I1 = 300; // in amperes
+Ra = 0.15; // in ohms
+R2 = 0.25; // in ohms
+
+//calculation:
+//Reading on the ammeter,
+I2 = I1*(N1/N2)
+//P.d. across the ammeter = I2*RA, where RA is the ammeter resistance
+pd = I2*Ra
+//Total resistance of secondary circuit
+Rt = Ra + R2
+//Induced e.m.f. in secondary
+V2 = I2*Rt
+//Total load on secondary
+S = V2*I2
+
+printf("\n\n Result \n\n")
+printf("\n (a)the reading on the ammeter is %.0f A ",I2)
+printf("\n (b)potential difference across the ammeter is %.2f V ",pd)
+printf("\n (c)total load (in VA) on the secondary is %.0f VA ",S)
\ No newline at end of file diff --git a/608/CH21/EX21.01/21_01.sce b/608/CH21/EX21.01/21_01.sce new file mode 100755 index 000000000..17f71869d --- /dev/null +++ b/608/CH21/EX21.01/21_01.sce @@ -0,0 +1,15 @@ +//Problem 21.01: An 8-pole, wave-connected armature has 600 conductors and is driven at 625 rev/min. If the flux per pole is 20 mWb, determine the generated e.m.f.
+
+//initializing the variables:
+Z = 600; // no. of conductors
+c = 2; // for a wave winding
+p = 4; // no. of pairs
+n = 625/60; // in rev/sec
+Phi = 20E-3; // in Wb
+
+//calculation:
+//Generated e.m.f., E = 2*p*Phi*n*Z/c
+E = 2*p*Phi*n*Z/c
+
+printf("\n\n Result \n\n")
+printf("\n the generated e.m.f is %.0f V ",E)
\ No newline at end of file diff --git a/608/CH21/EX21.02/21_02.sce b/608/CH21/EX21.02/21_02.sce new file mode 100755 index 000000000..7e615564f --- /dev/null +++ b/608/CH21/EX21.02/21_02.sce @@ -0,0 +1,16 @@ +//Problem 21.02: A 4-pole generator has a lap-wound armature with 50 slots with 16 conductors per slot. The useful flux per pole is 30 mWb. Determine the speed at which the machine must be driven to generate an e.m.f. of 240 V.
+
+//initializing the variables:
+Z = 50*16; // no. of conductors
+p = 1; // let no. of pairs
+c = 2*p; // for a lap winding
+Phi = 30E-3; // in Wb
+E = 240; // in Volts
+
+//calculation:
+//Generated e.m.f., E = 2*p*Phi*n*Z/c
+//Rearranging gives, speed
+n = E*c/(2*p*Phi*Z)
+
+printf("\n\n Result \n\n")
+printf("\n the speed at which the machine must be driven is %.0f rev/sec ",n)
\ No newline at end of file diff --git a/608/CH21/EX21.03/21_03.sce b/608/CH21/EX21.03/21_03.sce new file mode 100755 index 000000000..e3c07c47d --- /dev/null +++ b/608/CH21/EX21.03/21_03.sce @@ -0,0 +1,15 @@ +//Problem 21.03: An 8-pole, lap-wound armature has 1200 conductors and a flux per pole of 0.03 Wb. Determine the e.m.f. generated when running at 500 rev/min.
+
+//initializing the variables:
+Z = 1200; // no. of conductors
+p = 1; // let, no. of pairs
+c = 2*p; // for a lap winding
+Phi = 30E-3; // in Wb
+n = 500/60; // in rev/sec
+
+//calculation:
+//Generated e.m.f., E = 2*p*Phi*n*Z/c
+E = 2*p*Phi*n*Z/c
+
+printf("\n\n Result \n\n")
+printf("\n Generated e.m.f. is %.0f V ",E)
\ No newline at end of file diff --git a/608/CH21/EX21.04/21_04.sce b/608/CH21/EX21.04/21_04.sce new file mode 100755 index 000000000..9cdfcef05 --- /dev/null +++ b/608/CH21/EX21.04/21_04.sce @@ -0,0 +1,15 @@ +//Problem 21.04: Determine the generated e.m.f. in problem 21.03 if the armature is wave-wound.
+
+//initializing the variables:
+Z = 1200; // no. of conductors
+p = 4; // let, no. of pairs
+c = 2; // for a wave winding
+Phi = 30E-3; // in Wb
+n = 500/60; // in rev/sec
+
+//calculation:
+//Generated e.m.f., E = 2*p*Phi*n*Z/c
+E = 2*p*Phi*n*Z/c
+
+printf("\n\n Result \n\n")
+printf("\n Generated e.m.f. is %.0f V ",E)
\ No newline at end of file diff --git a/608/CH21/EX21.06/21_06.sce b/608/CH21/EX21.06/21_06.sce new file mode 100755 index 000000000..e3e816ffd --- /dev/null +++ b/608/CH21/EX21.06/21_06.sce @@ -0,0 +1,18 @@ +//Problem 21.06: A d.c. generator running at 30 rev/s generates an e.m.f. of 200 V. Determine the percentage increase in the flux per pole required to generate 250 V at 20 rev/s.
+
+//initializing the variables:
+n1 = 30; // in rev/sec
+E1 = 200; // in Volts
+n2 = 20; // in rev/sec
+E2 = 250; // in Volts
+
+//calculation:
+//generated e.m.f., E proportional to phi*w and since w = 2*pi*n, then
+// E proportional to phi*n
+// E1/E2 = Phi1*n1/(Phi2*n2)
+// let Phi2/Phi1 = Phi
+Phi = E2*n1/(E1*n2)
+Phi_inc = (Phi - 1)*100 ///in percent
+
+printf("\n\n Result \n\n")
+printf("\n percentage increase in the flux per pole is %.1f percent ",Phi_inc)
\ No newline at end of file diff --git a/608/CH21/EX21.07/21_07.sce b/608/CH21/EX21.07/21_07.sce new file mode 100755 index 000000000..d5abd1c29 --- /dev/null +++ b/608/CH21/EX21.07/21_07.sce @@ -0,0 +1,14 @@ +//Problem 21.07: Determine the terminal voltage of a generator which develops an e.m.f. of 200 V and has an armature current of 30 A on load. Assume the armature resistance is 0.30 ohm.
+
+//initializing the variables:
+Ra = 0.30; // in ohms
+Ia = 30; // in Amperes
+E = 200; // in Volts
+
+//calculation:
+//terminal voltage,
+//V = E - Ia*Ra
+V = E - Ia*Ra
+
+printf("\n\n Result \n\n")
+printf("\n terminal voltage of a generator is %.0f V ",V)
\ No newline at end of file diff --git a/608/CH21/EX21.08/21_08.sce b/608/CH21/EX21.08/21_08.sce new file mode 100755 index 000000000..a1a8ebb50 --- /dev/null +++ b/608/CH21/EX21.08/21_08.sce @@ -0,0 +1,17 @@ +//Problem 21.08: A generator is connected to a 60 ohm load and a current of 8 A flows. If the armature resistance is 1 ohm determine (a) the terminal voltage, and (b) the generated e.m.f.
+
+//initializing the variables:
+RL = 60; // in ohms
+Ia = 8; // in Amperes
+Ra = 1; // in ohms
+
+//calculation:
+//terminal voltage,
+//V = Ia*RL
+V = Ia*RL
+//Generated e.m.f., E
+E = V + Ia*Ra
+
+printf("\n\n Result \n\n")
+printf("\n (a)terminal voltage is %.0f V ",V)
+printf("\n (b)generated e.m.f. is %.0f V ",E)
\ No newline at end of file diff --git a/608/CH21/EX21.09/21_09.sce b/608/CH21/EX21.09/21_09.sce new file mode 100755 index 000000000..440c1f519 --- /dev/null +++ b/608/CH21/EX21.09/21_09.sce @@ -0,0 +1,25 @@ +//Problem 21.09: A separately-excited generator develops a no-load e.m.f. of 150 V at an armature speed of 20 rev/s and a flux per pole of 0.10 Wb. Determine the generated e.m.f. when (a) the speed increases to 25 rev/s and the pole flux remains unchanged, (b) the speed remains at 20 rev/s and the pole flux is decreased to 0.08 Wb, and (c) the speed increases to 24 rev/s and the pole flux is decreased to 0.07 Wb.
+
+//initializing the variables:
+E1 = 150; // in Volts
+n1 = 20; // in rev/sec
+Phi1 = 0.10; // in Wb
+n2 = 25; // in rev/sec
+Phi2 = 0.10; // in Wb
+n3 = 20; // in rev/sec
+Phi3 = 0.08; // in Wb
+n4 = 24; // in rev/sec
+Phi4 = 0.07; // in Wb
+
+//calculation:
+//generated e.m.f., E proportional to phi*w and since w = 2*pi*n, then
+// E proportional to phi*n
+// E1/E2 = Phi1*n1/(Phi2*n2)
+E2 = E1*Phi2*n2/(Phi1*n1)
+E3 = E1*Phi3*n3/(Phi1*n1)
+E4 = E1*Phi4*n4/(Phi1*n1)
+
+printf("\n\n Result \n\n")
+printf("\n (a)the generated e.m.f is %.1f V ",E2)
+printf("\n (b)generated e.m.f. is %.0f V ",E3)
+printf("\n (c)generated e.m.f. is %.0f V ",E4)
\ No newline at end of file diff --git a/608/CH21/EX21.10/21_10.sce b/608/CH21/EX21.10/21_10.sce new file mode 100755 index 000000000..f1c07daba --- /dev/null +++ b/608/CH21/EX21.10/21_10.sce @@ -0,0 +1,26 @@ +//Problem 21.10: A shunt generator supplies a 20 kW load at 200 V through cables of resistance, R = 100 mohm. If the field winding resistance, Rf=D 50ohm and the armature resistance, Ra = 40 mohm, determine (a) the terminal voltage, and (b) the e.m.f. generated in the armature.
+
+//initializing the variables:
+Ps = 20000; // in Watts
+Vs = 200; // in Volts
+Rs = 0.1; // in ohms
+Rf = 50; // in ohms
+Ra = 0.04; // in ohms
+
+//calculation:
+//Load current, I
+Is = Ps/Vs
+//Volt drop in the cables to the load
+Vd = Is*Rs
+//Hence terminal voltage,
+V = Vs + Vd
+//Field current, If
+If = V/Rf
+//Armature current
+Ia = If + Is
+//Generated e.m.f. E
+E = V + Ia*Ra
+
+printf("\n\n Result \n\n")
+printf("\n (a)terminal voltage is %.0f V ",V)
+printf("\n (b)generated e.m.f. is %.2f V ",E)
\ No newline at end of file diff --git a/608/CH21/EX21.11/21_11.sce b/608/CH21/EX21.11/21_11.sce new file mode 100755 index 000000000..fe40f60b9 --- /dev/null +++ b/608/CH21/EX21.11/21_11.sce @@ -0,0 +1,23 @@ +//Problem 21.11: A short-shunt compound generator supplies 80 A at 200 V. If the field resistance, Rf = 40 ohm, the series resistance, Rse = 0.02ohms and the armature resistance, Ra = 0.04 ohm, determin the e.m.f. generated.
+
+//initializing the variables:
+Is = 80; // in amperes
+Vs = 200; // in Volts
+Rf = 40; // in ohms
+Rse = 0.02; // in ohms
+Ra = 0.04; // in ohms
+
+//calculation:
+//Volt drop in series winding
+Vse = Is*Rse
+//P.d. across the field winding = p.d. across armature
+V1 = Vs + Vse
+//Field current, If
+If = V1/Rf
+//Armature current
+Ia = If + Is
+//Generated e.m.f. E
+E = V1 + Ia*Ra
+
+printf("\n\n Result \n\n")
+printf("\n generated e.m.f. is %.0f V ",E)
\ No newline at end of file diff --git a/608/CH21/EX21.12/21_12.sce b/608/CH21/EX21.12/21_12.sce new file mode 100755 index 000000000..c68b19350 --- /dev/null +++ b/608/CH21/EX21.12/21_12.sce @@ -0,0 +1,22 @@ +//Problem 21.12: A 10 kW shunt generator having an armature circuit resistance of 0.75 ohm and a field resistance of 125 ohms , generates a terminal voltage of 250 V at full load. Determine the efficiency of the generator at full load, assuming the iron, friction and windage losses amount to 600 W.
+
+//initializing the variables:
+Ps = 10000; // in Watt
+Pl = 600; // in Watt
+Ra = 0.75; // in ohms
+Rf = 125; // in ohms
+V = 250; // in Volts
+
+//calculation:
+//Output power Ps = V*I
+//from which, load current I
+I = Ps/V
+//Field current, If
+If = V/Rf
+//Armature current
+Ia = If + I
+//Efficiency,
+eff = Ps*100/((V*I) + (Ia*Ia*Ra) + (If*V) + (Pl)) // in Percent
+
+printf("\n\n Result \n\n")
+printf("\n Efficiency is %.2f percent ",eff)
\ No newline at end of file diff --git a/608/CH21/EX21.13/21_13.sce b/608/CH21/EX21.13/21_13.sce new file mode 100755 index 000000000..518159c67 --- /dev/null +++ b/608/CH21/EX21.13/21_13.sce @@ -0,0 +1,13 @@ +//Problem 21.13: A d.c. motor operates from a 240 V supply. The armature resistance is 0.2 ohm. Determine the back e.m.f. when the armature current is 50 A.
+
+//initializing the variables:
+Ra = 0.2; // in ohms
+V = 240; // in Volts
+Ia = 50; // in Amperes
+
+//calculation:
+//For a motor, V = E + Ia*Ra
+E = V - Ia*Ra
+
+printf("\n\n Result \n\n")
+printf("\n back e.m.f. is %.0f V ",E)
\ No newline at end of file diff --git a/608/CH21/EX21.14/21_14.sce b/608/CH21/EX21.14/21_14.sce new file mode 100755 index 000000000..f98ecf87a --- /dev/null +++ b/608/CH21/EX21.14/21_14.sce @@ -0,0 +1,19 @@ +//Problem 21.14: The armature of a d.c. machine has a resistance of 0.25 ohm and is connected to a 300 V supply. Calculate the e.m.f. generated when it is running: (a) as a generator giving 100 A, and (b) as a motor taking 80 A.
+
+//initializing the variables:
+Ra = 0.25; // in ohms
+V = 300; // in Volts
+Ig = 100; // in Amperes
+Im = 80; // in Amperes
+
+//calculation:
+//As a generator, generated e.m.f.,
+// E = V + Ia*Ra
+Eg = V + Ig*Ra
+//For a motor, generated e.m.f. (or back e.m.f.),
+// E = V - Ia*Ra
+E = V - Im*Ra
+
+printf("\n\n Result \n\n")
+printf("\n (a)As a generator, generated e.m.f. is %.0f V ",Eg)
+printf("\n (b)back e.m.f. is %.0f V ",E)
\ No newline at end of file diff --git a/608/CH21/EX21.15/21_15.sce b/608/CH21/EX21.15/21_15.sce new file mode 100755 index 000000000..f4d840783 --- /dev/null +++ b/608/CH21/EX21.15/21_15.sce @@ -0,0 +1,15 @@ +//Problem 21.15: An 8-pole d.c. motor has a wave-wound armature with 900 conductors. The useful flux per pole is 25 mWb. Determine the torque exerted when a current of 30 A flows in each armature conductor.
+
+//initializing the variables:
+p = 4;
+c = 2; // for a wave winding
+Phi = 25E-3; // Wb
+Z = 900;
+Ia = 30; // in Amperes
+
+//calculation:
+//torque T = p*Phi*Z*Ia/(pi* c)
+T = p*Phi*Z*Ia/(1*%pi*c)
+
+printf("\n\n Result \n\n")
+printf("\n the torque exerted is %.1f Nm ",T)
\ No newline at end of file diff --git a/608/CH21/EX21.16/21_16.sce b/608/CH21/EX21.16/21_16.sce new file mode 100755 index 000000000..b71896dc3 --- /dev/null +++ b/608/CH21/EX21.16/21_16.sce @@ -0,0 +1,16 @@ +//Problem 21.16: Determine the torque developed by a 350 V d.c. motor having an armature resistance of 0.5 ohm and running at 15 rev/s. The armature current is 60 A.
+
+//initializing the variables:
+V = 350; // in Volts
+Ra = 0.5; // in ohms
+n = 15; // in rev/sec
+Ia = 60; // in Amperes
+
+//calculation:
+//Back e.m.f. E = V - Ia*Ra
+E = V - Ia*Ra
+//torque T = E*Ia/(2* n*pi)
+T = E*Ia/(2*n*%pi)
+
+printf("\n\n Result \n\n")
+printf("\n the torque exerted is %.1f Nm ",T)
\ No newline at end of file diff --git a/608/CH21/EX21.17/21_17.sce b/608/CH21/EX21.17/21_17.sce new file mode 100755 index 000000000..64ee296e9 --- /dev/null +++ b/608/CH21/EX21.17/21_17.sce @@ -0,0 +1,23 @@ +//Problem 21.17: A six-pole lap-wound motor is connected to a 250 V d.c. supply. The armature has 500 conductors and a resistance of 1 ohm. The flux per pole is 20 mWb. Calculate (a) the speed and (b) the torque developed when the armature current is 40 A
+
+//initializing the variables:
+p = 1; // let
+c = 2*p; // for a lap winding
+Phi = 20E-3; // Wb
+Z = 500;
+V = 250; // in Volts
+Ra = 1; // in ohms
+Ia = 40; // in Amperes
+
+//calculation:
+//Back e.m.f. E = V - Ia*Ra
+E = V - Ia*Ra
+//E.m.f. E = 2*p*Phi*n*Z/c
+// rearrange,
+n = E*c/(2*p*Phi*Z)
+//torque T = E*Ia/(2* n*pi)
+T = E*Ia/(2*n*%pi)
+
+printf("\n\n Result \n\n")
+printf("\n (a)speed n is %.0f rev/sec ",n)
+printf("\n (b)the torque exerted is %.2f Nm ",T)
\ No newline at end of file diff --git a/608/CH21/EX21.18/21_18.sce b/608/CH21/EX21.18/21_18.sce new file mode 100755 index 000000000..d88fa62c6 --- /dev/null +++ b/608/CH21/EX21.18/21_18.sce @@ -0,0 +1,18 @@ +//Problem 21.18: The shaft torque of a diesel motor driving a 100 V d.c. shunt-wound generator is 25 Nm. The armature current of the generator is 16 A at this value of torque. If the shunt field regulator is adjusted so that the flux is reduced by 15%, the torque increases to 35 Nm. Determine the armature current at this new value of torque.
+
+//initializing the variables:
+T1 = 25; // in Nm
+T2 = 35; // in Nm
+Ia1 = 16; // in Amperes
+V = 100; // in Volts
+x = 0.15;
+
+//calculation:
+//the shaft torque T of a generator is proportional to (phi*Ia), where Phi is the flux and Ia is the armature current. Thus, T = k*Phi*Ia, where k is a constant.
+//The torque at flux phi1 and armature current Ia1 is T1 = k*Phi1*Ia1.
+//similarly T2 = k*Phi2*Ia2
+Phi2 = (1 - x)*Phi1
+Ia2 = T2*Ia1*Phi1/(Phi2*T1)
+
+printf("\n\n Result \n\n")
+printf("\n armature current at the new value of torque is %.2f A ",Ia2)
\ No newline at end of file diff --git a/608/CH21/EX21.19/21_19.sce b/608/CH21/EX21.19/21_19.sce new file mode 100755 index 000000000..5e78753a0 --- /dev/null +++ b/608/CH21/EX21.19/21_19.sce @@ -0,0 +1,20 @@ +//Problem 21.19: A 100 V d.c. generator supplies a current of 15 A when running at 1500 rev/min. If the torque on the shaft driving the generator is 12 Nm, determine (a) the efficiency of the generator and (b) the power loss in the generator.
+
+//initializing the variables:
+T = 12; // in Nm
+I = 15; // in Amperes
+V = 100; // in Volts
+n = 1500/60; // in rev/sec
+
+//calculation:
+//the efficiency of a generator = (output power/input power)*100 %
+//The output power is the electrical output, i.e. VI watts. The input power to a generator is the mechanical power in the shaft driving the generator, i.e. T*w or T(2*pi*n) watts, where T is the torque in Nm and n is speed of rotation in rev/s. Hence, for a generator
+//efficiency = V*I*100/(T*2*pi* n) %
+eff = V*I*100/(T*2*%pi*n) // in Percent
+//The input power = output power + losses
+// hence, T*2*%pi*n = V*I + losses
+Pl = T*2*%pi*n - V*I
+
+printf("\n\n Result \n\n")
+printf("\n (a) efficiency is %.2f percent ",eff)
+printf("\n (b) power loss is %.0f W ",Pl)
\ No newline at end of file diff --git a/608/CH21/EX21.20/21_20.sce b/608/CH21/EX21.20/21_20.sce new file mode 100755 index 000000000..4f5124520 --- /dev/null +++ b/608/CH21/EX21.20/21_20.sce @@ -0,0 +1,20 @@ +//Problem 21.20: A 240 V shunt motor takes a total current of 30 A. If the field winding resistance Rf = 150 ohm and the armature resistance Ra = 0.4 ohm. determine (a) the current in the armature, and (b) the back e.m.f.
+
+//initializing the variables:
+Rf = 150; // in Ohms
+Ra = 0.4; // in Ohms
+I = 30; // in Amperes
+V = 240; // in Volts
+
+//calculation:
+//Field current If
+If = V/Rf
+//Supply current I = Ia + If
+//Hence armature current, Ia
+Ia = I - If
+//Back e.m.f. E = V - Ia*Ra
+E = V - (Ia*Ra)
+
+printf("\n\n Result \n\n")
+printf("\n (a) current in the armature is %.1f A ",Ia)
+printf("\n (b) Back e.m.f. E is %.2f V ",E)
\ No newline at end of file diff --git a/608/CH21/EX21.21/21_21.sce b/608/CH21/EX21.21/21_21.sce new file mode 100755 index 000000000..2cc2c0368 --- /dev/null +++ b/608/CH21/EX21.21/21_21.sce @@ -0,0 +1,20 @@ +//Problem 21.21: A 200 V, d.c. shunt-wound motor has an armature resistance of 0.4 ohm and at a certain load has an armature current of 30 A and runs at 1350 rev/min. If the load on the shaft of the motor is increased so that the armature current increases to 45 A, determine the speed of the motor, assuming the flux remains constant.
+
+//initializing the variables:
+Ia1 = 30; // in Amperes
+Ia2 = 45; // in Amperes
+Ra = 0.4; // in ohm
+n1 = 1350/60; // in Rev/sec
+V = 200; // in Volts
+
+//calculation:
+//The relationship E proportional to (Phi*n) applies to both generators and motors. For a motor,
+//E = V - (Ia*Ra)
+E1 = V - (Ia1*Ra)
+E2 = V - (Ia2*Ra)
+//The relationship, E1/E2 = Phi1*n1/Phi2*n2, applies to both generators and motors. Since the flux is constant, Phi1 = Phi2
+Phi2 = Phi1
+n2 = E2*Phi1*n1/(Phi2*E1)
+
+printf("\n\n Result \n\n")
+printf("\n the speed of the motor is %.2f rev/sec ",n2)
\ No newline at end of file diff --git a/608/CH21/EX21.22/21_22.sce b/608/CH21/EX21.22/21_22.sce new file mode 100755 index 000000000..e6fc14636 --- /dev/null +++ b/608/CH21/EX21.22/21_22.sce @@ -0,0 +1,26 @@ +//Problem 21.22: A 220 V, d.c. shunt-wound motor runs at 800 rev/min and the armature current is 30 A. The armature circuit resistance is 0.4 ohm. Determine (a) the maximum value of armature current if the flux is suddenly reduced by 10% and (b) the steady state value of the armature current at the new value of flux, assuming the shaft torque of the motor remains constant.
+
+//initializing the variables:
+Ia1 = 30; // in Amperes
+Ra = 0.4; // in ohm
+n = 800/60; // in Rev/sec
+V = 220; // in Volts
+x= 0.1;
+
+//calculation:
+//For a d.c. shunt-wound motor, E = V - (Ia*Ra),Hence initial generated e.m.f.,
+E1 = V - (Ia1*Ra)
+//The generated e.m.f. is also such that E proportional to (Phi*n) so at the instant the flux is reduced, the speed has not had time to change, and
+E = E1*(1-x)
+//Hence, the voltage drop due to the armature resistance is
+Vd = V - E
+//The instantaneous value of the current is
+Ia = Vd/Ra
+//T proportional to (Phi*Ia), since the torque is constant,
+//Phi1*Ia1 = Phi2*Ia2, The flux 8 is reduced by 10%, hence
+Phi2 = (1-x)*Phi1
+Ia2 = Phi1*Ia1/Phi2
+
+printf("\n\n Result \n\n")
+printf("\n (a)instantaneous value of the current %.0f A ",Ia)
+printf("\n (b)steady state value of armature current, %.2f A ",Ia2)
\ No newline at end of file diff --git a/608/CH21/EX21.23/21_23.sce b/608/CH21/EX21.23/21_23.sce new file mode 100755 index 000000000..ebf977d2b --- /dev/null +++ b/608/CH21/EX21.23/21_23.sce @@ -0,0 +1,24 @@ +//Problem 21.23: A series motor has an armature resistance of 0.2 ohm and a series field resistance of 0.3 ohm. It is connected to a 240 V supply and at a particular load runs at 24 rev/s when drawing 15 A from the supply. (a) Determine the generated e.m.f. at this load. (b) Calculate the speed of the motor when the load is changed such that the current is increased to 30 A. Assume that this causes a doubling of the flux.
+
+//initializing the variables:
+Ia1 = 15; // in Amperes
+Ia2 = 30; // in Amperes
+Rf = 0.3; // in ohms
+Ra = 0.2; // in ohm
+n1 = 24; // in Rev/sec
+V = 240; // in Volts
+x= 2;
+
+//calculation:
+//generated e.m.f., E, at initial load, is given by
+E1 = V - Ia1*(Ra + Rf)
+//When the current is increased to 30 A, the generated e.m.f. is given by:
+E2 = V - Ia2*(Ra + Rf)
+//E proportional to (Phi*n)
+//E1/E2 = Phi1*n1/Phi2*n2
+Phi2 = x*Phi1
+n2 = E2*Phi1*n1/(Phi2*E1)
+
+printf("\n\n Result \n\n")
+printf("\n (a)generated e.m.f., E is %.1f V ",E1)
+printf("\n (b)speed of motor, n2, %.2f A ",n2)
\ No newline at end of file diff --git a/608/CH21/EX21.24/21_24.sce b/608/CH21/EX21.24/21_24.sce new file mode 100755 index 000000000..e4a08d829 --- /dev/null +++ b/608/CH21/EX21.24/21_24.sce @@ -0,0 +1,20 @@ +//Problem 21.24: A 320 V shunt motor takes a total current of 80 A and runs at 1000 rev/min. If the iron, friction and windage losses amount to 1.5 kW, the shunt field resistance is 40 ohm and the armature resistance is 0.2 ohm , determine the overall efficiency of the motor.
+
+//initializing the variables:
+I = 80; // in Amperes
+C = 1500; // in Watt
+Rf = 40; // in ohms
+Ra = 0.2; // in ohm
+n = 1000/60; // in Rev/sec
+V = 320; // in Volts
+
+//calculation:
+//Field current, If
+If = V/Rf
+//Armature current Ia
+Ia = I - If
+//Efficiency =((V*I - Ia*Ia*Ra - If*V - C)/(V*I))*100%
+eff = ((V*I - (Ia*Ia*Ra) - (If*V) - C)/(V*I))*100 // in percent
+
+printf("\n\n Result \n\n")
+printf("\n efficiency is %.1f",eff)
\ No newline at end of file diff --git a/608/CH21/EX21.25/21_25.sce b/608/CH21/EX21.25/21_25.sce new file mode 100755 index 000000000..60e1dee74 --- /dev/null +++ b/608/CH21/EX21.25/21_25.sce @@ -0,0 +1,16 @@ +//Problem 21.25: A 250 V series motor draws a current of 40 A. The armature resistance is 0.15 ohm and the field resistance is 0.05ohm . Determine the maximum efficiency of the motor.
+
+//initializing the variables:
+I = 40; // in Amperes
+Rf = 0.05; // in ohms
+Ra = 0.15; // in ohm
+V = 250; // in Volts
+
+//calculation:
+//However for a series motor, If = 0 and the Ia*Ia*Ra loss needs to be I*I*(Ra + Rf)
+//For maximum efficiency I*I*(Ra + Rf) = C
+//Efficiency =((V*I -2*Ia*Ia*Ra)/(V*I))*100%
+eff = ((V*I - (2*I*I*(Ra + Rf)))/(V*I))*100 // in percent
+
+printf("\n\n Result \n\n")
+printf("\n efficiency is %.1f",eff)
\ No newline at end of file diff --git a/608/CH21/EX21.26/21_26.sce b/608/CH21/EX21.26/21_26.sce new file mode 100755 index 000000000..281faeca1 --- /dev/null +++ b/608/CH21/EX21.26/21_26.sce @@ -0,0 +1,16 @@ +//Problem 21.26: A 200 V d.c. motor develops a shaft torque of 15 Nm at 1200 rev/min. If the efficiency is 80%, determine the current supplied to the motor.
+
+//initializing the variables:
+T = 15; // in Nm
+n = 1200/60; // in rev/sec
+eff = 0.8;
+V = 200; // in Volts
+
+//calculation:
+//The efficiency of a motor = (output power/input power)*100 %
+//The output power of a motor is the power available to do work at its shaft and is given by Tw or T proportional to (2*pi*n) watts, where T is the torque in Nm and n is the speed of rotation in rev/s. The input power is the electrical power in watts supplied to the motor, i.e. VI watts.
+//Thus for a motor, efficiency =(T*2*pi* n/(V*I))%
+I = T*2*%pi*n/(V*eff)
+
+printf("\n\n Result \n\n")
+printf("\n current supplied, I is %.1f A",I)
\ No newline at end of file diff --git a/608/CH21/EX21.27/21_27.sce b/608/CH21/EX21.27/21_27.sce new file mode 100755 index 000000000..71764dacd --- /dev/null +++ b/608/CH21/EX21.27/21_27.sce @@ -0,0 +1,15 @@ +//Problem 21.27: A d.c. series motor drives a load at 30 rev/s and takes a current of 10 A when the supply voltage is 400 V. If the total resistance of the motor is 2 ohm and the iron, friction and windage losses amount to 300 W, determine the efficiency of the motor.
+
+//initializing the variables:
+R = 2; // in ohm
+n = 30; // in rev/sec
+I = 10; // in A
+C = 300; // in Watt
+V = 400; // in Volts
+
+//calculation:
+//Efficiency =((V*I - I*I*R - C)/(V*I))*100%
+eff = ((V*I - (I*I*R) - C)/(V*I))*100 // in percent
+
+printf("\n\n Result \n\n")
+printf("\n efficiency is %.1f percent",eff)
\ No newline at end of file diff --git a/608/CH21/EX21.28/21_28.sce b/608/CH21/EX21.28/21_28.sce new file mode 100755 index 000000000..d0bef77dc --- /dev/null +++ b/608/CH21/EX21.28/21_28.sce @@ -0,0 +1,27 @@ +//Problem 21.28: A 500 V shunt motor runs at its normal speed of 10 rev/s when the armature current is 120 A. The armature resistance is 0.2 ohm. (a) Determine the speed when the current is 60 A and a resistance of 0.5 ohm is connected in series with the armature, the shunt field remaining constant. (b) Determine the speed when the current is 60 A and the shunt field is reduced to 80% of its normal value by increasin resistance in the field circuit.
+
+//initializing the variables:
+Ia1 = 120; // in A
+Ia2 = 60; // in A
+Ra = 0.2; // in ohm
+n1 = 10; // in rev/sec
+R = 0.5; // in ohm
+x = 0.8;
+V = 500; // in Volts
+
+//calculation:
+//back e.m.f. at Ia1
+E1 = V - Ia1*Ra
+//at Ia2
+E2 = V - Ia2*(Ra + R)
+//E1/E2 = Phi1*n1/Phi2*n2
+Phi2 = Phi1
+n2 = Phi1*n1*E2/(Phi2*E1)
+//Back e.m.f. when Ia2
+E3 = V - Ia2*Ra
+Phi3 = x*Phi1
+n3 = Phi1*n1*E3/(Phi3*E1)
+
+printf("\n\n Result \n\n")
+printf("\n (a)speed n2 is %.2f rev/sec", n2)
+printf("\n (b)speed n3 is %.2f rev/sec", n3)
\ No newline at end of file diff --git a/608/CH21/EX21.29/21_29.sce b/608/CH21/EX21.29/21_29.sce new file mode 100755 index 000000000..ef2a53f2e --- /dev/null +++ b/608/CH21/EX21.29/21_29.sce @@ -0,0 +1,28 @@ +//Problem 21.29: On full-load a 300 V series motor takes 90 A and runs at 15 rev/s. The armature resistance is 0.1 ohm and the series winding resistance is 50 mohm. Determine the speed when developing full load torque but with a 0.2 ohm diverter in parallel with the field winding. (Assume that the flux is proportional to the field current.)
+
+//initializing the variables:
+Ia1 = 90; // in Amperes
+Ra = 0.1; // in ohm
+Rse = 0.05; // in ohm
+Rd = 0.2; // in Ohm
+n1 = 15; // in rev/sec
+V = 300; // in Volts
+
+//calculation:
+//e.m.f. E1
+E1 = V - Ia1*(Ra + Rse)
+//With the Rd diverter in parallel with Rse
+//equivalent resistance, Re
+Re = Rd*Rse/(Rd+Rse)
+//Torque, T proprtional to Ia*Phi and for full load torque, Ia1*Phi1 = Ia2*Phi2
+//Since flux is proportional to field current Phi1 proportional to Ta1 and Phi2 Proportional to I1
+I1 = (Ia1*Ia1*0.8)^0.5
+//By current division, current I1
+Ia2 = I1/(Rd/(Rd + Rse))
+//Hence e.m.f. E2
+E2 = V - Ia2*(Ra + Re)
+//E1/E2 = Phi1*n1/Phi2*n2
+n2 = E2*Ia1*n1/(I1*E1)
+
+printf("\n\n Result \n\n")
+printf("\n speed n2 is %.2f rev/sec", n2)
\ No newline at end of file diff --git a/608/CH21/EX21.30/21_30.sce b/608/CH21/EX21.30/21_30.sce new file mode 100755 index 000000000..a7007d5c6 --- /dev/null +++ b/608/CH21/EX21.30/21_30.sce @@ -0,0 +1,21 @@ +//Problem 21.30: A series motor runs at 800 rev/min when the voltag is 400 V and the current is 25 A. The armature resistance is 0.4 ohm and the series field resistance is 0.2 ohm. Determine the resistance to be connected in series to reduce the speed to 600 rev/min with the same current.
+
+//initializing the variables:
+Ia1 = 25; // in Amperes
+Ra = 0.4; // in ohm
+Rse = 0.2; // in ohm
+n1 = 800/60; // in rev/sec
+n2 = 600/60; // in rev/sec
+V = 400; // in Volts
+
+//calculation:
+//e.m.f. E1
+E1 = V - Ia1*(Ra + Rse)
+//At n2, since the current is unchanged, the flux is unchanged.
+//E1/E2 = n1/n2
+E2 = E1*n2/n1
+//and E2 = V - Ia1(Ra + Rse + R)
+R = (V - E2)/Ia1 - Ra - Rse
+
+printf("\n\n Result \n\n")
+printf("\n Resistance is %.2f ohm", R)
\ No newline at end of file diff --git a/608/CH22/EX22.01/22_01.sce b/608/CH22/EX22.01/22_01.sce new file mode 100755 index 000000000..916af97c2 --- /dev/null +++ b/608/CH22/EX22.01/22_01.sce @@ -0,0 +1,13 @@ +//Problem 22.01: A three-phase two-pole induction motor is connected to a 50 Hz supply. Determine the synchronous speed of the motor in rev/min.
+
+//initializing the variables:
+f = 50; // in Hz
+p = 2/2; // number of pairs of poles
+
+//calculation:
+//ns is the synchronous speed, f is the frequency in hertz of the supply to the stator and p is the number of pairs of poles.
+ns = f/p
+nsrpm = ns*60
+
+printf("\n\n Result \n\n")
+printf("\nsynchronous speed of the motor is %.0f rev/min", nsrpm)
\ No newline at end of file diff --git a/608/CH22/EX22.02/22_02.sce b/608/CH22/EX22.02/22_02.sce new file mode 100755 index 000000000..9a8679c44 --- /dev/null +++ b/608/CH22/EX22.02/22_02.sce @@ -0,0 +1,13 @@ +//Problem 22.02: A stator winding supplied from a three-phase 60 Hz system is required to produce a magnetic flux rotating at 900 rev/min. Determine the number of poles.
+
+//initializing the variables:
+f = 60; // in Hz
+ns = 900/60; // in rev/sec
+
+//calculation:
+//ns is the synchronous speed, f is the frequency in hertz of the supply to the stator and p is the number of pairs of poles.
+p = f/ns
+np = p*2
+
+printf("\n\n Result \n\n")
+printf("\nnumber of poles is %.0f", np)
\ No newline at end of file diff --git a/608/CH22/EX22.03/22_03.sce b/608/CH22/EX22.03/22_03.sce new file mode 100755 index 000000000..9f94cae36 --- /dev/null +++ b/608/CH22/EX22.03/22_03.sce @@ -0,0 +1,12 @@ +//Problem 22.03: A three-phase 2-pole motor is to have a synchronous speed of 6000 rev/min. Calculate the frequency of the supply voltage.
+
+//initializing the variables:
+p = 2/2; // number of pairs of poles
+ns = 6000/60; // in rev/sec
+
+//calculation:
+//ns is the synchronous speed, f is the frequency in hertz of the supply to the stator and p is the number of pairs of poles.
+f = p*ns
+
+printf("\n\n Result \n\n")
+printf("\nfrequency is %.0f Hz",f)
\ No newline at end of file diff --git a/608/CH22/EX22.04/22_04.sce b/608/CH22/EX22.04/22_04.sce new file mode 100755 index 000000000..c9bc45ad9 --- /dev/null +++ b/608/CH22/EX22.04/22_04.sce @@ -0,0 +1,16 @@ +//Problem 22.04: The stator of a 3-phase, 4-pole induction motor is connected to a 50 Hz supply. The rotor runs at 1455 rev/min at full load. Determine (a) the synchronous speed and (b) the slip at full load.
+
+//initializing the variables:
+p = 4/2; // number of pairs of poles
+f = 50; // in Hz
+nr = 1455/60; // in rev/sec
+
+//calculation:
+//ns is the synchronous speed, f is the frequency in hertz of the supply to the stator and p is the number of pairs of poles.
+ns = f/p
+//The slip, s
+s = ((ns - nr)/ns)*100 // in percent
+
+printf("\n\n Result \n\n")
+printf("\n(a) synchronous speed is %.0f rev/sec",ns)
+printf("\n(b) slip is %.0f percent",s)
\ No newline at end of file diff --git a/608/CH22/EX22.05/22_05.sce b/608/CH22/EX22.05/22_05.sce new file mode 100755 index 000000000..c53f41c75 --- /dev/null +++ b/608/CH22/EX22.05/22_05.sce @@ -0,0 +1,19 @@ +//Problem 22.05: A 3-phase, 60 Hz induction motor has 2 poles. If the slip is 2% at a certain load, determine (a) the synchronous speed, (b) the speed of the rotor and (c) the frequency of the induced e.m.f.’s in the rotor.
+
+//initializing the variables:
+p = 2/2; // number of pairs of poles
+f = 60; // in Hz
+s = 0.02; // slip
+
+//calculation:
+//ns is the synchronous speed, f is the frequency in hertz of the supply to the stator and p is the number of pairs of poles.
+ns = f/p
+//The the rotor runs at
+nr = ns*(1 - s)
+//frequency of the e.m.f.’s induced in the rotor bars is
+fr = ns - nr
+
+printf("\n\n Result \n\n")
+printf("\n(a) synchronous speed is %.0f rev/sec",ns)
+printf("\n(b) rotor speed is %.1f rev/sec",nr)
+printf("\n(c) frequency of the e.m.f.’s induced in the rotor bars is is %.1f Hz",fr)
\ No newline at end of file diff --git a/608/CH22/EX22.06/22_06.sce b/608/CH22/EX22.06/22_06.sce new file mode 100755 index 000000000..c91a882bf --- /dev/null +++ b/608/CH22/EX22.06/22_06.sce @@ -0,0 +1,13 @@ +//Problem 22.06: A three-phase induction motor is supplied from a 50 Hz supply and runs at 1200 rev/min when the slip is 4%. Determine the synchronous speed.
+
+//initializing the variables:
+f = 50; // in Hz
+nr = 1200/60; // in rev/min
+s = 0.04; // slip
+
+//calculation:
+//the synchronous speed.
+ns = nr/(1 - s)
+nsrpm = ns*60
+printf("\n\n Result \n\n")
+printf("\n synchronous speed is %.0f rev/min",nsrpm)
\ No newline at end of file diff --git a/608/CH22/EX22.07/22_07.sce b/608/CH22/EX22.07/22_07.sce new file mode 100755 index 000000000..379b63e03 --- /dev/null +++ b/608/CH22/EX22.07/22_07.sce @@ -0,0 +1,19 @@ +//Problem 22.07: The frequency of the supply to the stator of an 8- pole induction motor is 50 Hz and the rotor frequency is 3 Hz. Determine (a) the slip, and (b) the rotor speed.
+
+//initializing the variables:
+p = 8/2; // number of pairs of poles
+f = 50; // in Hz
+fr = 3; // in Hz
+
+//calculation:
+//ns is the synchronous speed, f is the frequency in hertz of the supply to the stator and p is the number of pairs of poles.
+ns = f/p
+//fr = s*f
+s = (fr/f)
+//the rotor speed.
+nr = ns*(1 - s)
+nrrpm = nr*60
+
+printf("\n\n Result \n\n")
+printf("\n(a) slip is %.0f percent",s*100)
+printf("\n (b) rotor speed is %.0f rev/min",nrrpm)
\ No newline at end of file diff --git a/608/CH22/EX22.08/22_08.sce b/608/CH22/EX22.08/22_08.sce new file mode 100755 index 000000000..b6a858ad7 --- /dev/null +++ b/608/CH22/EX22.08/22_08.sce @@ -0,0 +1,25 @@ +//Problem 22.08: The power supplied to a three-phase induction motor is 32 kW and the stator losses are 1200 W. If the slip is 5%, determine (a) the rotor copper loss, (b) the total mechanical power developed by the rotor, (c) the output power of the motor if friction and windage losses are 750 W, and (d) the efficiency of the motor, neglecting rotor iron loss.
+
+//initializing the variables:
+Psi = 32000; // in Watts
+Psl = 1200; // in Watts
+s = 0.05; // slip
+Pfl = 750; // in Watts
+
+//calculation:
+//Input power to rotor = stator input power - stator losses
+Pi = Psi - Psl
+//slip = rotor copper loss/rotor input
+Pl = s*Pi
+//Total mechanical power developed by the rotor = rotor input power - rotor losses
+Pr = Pi - Pl
+//Output power of motor = power developed by the rotor - friction and windage losses
+Po = Pr - Pfl
+//Efficiency of induction motor = (output power/input power)*100
+eff = (Po/Psi)*100 // in percent
+
+printf("\n\n Result \n\n")
+printf("\n(a) rotor copper loss is %.0f Watt",Pl)
+printf("\n(b) Total mechanical power developed by the rotor is %.0f W",Pr)
+printf("\n(c) Output power of motor is %.0f Watt",Po)
+printf("\n(d) efficiency of induction motor is %.2f percent",eff)
\ No newline at end of file diff --git a/608/CH22/EX22.09/22_09.sce b/608/CH22/EX22.09/22_09.sce new file mode 100755 index 000000000..715fdcc5f --- /dev/null +++ b/608/CH22/EX22.09/22_09.sce @@ -0,0 +1,26 @@ +//Problem 22.09: The speed of the induction motor of Problem 22.08 is reduced to 35% of its synchronous speed by using external rotor resistance. If the torque and stator losses are unchanged, determine (a) the rotor copper loss, and (b) the efficiency of the motor.
+
+//initializing the variables:
+Psi = 32000; // in Watts
+Psl = 1200; // in Watts
+Pfl = 750; // in Watts
+x = 0.35;
+
+//calculation:
+nr = x*ns
+//The slip, s
+s = ((ns - nr)/ns)
+//Input power to rotor = stator input power - stator losses
+Pi = Psi - Psl
+//slip = rotor copper loss/rotor input
+Pl = s*Pi
+//Total mechanical power developed by the rotor = rotor input power - rotor losses
+Pr = Pi - Pl
+//Output power of motor = power developed by the rotor - friction and windage losses
+Po = Pr - Pfl
+//Efficiency of induction motor = (output power/input power)*100
+eff = (Po/Psi)*100 // in percent
+
+printf("\n\n Result \n\n")
+printf("\n(a) rotor copper loss is %.0f Watt",Pl)
+printf("\n(b) efficiency of induction motor is %.2f percent",eff)
\ No newline at end of file diff --git a/608/CH22/EX22.10/22_10.sce b/608/CH22/EX22.10/22_10.sce new file mode 100755 index 000000000..f6c34894e --- /dev/null +++ b/608/CH22/EX22.10/22_10.sce @@ -0,0 +1,47 @@ +//Problem 22.10: A 415 V, three-phase, 50 Hz, 4 pole, star-connected induction motor runs at 24 rev/s on full load. The rotor resistance and reactance per phase are 0.35 ohm and 3.5 ohm respectively, and the effective rotor-stator turns ratio is 0.85:1. Calculate (a) the synchronous speed, (b) the slip, (c) the full load torque, (d) the power output if mechanical losses amount to 770 W, (e) the maximum torque, (f) the speed at which maximum torque occurs and (g) the starting torque.
+
+//initializing the variables:
+V = 415; // in Volts
+f = 50 ; // in Hz
+nr = 24; // in rev/sec
+p = 4/2; // no. of pole pairs
+R2 = 0.35; // in Ohms
+X2 = 3.5; // in Ohms
+tr = 0.85; // turn ratio N2/N1
+Pl = 770; // in Watt
+m = 3; // no. of phases
+
+//calculation:
+//ns is the synchronous speed, f is the frequency in hertz of the supply to the stator and p is the number of pairs of poles.
+ns = f/p
+//The slip, s
+s = ((ns - nr)/ns)*100 // in percent
+//Phase voltage, E1 = V/(3^0.5)
+E1 = V/(3^0.5)
+//Full load torque
+T = [m*(tr^2)/(2*%pi*ns)]*[(s/100)*E1*E1*R2/(R2*R2 + (X2*(s/100))^2)]
+//Output power, including friction losses
+Pm = 2*%pi*nr*T
+//power output
+Po = Pm - Pl
+//Maximum torque occurs when R2 = Xr = 0.35 ohm
+//Slip
+sm = R2/X2
+//maximum torque, Tm
+Tm = [m*(tr^2)/(2*%pi*ns)]*[sm*E1*E1*R2/(R2*R2 + (X2*sm)^2)]
+//speed at which maximum torque occurs
+nrm = ns*(1 - sm)
+nrmrpm = nrm*60
+//At the start, i.e., at standstill, slip, s=1
+ss = 1
+//starting torque
+Ts = [m*(tr^2)/(2*%pi*ns)]*[ss*E1*E1*R2/(R2*R2 + (X2*ss)^2)]
+
+printf("\n\n Result \n\n")
+printf("\n(a)Synchronous speed is %.0f rev/sec",ns)
+printf("\n(b)Slip is %.0f percent",s)
+printf("\n(c)Full load torque is %.2f Nm",T)
+printf("\n(d)power output is %.2E W",Po)
+printf("\n(e)maximum torque is %.2f Nm",Tm)
+printf("\n(f)speed at which maximum torque occurs is %.0frev/min",nrmrpm)
+printf("\n(g)starting torque is %.2f Nm",Ts)
\ No newline at end of file diff --git a/608/CH22/EX22.11/22_11.sce b/608/CH22/EX22.11/22_11.sce new file mode 100755 index 000000000..346180c82 --- /dev/null +++ b/608/CH22/EX22.11/22_11.sce @@ -0,0 +1,31 @@ +//Problem 22.11: Determine for the induction motor in problem 22.10 at full load, (a) the rotor current, (b) the rotor copper loss, and (c) the starting current.
+
+//initializing the variables:
+V = 415; // in Volts
+f = 50 ; // in Hz
+nr = 24; // in rev/sec
+p = 4/2; // no. of pole pairs
+R2 = 0.35; // in Ohms
+X2 = 3.5; // in Ohms
+tr = 0.85; // turn ratio N2/N1
+m = 3; // no. of phases
+
+//calculation:
+//ns is the synchronous speed, f is the frequency in hertz of the supply to the stator and p is the number of pairs of poles.
+ns = f/p
+//The slip, s
+s = ((ns - nr)/ns)*100 // in percent
+//Phase voltage, E1 = V/(3^0.5)
+E1 = V/(3^0.5)
+//rotor current,
+Ir = (s/100)*E1*tr/((R2^2 + (X2*(s/100))^2)^0.5)
+//Rotor copper loss
+Pcl = m*R2*(Ir^2)
+//starting current,
+ss =1
+I2 = ss*tr*E1/((R2^2 + (X2*ss)^2)^0.5)
+
+printf("\n\n Result \n\n")
+printf("\n(a)rotor current is %.2f A",Ir)
+printf("\n(b)Total copper loss is %.2f W",Pcl)
+printf("\n(c)starting current is %.2f A",I2)
\ No newline at end of file diff --git a/608/CH22/EX22.12/22_12.sce b/608/CH22/EX22.12/22_12.sce new file mode 100755 index 000000000..f1e9dffc9 --- /dev/null +++ b/608/CH22/EX22.12/22_12.sce @@ -0,0 +1,24 @@ +//Problem 22.12: For the induction motor in problems 22.10 and 22.11, if the stator losses are 650 W, determine (a) the power input at full load, (b) the efficiency of the motor at full load and (c) the current taken from the supply at full load, if the motor runs at a power factor of 0.87 lagging.
+
+//initializing the variables:
+V = 415; // in Volts
+Psl = 650; // in Watt
+pf = 0.87; // power factor
+
+//calculation:
+Pm = 11770; // watts from part (d), Problem 22.10
+Pcl = 490.35; // watts, Rotor copper loss, from part (b), Problem 22.11
+//Stator input power
+P1 = Pm + Pcl + Psl
+Po = 11000 // watts, Net power output, from part (d), Problem 22.10
+//efficiency = (output/input) *100
+eff = (Po/P1)*100 // in percent
+//Power input, P1 = (3^0.5)*VL*IL*cos(phi)
+// pf = cos(phi)
+//supply current, IL
+I = P1/((3^0.5)*V*pf)
+
+printf("\n\n Result \n\n")
+printf("\n(aStator input power is %.2E W",P1)
+printf("\n(b)efficiency is %.2f percent",eff)
+printf("\n(c)supply current is %.2f A",I)
\ No newline at end of file diff --git a/608/CH22/EX22.13/22_13.sce b/608/CH22/EX22.13/22_13.sce new file mode 100755 index 000000000..0020438dc --- /dev/null +++ b/608/CH22/EX22.13/22_13.sce @@ -0,0 +1,19 @@ +//Problem 22.13: For the induction motor of Problems 22.10 to 22.12, determine the resistance of the rotor winding required for maximum starting torque.
+
+//initializing the variables:
+V = 415; // in Volts
+f = 50 ; // in Hz
+nr = 24; // in rev/sec
+p = 4/2; // no. of pole pairs
+R2 = 0.35; // in Ohms
+X2 = 3.5; // in Ohms
+
+//calculation:
+//At the moment of starting, slip,
+s = 1
+//Maximum torque occurs when rotor reactance equals rotor resistance
+//for maximum torque
+R2 = s*X2
+
+printf("\n\n Result \n\n")
+printf("\nresistance of the rotor is %.1f Ohm",R2)
\ No newline at end of file diff --git a/608/CH23/EX23.01/23_01.sce b/608/CH23/EX23.01/23_01.sce new file mode 100755 index 000000000..0bf483fa0 --- /dev/null +++ b/608/CH23/EX23.01/23_01.sce @@ -0,0 +1,14 @@ +//Problem 23.01: In an electrical circuit the total impedance ZT is given by ZT = (Z1*Z2/(Z1 + Z2))+ Z3. Determine ZT in (a + jb) form, correct to two decimal places, when Z1 = 5 - j3, Z2 = 4 - i7 and Z3 = 3.9 - i6.7.
+
+//initializing the variables:
+Z1 = 5 - 3*%i;
+Z2 = 4 + 7*%i;
+Z3 = 3.9 - 6.7*%i;
+
+//calculation:
+ZT = (Z1*Z2/(Z1 + Z2))+ Z3
+y = imag(ZT)
+x = real(ZT)
+
+printf("\n\n Result \n\n")
+printf("\n ZT is %.2f + (%.2f)i", x,y)
\ No newline at end of file diff --git a/608/CH23/EX23.02/23_02.sce b/608/CH23/EX23.02/23_02.sce new file mode 100755 index 000000000..d0dfaac72 --- /dev/null +++ b/608/CH23/EX23.02/23_02.sce @@ -0,0 +1,26 @@ +//Problem 23.02: Given Z1 = 3 + i4, Z2 = 2 - i5 determine in cartesian form correct to three decimal places:
+//(a)1/Z1, (b)1/Z2, (c)1/Z1 + 1/Z2, (d)1/(1/Z1 + 1/Z2)
+
+//initializing the variables:
+Z1 = 3 + 4*%i;
+Z2 = 2 - 5*%i;
+
+//calculation:
+za = 1/Z1
+zb = 1/Z2
+zc = za + zb
+zd = 1/zc
+zax = real(za)
+zay = imag(za)
+zbx = real(zb)
+zby = imag(zb)
+zcx = real(zc)
+zcy = imag(zc)
+zdx = real(zd)
+zdy = imag(zd)
+
+printf("\n\n Result \n\n")
+printf("\n (a)1/Z1 is %.3f + (%.3f)i", zax,zay)
+printf("\n (b)1/Z2 is %.3f + (%.3f)i", zbx,zby)
+printf("\n (c)1/Z1 + 1/Z2 is %.3f + (%.3f)i", zcx,zcy)
+printf("\n (d)1/(1/Z1 + 1/Z2) is %.3f + (%.3f)i", zdx,zdy)
\ No newline at end of file diff --git a/608/CH23/EX23.03/23_03.sce b/608/CH23/EX23.03/23_03.sce new file mode 100755 index 000000000..b2e71a6e1 --- /dev/null +++ b/608/CH23/EX23.03/23_03.sce @@ -0,0 +1,25 @@ +//Problem 23.03: Solve the following complex equations:
+//(a) 3(a + ib) = 9-i2
+//(b) (2+i)(-2+i) = x+iy
+//(c) (a-i(2b))+(b-i3a) = 5+i2
+
+//initializing the variables:
+Z1 = 9 - 2*%i;
+Z2 = 2 + 1*%i;
+Z3 = -2 + 1*%i;
+Z4 = 5 + 2*%i;
+
+//calculation:
+za = Z1/3
+zb = Z2*Z3
+zca = (2*real(Z4) + imag(Z4))/-1
+zcb = real(Z4) - zca
+zaa = real(za)
+zab = imag(za)
+zbx = real(zb)
+zby = imag(zb)
+
+printf("\n\n Result \n\n")
+printf("\n (a)a and b are %.0f and %.2f resp.", zaa,zab)
+printf("\n (b)x and y are %.0f and %.0f resp.", zbx,zby)
+printf("\n (c)a and b are %.0f and %.0f resp.", zca,zcb)
\ No newline at end of file diff --git a/608/CH23/EX23.05/23_05.sce b/608/CH23/EX23.05/23_05.sce new file mode 100755 index 000000000..d2a7ea51a --- /dev/null +++ b/608/CH23/EX23.05/23_05.sce @@ -0,0 +1,13 @@ +//Problem 23.05: Convert (5,-132°) into a + ib form correct to four significant figures.
+
+//initializing the variables:
+r = 5; // magnitude
+theta = -132; // in degree
+
+//calculation:
+x = r*sin(theta*%pi/180)
+y = r*cos(theta*%pi/180)
+z = x+%i*y
+
+printf("\n\n Result \n\n")
+printf("\n Z is %.3f + (%.3f)i", x,y)
\ No newline at end of file diff --git a/608/CH23/EX23.06/23_06.sce b/608/CH23/EX23.06/23_06.sce new file mode 100755 index 000000000..5f5e3a34b --- /dev/null +++ b/608/CH23/EX23.06/23_06.sce @@ -0,0 +1,23 @@ +//Problem 23.06: Two impedances in an electrical network are given by Z1 = (4.7,35°) and Z2 = (7.3, -48)°. Determine in polar form the total impedance ZT given that ZT = Z1*Z2/(Z1 + Z2)
+
+//initializing the variables:
+r1 = 4.7; // magnitude
+theta1 = 35; // in degree
+r2 = 7.3; // magnitude
+theta2 = -48; // in degree
+
+//calculation:
+x1 = r1*cos(theta1*%pi/180)
+y1 = r1*sin(theta1*%pi/180)
+z1 = x1+%i*y1
+x2 = r2*cos(theta2*%pi/180)
+y2 = r2*sin(theta2*%pi/180)
+z2 = x2+%i*y2
+z3 = z1*z2/(z1 + z2)
+x3 = real(z3)
+y3 = imag(z3)
+r3 = (x3^2 + y3^2)^0.5
+theta3 = atan(y3/x3)*180/%pi
+
+printf("\n\n Result \n\n")
+printf("\n ZT is (%.2f/_%.2f°)", r3,theta3)
\ No newline at end of file diff --git a/608/CH23/EX23.07/23_07.sce b/608/CH23/EX23.07/23_07.sce new file mode 100755 index 000000000..3788919c9 --- /dev/null +++ b/608/CH23/EX23.07/23_07.sce @@ -0,0 +1,20 @@ +//Problem 23.07: Determine (-2 + i3)^5 in polar and in cartesian form.
+
+//initializing the variables:
+z = -2 + %i*3;
+
+//calculation:
+zc = z^5
+x = real(zc)
+y = imag(zc)
+r = (x^2 + y^2)^0.5
+theta = atan(y/x)*180/%pi
+if ((x<0)&(y<0)) then
+ theta = theta -180;
+elseif ((x<0)&(y>0)) then
+ theta = theta +180;
+end
+
+printf("\n\n Result \n\n")
+printf("\n Z is %.0f + (%.0f)i", x,y)
+printf("\n ZT is (%.1f/_%.2f°)", r,theta)
\ No newline at end of file diff --git a/608/CH23/EX23.08/23_08.sce b/608/CH23/EX23.08/23_08.sce new file mode 100755 index 000000000..505867360 --- /dev/null +++ b/608/CH23/EX23.08/23_08.sce @@ -0,0 +1,34 @@ +//Problem 23.08: Determine the two square roots of the complex number (12 + i5) in cartesian and polar form, correct to three significant figures. Show the roots on an Argand diagram.
+
+//initializing the variables:
+z = 12 + %i*5;
+
+//calculation:
+x = real(z)
+y = imag(z)
+r = (x^2 + y^2)^0.5
+theta1 = atan(y/x)*180/%pi
+if ((x<0)&(y<0)) then
+ theta1 = theta1 -180;
+elseif ((x<0)&(y>0)) then
+ theta1 = theta1 +180;
+end
+theta2 = theta1 + 360
+rtheta1 = theta1/2
+rtheta2 = theta2/2
+if (rtheta2 > 180) then
+ rtheta2 = rtheta2 -360;
+elseif ((x<0)&(y>0)) then
+ rtheta2 = rtheta2 +360;
+end
+rr = r^0.5
+x1 = rr*cos(rtheta1*%pi/180)
+y1 = rr*sin(rtheta1*%pi/180)
+z1 = x1 + %i*y1
+x2 = rr*cos(rtheta2*%pi/180)
+y2 = rr*sin(rtheta2*%pi/180)
+z2 = x2 + %i*y2
+
+printf("\n\n Result \n\n")
+printf("\n two roots are (%.2f + (%.2f)i) and (%.2f + (%.2f)i)", x1,y1,x2,y2)
+printf("\n two roots are (%.1f/_%.2f°) and (%.1f/_%.2f°)", rr,rtheta1, rr,rtheta2)
\ No newline at end of file diff --git a/608/CH24/EX24.01/24_01.sce b/608/CH24/EX24.01/24_01.sce new file mode 100755 index 000000000..da0743913 --- /dev/null +++ b/608/CH24/EX24.01/24_01.sce @@ -0,0 +1,34 @@ +//Problem 24.01: Determine the values of the resistance and the series-connected inductance or capacitance for each of the following impedances:(a)(12 + i5)ohm (b)-i40 ohm (c)30/_60° ohm (d)2.20 x 10^6 /_-30° ohm. Assume for each a frequency of 50 Hz.
+
+//initializing the variables:
+z1 = 12 + %i*5;
+z2 = -40*%i;
+r3 = 30;
+theta3 = 60; // in degrees
+r4 = 2.20E6;
+theta4 = -30; // in degrees
+f = 50; // in Hz
+
+//calculation:
+//for an R–L series circuit, impedance
+// Z = R + iXL
+Ra = real(z1)
+XLa = imag(z1)
+La = XLa/(2*%pi*f)
+//for a purely capacitive circuit, impedance Z = -iXc
+Xcb = abs(imag(z2))
+Cb = 1/(2*%pi*f*Xcb)
+z3 = r3*cos(theta3*%pi/180) + %i*(r3*sin(theta3*%pi/180))
+Rc = real(z3)
+XLc = imag(z3)
+Lc = XLc/(2*%pi*f)
+z4 = r4*cos(theta4*%pi/180) + %i*(r4*sin(theta4*%pi/180))
+Rd = real(z4)
+Xcd = abs(imag(z4))
+Cd = 1/(2*%pi*f*Xcd)
+
+printf("\n\n Result \n\n")
+printf("\n (a)an impedance (12 + i5)ohm represents a resistance of %.0f ohm in series with an inductance of %.2E", Ra,La)
+printf("\n (b)an impedance -i40 ohm represents a pure capacitor of capacitance %.2E", Cb)
+printf("\n (c)an impedance 30/_60° ohm represents a resistance of %.0f ohm in series with an inductance of %.2E", Rc,Lc)
+printf("\n (d)an impedance 2.20 x 10^6 /_-30° ohm represents a resistance of %.2E ohm in series with a capacitor of capacitance %.2E",Rd, Cd)
\ No newline at end of file diff --git a/608/CH24/EX24.02/24_02.sce b/608/CH24/EX24.02/24_02.sce new file mode 100755 index 000000000..cd3f7cde2 --- /dev/null +++ b/608/CH24/EX24.02/24_02.sce @@ -0,0 +1,27 @@ +//Problem 24.02: Determine, in polar and rectangular forms, the current flowing in an inductor of negligible resistance and inductance 159.2 mH when it is connected to a 250 V, 50 Hz supply.
+
+//initializing the variables:
+L = 0.1592 ; // in Henry
+V = 250; // in Volts
+f = 50; // in Hz
+R = 0; // in ohms
+
+//calculation:
+//for an R–L series circuit, impedance
+// Z = R + iXL
+XL = 2*%pi*f*L
+Z = R + %i*XL
+I = V/Z
+x = real(I)
+y = imag(I)
+r = (x^2 + y^2)^0.5
+if ((x==0)&(y<0)) then
+ theta = -90
+elseif ((x==0)&(y>0)) then
+ theta = +90
+else
+ theta = atan(y/x)*180/%pi
+end
+
+printf("\n\n Result \n\n")
+printf("\n current is (%.0f/_%.0f°) A", r, theta)
\ No newline at end of file diff --git a/608/CH24/EX24.03/24_03.sce b/608/CH24/EX24.03/24_03.sce new file mode 100755 index 000000000..aeed42f53 --- /dev/null +++ b/608/CH24/EX24.03/24_03.sce @@ -0,0 +1,20 @@ +//Problem 24.03: A 3 μF capacitor is connected to a supply of frequency 1 kHz and a current of 2.83/_90° A flows. Determine the value of the supply p.d.
+
+//initializing the variables:
+C = 3E-6 ; // in farad
+f = 1000; // in Hz
+ri = 2.83;
+thetai = 90; // in degrees
+
+//calculation:
+//Capacitive reactance Xc
+Xc = 1/(2*%pi*f*C)
+// circuit impedance Z
+Z = -1*%i*Xc
+I = ri*cos(thetai*%pi/180) + %i*ri*sin(thetai*%pi/180)
+V = I*Z
+x = real(V)
+y = imag(V)
+
+printf("\n\n Result \n\n")
+printf("\n supply p.d. is %.0f + (%.0f) V", x,y)
\ No newline at end of file diff --git a/608/CH24/EX24.04/24_04.sce b/608/CH24/EX24.04/24_04.sce new file mode 100755 index 000000000..f59dd5524 --- /dev/null +++ b/608/CH24/EX24.04/24_04.sce @@ -0,0 +1,34 @@ +//Problem 24.04: The impedance of an electrical circuit is (30 - i50) ohms.Determine (a) the resistance, (b) the capacitance, (c) the modulus of the impedance, and (d) the current flowing and its phase angle, when the circuit is connected to a 240 V, 50 Hz supply.
+
+//initializing the variables:
+V = 240; // in Volts
+f = 50; // in Hz
+Z = 30 - %i*50;
+
+//calculation:
+//Since impedance Z = 30 - i50,
+//resistance
+R = real(Z)
+//capacitive reactance
+Xc = abs(imag(Z))
+//capacitance
+C = 1/(2*%pi*f*Xc)
+//modulus of impedance
+modZ = (R^2 + Xc^2)^0.5
+I = V/Z
+x = real(I)
+y = imag(I)
+r = (x^2 + y^2)^0.5
+if ((x==0)&(y<0)) then
+ theta = -90
+elseif ((x==0)&(y>0)) then
+ theta = +90
+else
+ theta = atan(y/x)*180/%pi
+end
+
+printf("\n\n Result \n\n")
+printf("\n (a)resistance is %.0f ohm", R)
+printf("\n (b)capacitance is %.2E Farad", C)
+printf("\n (c)modulus of impedance is %.2f ohm", modZ)
+printf("\n (d)current flowing and its phase angle is (%.2f/_%.2f°) A", r, theta)
\ No newline at end of file diff --git a/608/CH24/EX24.05/24_05.sce b/608/CH24/EX24.05/24_05.sce new file mode 100755 index 000000000..34b8162d8 --- /dev/null +++ b/608/CH24/EX24.05/24_05.sce @@ -0,0 +1,43 @@ +//Problem 24.05: A 200 V, 50 Hz supply is connected across a coil of negligible resistance and inductance 0.15 H connected in series with a 32 ohm resistor. Determine (a) the impedance of the circuit, (b) the current and circuit phase angle, (c) the p.d. across the 32 ohm resistor, and (d) the p.d. across the coil.
+
+//initializing the variables:
+V = 200; // in Volts
+f = 50; // in Hz
+R = 32; // in ohms
+L = 0.15; // in Henry
+
+//calculation:
+//Inductive reactance XL
+XL = 2*%pi*f*L
+//impedance, Z
+Z = R + %i*XL
+//Current I
+I = V/Z
+xi = real(I)
+yi = imag(I)
+ri = (xi^2 + yi^2)^0.5
+if ((xi==0)&(yi<0)) then
+ thetai = -90
+elseif ((xi==0)&(yi>0)) then
+ thetai = +90
+else
+ thetai = atan(yi/xi)*180/%pi
+end
+//P.d. across the resistor
+VR = I*R
+xr = real(VR)
+yr = imag(VR)
+rr = (xr^2 + yr^2)^0.5
+thetar = atan(yr/xr)*180/%pi
+//P.d. across the coil, VL
+VL = I*%i*XL
+xl = real(VL)
+yl = imag(VL)
+rl = (xl^2 + yl^2)^0.5
+thetal = atan(yl/xl)*180/%pi
+
+printf("\n\n Result \n\n")
+printf("\n (a)impedance is %.0f + (%.1f)i ohm", real(Z), imag(Z))
+printf("\n (b)current flowing and its phase angle is (%.2f/_%.2f°) A", ri, thetai)
+printf("\n (c)P.d. across the resistor is (%.2f/_%.2f°) V", rr,thetar)
+printf("\n (d)P.d. across the coil, VL is (%.2f/_%.2f°) V", rl, thetal)
\ No newline at end of file diff --git a/608/CH24/EX24.06/24_06.sce b/608/CH24/EX24.06/24_06.sce new file mode 100755 index 000000000..0218f0b51 --- /dev/null +++ b/608/CH24/EX24.06/24_06.sce @@ -0,0 +1,21 @@ +//Problem 24.06: Determine the value of impedance if a current of (7+i16)A flows in a circuit when the supply voltage is (120+i200)V. If the frequency of the supply is 5 MHz, determine the value of the components forming the series circuit.
+
+//initializing the variables:
+V = 120 + %i*200; // in Volts
+f = 5E6; // in Hz
+I = 7 + %i*16; // in amperes
+
+//calculation:
+//impedance, Z
+Z = V/I
+R = real(Z)
+X = imag(Z)
+if ((R>0)&(X<0)) then
+ printf("\n\n Result \n\n")
+ C = -1/(2*%pi*f*X)
+ printf("\n The series circuit thus consists of a resistor of resistance %.2f ohm and a capacitor of capacitance %.2E Farad\n",R,C)
+elseif ((R>0)&(X>0)) then
+ printf("\n\n Result \n\n")
+ L = 2*%pi*f*X
+ printf("\n The series circuit thus consists of a resistor of resistance %.2f ohm and a inductor of insuctance %.2E Henry\n",R,L)
+end
\ No newline at end of file diff --git a/608/CH24/EX24.07/24_07.sce b/608/CH24/EX24.07/24_07.sce new file mode 100755 index 000000000..862ea67d5 --- /dev/null +++ b/608/CH24/EX24.07/24_07.sce @@ -0,0 +1,24 @@ +//Problem 24.07: For the circuit shown in Figure 24.11, determine the value of impedance Z2.
+
+//initializing the variables:
+rv = 70; // in volts
+thetav = 30; // in degrees
+ri = 3.5; // in amperes
+thetai = -20; // in degrees
+//z1 consist of two resistance
+R1 = 4.36; // in ohms
+R2 = -2.1*%i; // in ohms
+
+//calculation:
+V = rv*cos(thetav*%pi/180) + %i*rv*sin(thetav*%pi/180)
+I = ri*cos(thetai*%pi/180) + %i*ri*sin(thetai*%pi/180)
+//impedance, Z
+Z = V/I
+//Total impedance Z = z1 + z2
+Z1 = R1 + R2
+Z2 = Z - Z1
+x = real(Z2)
+y = imag(Z2)
+
+printf("\n\n Result \n\n")
+printf("\n impedance Z2 is %.2f + (%.2f) ohm\n",x,y)
\ No newline at end of file diff --git a/608/CH24/EX24.08/24_08.sce b/608/CH24/EX24.08/24_08.sce new file mode 100755 index 000000000..ae16b15bd --- /dev/null +++ b/608/CH24/EX24.08/24_08.sce @@ -0,0 +1,29 @@ +//Problem 24.08: A circuit comprises a resistance of 90 ohm in series with an inductor of inductive reactance 150 ohm. If the supply current is(1.35/_0°)A, determine (a) the supply voltage, (b) the voltage across the 90 ohm resistance, (c) the voltage across the inductance, and (d) the circuit phase angle. Draw the phasor diagram.
+
+//initializing the variables:
+R = 90; // in ohms
+XL = 150; // in ohms
+ri = 1.35; // in amperes
+thetai = 0; // in degrees
+
+//calculation:
+I = ri*cos(thetai*%pi/180) + %i*ri*sin(thetai*%pi/180)
+//Circuit impedance Z
+Z = R + %i*XL
+//Supply voltage, V
+V = I*Z
+//Voltage across 90 ohm resistor
+VR = real(V)
+//Voltage across inductance, VL
+VL = imag(V)
+xv = real(V)
+yv = imag(V)
+rv = (xv^2 + yv^2)^0.5
+thetav = atan(yv/xv)*180/%pi
+phi = thetav - thetai
+
+printf("\n\n Result \n\n")
+printf("\n (a)Supply voltage, V is %.2f + (%.2f)i V",xv,yv)
+printf("\n (b)Voltage across 90 ohm resistor, VR is %.2f V",VR)
+printf("\n (c)Voltage across inductance, VL is %.2f V",VL)
+printf("\n (d)Circuit phase angle is %.2f°",phi)
\ No newline at end of file diff --git a/608/CH24/EX24.09/24_09.sce b/608/CH24/EX24.09/24_09.sce new file mode 100755 index 000000000..f54b72242 --- /dev/null +++ b/608/CH24/EX24.09/24_09.sce @@ -0,0 +1,28 @@ +//Problem 24.09: A coil of resistance 25 ohm and inductance 20 mH has an alternating voltage given by v = 282.8sin(628.4t + pi/3) volts applied across it. Determine (a) the rms value of voltage (in polar form), (b) the circuit impedance, (c) the rms current flowing, and (d) the circuit phase angle.
+
+//initializing the variables:
+R = 25; // in ohms
+L = 0.02; // in henry
+Vm = 282.8; // in volts
+w = 628.4; // in rad/sec
+phiv = %pi/3; // phase angle
+
+//calculation:
+//rms voltage
+Vrms = 0.707*Vm*cos(phiv) + %i*0.707*Vm*sin(phiv)
+//frequency
+f = w/(2*%pi)
+//Inductive reactance XL
+XL = 2*%pi*f*L
+//Circuit impedance Z
+Z = R + %i*XL
+//Rms current
+Irms = Vrms/Z
+phii = atan(imag(Irms)/real(Irms))*180/%pi
+phi = phiv*180/%pi - phii
+
+printf("\n\n Result \n\n")
+printf("\n (a)the rms value of voltage is %.2f + (%.2f)i V ",real(Vrms), imag(Vrms))
+printf("\n (b)the circuit impedance is %.2f + (%.2f)i ohm ",R, XL)
+printf("\n (c)the rms current flowing is %.2f + (%.2f)i A ",real(Irms), imag(Irms))
+printf("\n (d)Circuit phase angle is %.2f° ",phi)
\ No newline at end of file diff --git a/608/CH24/EX24.10/24_10.sce b/608/CH24/EX24.10/24_10.sce new file mode 100755 index 000000000..8114102c1 --- /dev/null +++ b/608/CH24/EX24.10/24_10.sce @@ -0,0 +1,24 @@ +//Problem 24.10: A 240 V, 50 Hz voltage is applied across a series circuit comprising a coil of resistance 12 ohm and inductance 0.10 H, and 120 μF capacitor. Determine the current flowing in the circuit.
+
+//initializing the variables:
+R = 12; // in ohms
+L = 0.10; // in henry
+C = 120E-6; // in Farads
+f = 50; // in Hz
+V = 240; // in volts
+
+//calculation:
+//Inductive reactance, XL
+XL = 2*%pi*f*L
+//Capacitive reactance, Xc
+Xc = 1/(2*%pi*f*C)
+//Circuit impedance Z
+Z = R + %i*(XL - Xc)
+I = V/Z
+phii = atan(imag(I)/real(I))*180/%pi
+phiv = 0 // in degrees
+phi = phiv - phii
+
+printf("\n\n Result \n\n")
+printf("\n the current flowing is %.2f + (%.2f)i A\n",real(I), imag(I))
+printf("and Circuit phase angle is %.2f°\n",phi)
\ No newline at end of file diff --git a/608/CH24/EX24.11/24_11.sce b/608/CH24/EX24.11/24_11.sce new file mode 100755 index 000000000..73e0e273f --- /dev/null +++ b/608/CH24/EX24.11/24_11.sce @@ -0,0 +1,29 @@ +//Problem 24.11:A coil of resistance R ohms and inductance L henrys is connected in series with a 50 μF capacitor. If the supply voltage is 225 V at 50 Hz and the current flowing in the circuit is 1.56/_-30° A, determine the values of R and L. Determine also the voltage across the coil and the voltage across the capacitor.
+
+//initializing the variables:
+C = 50E-6; // in Farads
+f = 50; // in Hz
+V = 225; // in volts
+ri = 1.5; // in Amperes
+thetai = -30; // in degrees
+
+//calculation:
+I = ri*cos(thetai*%pi/180) + %i*ri*sin(thetai*%pi/180)
+//Capacitive reactance, Xc
+Xc = 1/(2*%pi*f*C)
+//Circuit impedance Z
+Z = V/I
+R = real(Z)
+XL = imag(Z) + Xc
+//inductance L
+L = XL/(2*%pi*f)
+//Voltage across coil
+Zcoil = R + %i*XL
+Vcoil = I*Zcoil
+//Voltage across capacitor,
+Vc = -1*I*Xc*%i
+
+printf("\n\n Result \n\n")
+printf("\n (a)resistance is %.2f ohm and inductance is %.2f H ",R, L)
+printf("\n (b)voltage across the coil is %.2f + (%.2f)i V ",real(Vcoil), imag(Vcoil))
+printf("\n (c)voltage across the capacitor is %.2f + (%.2f)i V ",real(Vc), imag(Vc))
\ No newline at end of file diff --git a/608/CH24/EX24.12/24_12.sce b/608/CH24/EX24.12/24_12.sce new file mode 100755 index 000000000..7bfc78088 --- /dev/null +++ b/608/CH24/EX24.12/24_12.sce @@ -0,0 +1,33 @@ +//Problem 24.12: For the circuit shown in Figure 24.17, determine the values of voltages V1 and V2 if the supply frequency is 4 kHz. Determine also the value of the supply voltage V and the circuit phase angle. Draw the phasor diagram.
+
+//initializing the variables:
+C = 2.653E-6; // in Farads
+R1 = 8; // in ohms
+R2 = 5; // in ohms
+L = 0.477E-3; // in Henry
+f = 4000; // in Hz
+ri = 6; // in Amperes
+thetai = 0; // in degrees
+
+//calculation:
+I = ri*cos(thetai*%pi/180) + %i*ri*sin(thetai*%pi/180)
+//Capacitive reactance, Xc
+Xc = 1/(2*%pi*f*C)
+//impedance Z1
+Z1 = R1 - %i*Xc
+//inductive reactance XL
+XL = 2*%pi*f*L
+//impedance Z2,
+Z2 = R2 + %i*XL
+//voltage V1
+V1 = I*Z1
+//voltage V2
+V2 = I*Z2
+//Supply voltage, V
+V = V1 + V2
+phiv = atan(imag(V)/real(V))*180/%pi
+phi = phiv - thetai
+
+printf("\n\n Result \n\n")
+printf("\n supply voltage is %.2f + (%.2f)i V\n",real(V), imag(V))
+printf("and Circuit phase angle is %.2f° \n",phi)
\ No newline at end of file diff --git a/608/CH25/EX25.01/25_01.sce b/608/CH25/EX25.01/25_01.sce new file mode 100755 index 000000000..1bccfcc55 --- /dev/null +++ b/608/CH25/EX25.01/25_01.sce @@ -0,0 +1,41 @@ +//Problem 25.01: Determine the admittance, conductance and susceptance of the following impedances: (a)-i5 ohm, (b)25+i40 ohm, (c)3-i2 ohm, (d)50/_40°ohm.
+
+//initializing the variables:
+Z1 = 0 - %i*5; // in ohms
+Z2 = 25 + %i*40; // in ohms
+Z3 = 3 - %i*2; // in ohms
+r4 = 50; // in ohms
+theta4 = 40; // in degrees
+
+//calculation:
+//admittance Y
+Y1 = 1/Z1
+//conductance, G
+G1 = real(Y1)
+//Suspectance, Bc
+Bc1 = abs(imag(Y1))
+//admittance Y
+Y2 = 1/Z2
+//conductance, G
+G2 = real(Y2)
+//Suspectance, Bc
+Bc2 = abs(imag(Y2))
+//admittance Y
+Y3 = 1/Z3
+//conductance, G
+G3 = real(Y3)
+//Suspectance, Bc
+Bc3 = abs(imag(Y3))
+Z4 = r4*cos(theta4*%pi/180) + %i*r4*sin(theta4*%pi/180)
+//admittance Y
+Y4 = 1/Z4
+//conductance, G
+G4 = real(Y4)
+//Suspectance, Bc
+Bc4 = abs(imag(Y4))
+
+printf("\n\n Result \n\n")
+printf("\n (a)admittance Y is (%.0f + (%.1f)i) S,conductance, G is %.0f S, susceptance,Bc is %.1f S ",real(Y1), imag(Y1),G1,Bc1)
+printf("\n (b)admittance Y is (%.4f + (%.4f)i) S, conductance, G is %.4f S, susceptance,Bc is %.4f S ",real(Y2), imag(Y2),G2, Bc2)
+printf("\n (c)admittance Y is (%.3f + (%.3f)i) S, conductance, G is %.3f S, susceptance,Bc is %.3f S ",real(Y3), imag(Y3),G3,Bc3)
+printf("\n (d)admittance Y is (%.4f + (%.4f)i) S, conductance, G is %.4f S, susceptance,Bc is %.4f S ",real(Y4), imag(Y4),G4,Bc4)
\ No newline at end of file diff --git a/608/CH25/EX25.02/25_02.sce b/608/CH25/EX25.02/25_02.sce new file mode 100755 index 000000000..40de96e79 --- /dev/null +++ b/608/CH25/EX25.02/25_02.sce @@ -0,0 +1,19 @@ +//Problem 25.02: Determine expressions for the impedance of the following admittances: (a)0.004/_30°S (b) (0.001-i0.002)S (c)(0.05 + i0.08)S
+
+//initializing the variables:
+Y2 = 0.001 - %i*0.002; // in S
+Y3 = 0.05 + %i*0.08; // in S
+r1 = 0.004; // in S
+theta1 = 30; // in degrees
+
+//calculation:
+//impedance, Z
+Z2 = 1/Y2
+Z3 = 1/Y3
+Y1 = r1*cos(theta1*%pi/180) + %i*r1*sin(theta1*%pi/180)
+Z1 = 1/Y1
+
+printf("\n\n Result \n\n")
+printf("\n (a)Impedance,Z is (%.1f + (%.0f)i) ohm ",real(Z1), imag(Z1))
+printf("\n (b)Impedance,Z is (%.0f + (%.0f)i) ohm ",real(Z2), imag(Z2))
+printf("\n (c)Impedance,Z is (%.2f + (%.2f)i) ohm ",real(Z3), imag(Z3))
\ No newline at end of file diff --git a/608/CH25/EX25.03/25_03.sce b/608/CH25/EX25.03/25_03.sce new file mode 100755 index 000000000..fc8cf8df1 --- /dev/null +++ b/608/CH25/EX25.03/25_03.sce @@ -0,0 +1,26 @@ +//Problem 25.03: The admittance of a circuit is (0.040 + i0.025) S. Determine the values of the resistance and the capacitive reactance of the circuit if they are connected (a) in parallel, (b) in series. Draw the phasor diagram for each of the circuits.
+
+//initializing the variables:
+Y = 0.040 - %i*0.025; // in S
+
+//calculation:
+//impedance, Z
+Z = 1/Y
+//conductance, G
+G = real(Y)
+//Suspectance, Bc
+Bc = abs(imag(Y))
+//parallrl
+//resistance, R
+Rp = 1/G
+//capacitive reactance
+Xcp = 1/Bc
+//series
+//resistance, R
+Rs = real(Z)
+//capacitive reactance
+Xcs = abs(imag(Z))
+
+printf("\n\n Result \n\n")
+printf("\n (a)for parallel, resistance,R is %.0f ohm and capacitive reactance, Xc is %.0f ohm ",Rp,Xcp)
+printf("\n (b)forseries, resistance,R is %.2f ohm and capacitive reactance, Xc is %.2f ohm ",Rs,Xcs)
\ No newline at end of file diff --git a/608/CH25/EX25.04/25_04.sce b/608/CH25/EX25.04/25_04.sce new file mode 100755 index 000000000..03af96e1d --- /dev/null +++ b/608/CH25/EX25.04/25_04.sce @@ -0,0 +1,23 @@ +//Problem 25.04: Determine the values of currents I, I1 and I2 shown in the network of Figure 25.5.
+
+//initializing the variables:
+R1 = 8; // in ohm
+R = 5; // in ohm
+R2 = 6; // ohm
+rv = 50; // in volts
+thetav = 0; // in degrees
+
+//calculation:
+//voltage,V
+V = rv*cos(thetav*%pi/180) + %i*rv*sin(thetav*%pi/180)
+//circuit impedance, ZT
+ZT = R + (R1*%i*R2/(R1 + %i*R2))
+//Current I
+I = V/ZT
+//current,I1
+I1 = I*(%i*R2/(R1 + %i*R2))
+//current, I2
+I2 = I*(R1/(R1 + %i*R2))
+
+printf("\n\n Result \n\n")
+printf("\n current, I is (%.2f + (%.2f)i) A,current,I1 is (%.2f + (%.2f)i) A, current, I2 is (%.2f + (%.2f)i) A ",real(I), imag(I),real(I1), imag(I1),real(I2), imag(I2))
\ No newline at end of file diff --git a/608/CH25/EX25.05/25_05.sce b/608/CH25/EX25.05/25_05.sce new file mode 100755 index 000000000..0768563a7 --- /dev/null +++ b/608/CH25/EX25.05/25_05.sce @@ -0,0 +1,33 @@ +//Problem 25.05: For the parallel network shown in Figure 25.6, determine the value of supply current I and its phase relative to the 40 V supply.
+
+//initializing the variables:
+R1 = 5; // in ohm
+R2 = 3; // in ohm
+R3 = 8; // ohm
+Xc = 4; // in ohms
+XL = 12; // in Ohms
+V = 40; // in volts
+f = 50; // in Hz
+
+//calculation:
+Z1 = R1 + %i*XL
+Z2 = R2 - %i*Xc
+Z3 = R3
+//circuit admittance, YT = 1/ZT
+YT = (1/Z1) + (1/Z2) + (1/Z3)
+//Current I
+I = V*YT
+I1 = V/Z1
+I2 = V/Z2
+I3 = V/Z2
+thetav = 0
+thetai = atan(imag(I)/real(I))*180/%pi
+phi = thetav - thetai
+if (phi>0) then
+ a = "lagging"
+else
+ a = "leading"
+end
+
+printf("\n\n Result \n\n")
+printf("\n current, I is (%.2f + (%.2f)i) A,and its phase relative to the 40 V supply is %s by %.2f°\n",real(I), imag(I),a,abs(phi))
\ No newline at end of file diff --git a/608/CH25/EX25.06/25_06.sce b/608/CH25/EX25.06/25_06.sce new file mode 100755 index 000000000..52373c214 --- /dev/null +++ b/608/CH25/EX25.06/25_06.sce @@ -0,0 +1,43 @@ +//Problem 25.06: An a.c. network consists of a coil, of inductance 79.58 mH and resistance 18 ohm, in parallel with a capacitor of capacitance 64.96 μF. If the supply voltage is 250/_0° V at 50 Hz, determine (a) the total equivalent circuit impedance, (b) the supply current, (c) the circuit phase angle, (d) the current in the coil, and (e) the current in the capacitor.
+
+//initializing the variables:
+L = 0.07958; // in Henry
+R = 18; // in ohm
+C = 64.96E-6; // in Farad
+rv = 250; // in volts
+thetav = 0; // in degrees
+f = 50; // in Hz
+
+//calculation:
+//Inductive reactance
+XL = 2*%pi*f*L
+//capacitive reactance
+Xc = 1/(2*%pi*f*C)
+//impedance of the coil,
+Zcoil = R + %i*XL
+//impedance presented by the capacitor,
+Zc = -1*%i*Xc
+//Total equivalent circuit impedance,
+ZT = Zcoil*Zc/(Zcoil + Zc)
+//voltage
+V = rv*cos(thetav*%pi/180) + %i*rv*sin(thetav*%pi/180)
+//current, I
+I = V/ZT
+thetai = atan(imag(I)/real(I))*180/%pi
+phi = thetav - thetai
+if (phi>0) then
+ a = "lagging"
+else
+ a = "leading"
+end
+//Current in the coil, ICOIL
+Icoil = V/Zcoil
+//Current in the capacitor, IC
+Ic = V/Zc
+
+printf("\n\n Result \n\n")
+printf("\n (a)the circuit impedance is %.2f + (%.2f)i ohm ",real(ZT), imag(ZT))
+printf("\n (b)supply current, I is %.2f + (%.2f)i A ",real(I), imag(I))
+printf("\n (c)circuit phase relative is %s by %.2f° ",a,abs(phi))
+printf("\n (d)current in coil, Icoil is %.2f + (%.2f)i A ",real(Icoil), imag(Icoil))
+printf("\n (e)current in capacitor, Ic is %.2f + (%.2f)i A ",real(Ic), imag(Ic))
\ No newline at end of file diff --git a/608/CH25/EX25.07/25_07.sce b/608/CH25/EX25.07/25_07.sce new file mode 100755 index 000000000..a4c790679 --- /dev/null +++ b/608/CH25/EX25.07/25_07.sce @@ -0,0 +1,43 @@ +//Problem 25.07: (a) For the network diagram of Figure 25.8, determine the value of impedance Z1 (b) If the supply frequency is 5 kHz, determine the value of the components comprising impedance Z1.
+
+//initializing the variables:
+RL = %i*6; // in ohm
+R2 = 8; // in ohm
+Z3 = 10; // in ohm
+rv = 50; // in volts
+thetav = 30; // in degrees
+ri = 31.4; // in amperes
+thetai = 52.48; // in degrees
+f = 5000; // in Hz
+
+//calculation:
+//impedance, Z2
+Z2 = R2 + RL
+//voltage
+V = rv*cos(thetav*%pi/180) + %i*rv*sin(thetav*%pi/180)
+//current, I
+I = ri*cos(thetai*%pi/180) + %i*ri*sin(thetai*%pi/180)
+//Total circuit admittance,
+YT = I/V
+//admittance, Y3
+Y3 = 1/Z3
+//admittance, Y2
+Y2 = 1/Z2
+//admittance, Y1
+Y1 = YT - Y2 - Y3
+//impedance, Z1
+Z1 = 1/Y1
+
+printf("\n\n Result \n\n")
+printf("\n (a)the impedance Z1 is %.2f + (%.2f)i ohm",real(Z1), imag(Z1))
+
+//resistance, R1
+R1 = real(Z1)
+X1 = imag(Z1)
+if ((R1>0)&(X1<0)) then
+ C1 = -1/(2*%pi*f*X1)
+ printf("\n (b)The series circuit thus consists of a resistor of resistance %.2f ohm and a capacitor of capacitance %.2E Farad\n",R1,C1)
+elseif ((R1>0)&(X1>0)) then
+ L1 = 2*%pi*f*X1
+ printf("\n (b)The series circuit thus consists of a resistor of resistance %.2f ohm and a inductor of insuctance %.2E Henry\n",R1,L1)
+end
\ No newline at end of file diff --git a/608/CH25/EX25.08/25_08.sce b/608/CH25/EX25.08/25_08.sce new file mode 100755 index 000000000..e58941ec9 --- /dev/null +++ b/608/CH25/EX25.08/25_08.sce @@ -0,0 +1,49 @@ +//Problem 25.08: For the series-parallel arrangement shown in Figure 25.9, determine (a) the equivalent series circuit impedance, (b) the supply current I, (c) the circuit phase angle, (d) the values of voltages V1 and V2, and (e) the values of currents IA and IB.
+
+//initializing the variables:
+RL1 = %i*1.02; // in ohm
+R1 = 1.65; // in ohm
+RLa = %i*7; // in ohm
+Ra = 5; // in ohm
+Rcb = -1*%i*15; // in ohm
+Rb = 4; // in ohm
+rv = 91; // in volts
+thetav = 0; // in degree
+
+//calculation:
+//voltage
+V = rv*cos(thetav*%pi/180) + %i*rv*sin(thetav*%pi/180)
+//impedance, Z1
+Z1 = R1 + RL1
+//impedance, Za
+Za = Ra + RLa
+//impedance, Zb
+Zb = Rb + Rcb
+//impedance, Z, of the two branches connected in parallel
+Z = Za*Zb/(Za + Zb)
+//Total circuit impedance
+ZT = Z1 + Z
+//Supply current, I
+I = V/ZT
+thetai = atan(imag(I)/real(I))*180/%pi
+phi = thetav - thetai
+if (phi>0) then
+ a = "lagging"
+else
+ a = "leading"
+end
+//Voltage V1
+V1 = I*Z1
+//Voltage V2
+V2 = I*Z
+//current Ia
+Ia = V2/Za
+//Current Ib
+Ib = V2/Zb
+
+printf("\n\n Result \n\n")
+printf("\n (a)equivalent series circuit impedance is %.2f + (%.2f)i ohm ",real(ZT), imag(ZT))
+printf("\n (b)supply current, I is %.2f + (%.2f)i A ",real(I), imag(I))
+printf("\n (c)circuit phase relative is %s by %.2f° ",a,abs(phi))
+printf("\n (d)voltage, V1 is (%.2f + (%.2f)i) V and V2 is(%.2f + (%.2f)i) V ",real(V1), imag(V1),real(V2), imag(V2))
+printf("\n (e)current, Ia is (%.2f + (%.2f)i) A and Ib is(%.2f + (%.2f)i) A ",real(Ia), imag(Ia),real(Ib), imag(Ib))
\ No newline at end of file diff --git a/608/CH26/EX26.01/26_01.sce b/608/CH26/EX26.01/26_01.sce new file mode 100755 index 000000000..cecbf6111 --- /dev/null +++ b/608/CH26/EX26.01/26_01.sce @@ -0,0 +1,20 @@ +//Problem 26.01: A coil of resistance 5 ohm and inductive reactance 12 ohm is connected across a supply voltage of 526/_30volts. Determine the active power in the circuit.
+
+//initializing the variables:
+RL = %i*12; // in ohm
+R = 5; // in ohm
+rv = 52; // in volts
+thetav = 30; // in degree
+
+//calculation:
+//voltage
+V = rv*cos(thetav*%pi/180) + %i*rv*sin(thetav*%pi/180)
+//impedance, Z
+Z = R + RL
+//current
+I = V/Z
+//Active power, P
+Pa = real(V)*real(I) + imag(V)*imag(I)
+
+printf("\n\n Result \n\n")
+printf("\nthe active power in the circuit %.0f W\n",Pa)
\ No newline at end of file diff --git a/608/CH26/EX26.02/26_02.sce b/608/CH26/EX26.02/26_02.sce new file mode 100755 index 000000000..39e0249b9 --- /dev/null +++ b/608/CH26/EX26.02/26_02.sce @@ -0,0 +1,15 @@ +//Problem 26.02: A current of (15+i8)A flows in a circuit whose supply voltage is (120+i200)V.Determine (a) the active power, and (b) the reactive power.
+
+//initializing the variables:
+V = 120 + %i*200; // in volts
+I = 15 + %i*8; // in amperes
+
+//calculation:
+//Active power, P
+Pa = real(V)*real(I) + imag(V)*imag(I)
+//Reactive power, Q
+Q = imag(V)*real(I) - real(V)*imag(I)
+
+printf("\n\n Result \n\n")
+printf("\n (a) the active power in the circuit %.0f W",Pa)
+printf("\n (b) the reactive power in the circuit %.0f var ",Q)
\ No newline at end of file diff --git a/608/CH26/EX26.03/26_03.sce b/608/CH26/EX26.03/26_03.sce new file mode 100755 index 000000000..c01887201 --- /dev/null +++ b/608/CH26/EX26.03/26_03.sce @@ -0,0 +1,36 @@ +//Problem 26.03: A series circuit possesses resistance R and capacitance C. The circuit dissipates a power of 1.732 kW and has a power factor of 0.866 leading. If the applied voltage is given by v = 141.4*sin(10000t + pi/9) volts, determine (a) the current flowing and its phase, (b) the value of resistance R, and (c) the value of capacitance C.
+
+//initializing the variables:
+Vm = 141.4; // in volts
+w = 10000; // in rad/sec
+phiv = %pi/9; // in radian
+Pd = 1732; // in Watts
+pf = 0.866; // power fctr
+
+//calculation:
+//the rms voltage,
+Vrms = 0.707*Vm
+//Power P = V*I*cos(phi)
+//current magnitude, Irms
+Irms = Pd/(Vrms*pf)
+phid = acos(pf)
+//current phase angle
+phii = phiv + phid
+phiid = phii*180/%pi // in degrees
+//Voltage, V
+V = Vrms*cos(phiv) + %i*Vrms*sin(phiv)
+//current, I
+I = Irms*cos(phii) + %i*Irms*sin(phii)
+//Impedance, Z
+Z = V/I
+//resistance, R
+R = real(Z)
+//capacitive reactance, Xc
+Xc = abs(imag(Z))
+//capacitance, C
+C = 1/ (w*Xc)
+
+printf("\n\n Result \n\n")
+printf("\n (a)the current flowing and Circuit phase angle is %.0f/_%.2f° A ",Irms,phiid)
+printf("\n (b) the resistance is %.2f ohm ",R)
+printf("\n (c) the capacitance is %.2E farad ",C)
\ No newline at end of file diff --git a/608/CH26/EX26.04/26_04.sce b/608/CH26/EX26.04/26_04.sce new file mode 100755 index 000000000..c6295fe79 --- /dev/null +++ b/608/CH26/EX26.04/26_04.sce @@ -0,0 +1,35 @@ +//Problem 26.04:For the circuit shown in Figure 26.8, determine the active power developed between points (a) A and B, (b) C and D, (c) E and F.
+
+//initializing the variables:
+rv = 100; // in volts
+thetav = 0; // in degrees
+R = 5; // in ohm
+R1 = 3; // in ohms
+RL = %i*4; // in ohm
+Rc = -10*%i; // in ohms
+
+//calculation:
+//impedance, Z1
+Z1 = R1 + RL
+//impedance, Zc
+Zc = Rc
+//Circuit impedance, Z
+Z = R + (Z1*Zc/(Z1 + Zc))
+//voltage
+V = rv*cos(thetav*%pi/180) + %i*rv*sin(thetav*%pi/180)
+I = V/Z
+Imag = ((real(I))^2 + (imag(I))^2)^0.5
+//Active power developed between points A and B
+Pab = (Imag^2)*R
+//Active power developed between points C and D
+Pcd = (Imag^2)*real(Zc)
+//Current, I1
+I1 = I*Zc/(Zc + Z1)
+I1mag = ((real(I1))^2 + (imag(I1))^2)^0.5
+//active power developed between points E and F
+Pef = (I1mag^2)*real(Z1)
+
+printf("\n\n Result \n\n")
+printf("\n (a)Active power developed between points A and B is %.2f W ",Pab)
+printf("\n (b)Active power developed between points C and D is %.2f W ",Pcd)
+printf("\n (c)Active power developed between points E and F is %.2f W ",Pef)
\ No newline at end of file diff --git a/608/CH26/EX26.05/26_05.sce b/608/CH26/EX26.05/26_05.sce new file mode 100755 index 000000000..6b0268ef2 --- /dev/null +++ b/608/CH26/EX26.05/26_05.sce @@ -0,0 +1,31 @@ +//Problem 26.05:The circuit shown in Figure 26.9 dissipates an active power of 400 Wand has a power factor of 0.766 lagging. Determine (a) the apparent power, (b) the reactive power, (c) the value and phase of current I, and (d) the value of impedance Z.
+
+//initializing the variables:
+Pa = 400; // in Watts
+rv = 100; // in volts
+thetav = 30; // in degrees
+R = 4; // in ohm
+pf = 0.766; // power factor
+
+//calculation:
+V = rv*cos(thetav*%pi/180) + %i*rv*sin(thetav*%pi/180)
+//magnitude of apparent power,S = V*I
+S = Pa/pf
+phi = acos(pf)
+theta = phi*180/%pi // in degrees
+//Reactive power Q
+Q = S*sin(phi)
+//magnitude of current
+Imag = S/rv
+thetai = thetav - theta
+I = Imag*cos(thetai*%pi/180) + %i*Imag*sin(thetai*%pi/180)
+//Total circuit impedance ZT
+ZT = V/I
+//impedance Z
+Z = ZT - R
+
+printf("\n\n Result \n\n")
+printf("\n (a)apparent power is %.2f VA ",S)
+printf("\n (b)reactive power is %.2f var ",Q)
+printf("\n (c)the current flowing and Circuit phase angle is %.2f/_%.0f° A ",Imag,thetai)
+printf("\n (d)impedance, Z is %.2f + (%.2f)i ohm ",real(Z), imag(Z))
\ No newline at end of file diff --git a/608/CH26/EX26.06/26_06.sce b/608/CH26/EX26.06/26_06.sce new file mode 100755 index 000000000..1502ce9d7 --- /dev/null +++ b/608/CH26/EX26.06/26_06.sce @@ -0,0 +1,22 @@ +//Problem 26.06: A 300 kVA transformer is at full load with an overall power factor of 0.70 lagging. The power factor is improved by adding capacitors in parallel with the transformer until the overall power factor becomes 0.90 lagging. Determine the rating (in kilovars) of the capacitors required.
+
+//initializing the variables:
+S = 300000; // in VA
+pf1 = 0.70; // in power factor
+pf2 = 0.90; // in power factor
+
+//calculation:
+//active power, P
+Pa = S*pf1
+phi1 = acos(pf1)
+phi1d = phi1*180/%pi
+//Reactive power, Q
+Q = S*sin(phi1)
+phi2 = acos(pf2)
+phi2d = phi2*180/%pi
+//The capacitor rating needed to improve the power factor to 0.90
+//the capacitor rating,
+Pr = Q - (Pa*tan(phi2))
+
+printf("\n\n Result \n\n")
+printf("\n the rating (in kilovars) of the capacitors is %.1f kvar\n",(Pr/1E3))
\ No newline at end of file diff --git a/608/CH26/EX26.07/26_07.sce b/608/CH26/EX26.07/26_07.sce new file mode 100755 index 000000000..9b77a675d --- /dev/null +++ b/608/CH26/EX26.07/26_07.sce @@ -0,0 +1,37 @@ +//Problem 26.07: A circuit has an impedance Z = (3+i4)ohm and a source p.d. of 50/_30° V at a frequency of 1.5 kHz. Determine (a) the supply current, (b) the active, apparent and reactive power, (c) the rating of a capacitor to be connected in parallel with impedance Z to improve the power factor of the circuit to 0.966 lagging, and (d) the value of capacitance needed to improve the power factor to 0.966 lagging.
+
+//initializing the variables:
+Z = 3 + %i*4; // in ohms
+rv = 50; // in volts
+thetav = 30; // in Degrees
+f = 1500; // in Hz
+pf1 = 0.966; // in power factor
+
+//calculation:
+V = rv*cos(thetav*%pi/180) + %i*rv*sin(thetav*%pi/180)
+//Supply current, I
+I = V/Z
+Istr = real(I) - %i*imag(I)
+//Apparent power, S
+S = V*Istr
+//active power, Pa
+Pa = real(S)
+//reactive power, Q
+Q = abs(imag(S))
+//apparent power, S
+S = (real(S)^2 + imag(S)^2)^0.5
+phi1 = acos(pf1)
+phi1d = phi1*180/%pi
+//rating of the capacitor
+Pr = Q - Pa*tan(phi1)
+//Current in capacitor, Ic
+Ic = Pr/rv
+//Capacitive reactance, Xc
+Xc = rv/Ic
+C = 1/(2*%pi*f*Xc)
+
+printf("\n\n Result \n\n")
+printf("\n (a)supply current, I is %.2f + (%.2f)i A ",real(I), imag(I))
+printf("\n (b)active power is %.0f W, apparent power is %.0f W and reactive power is %.0f W ",Pa, S, Q)
+printf("\n (c)the rating of the capacitors is %.1f var\n",Pr)
+printf(" (d)value of capacitance needed to improve the power factor to 0.966 lagging is %.3E F\n", C)
\ No newline at end of file diff --git a/608/CH27/EX27.02/27_02.sce b/608/CH27/EX27.02/27_02.sce new file mode 100755 index 000000000..71de15264 --- /dev/null +++ b/608/CH27/EX27.02/27_02.sce @@ -0,0 +1,18 @@ +//Problem 27.02: For the Wien bridge shown in Figure 27.9, R2 = R3 = 30 kohm, R4 = 1 kohm and C2 = C3 = 1 nF. Determine, when the bridge is balanced, (a) the value of resistance R1, and (b) the frequency of the bridge.
+
+//initializing the variables:
+R2 = 30000; // in ohms
+R3 = 30000; // in ohms
+R4 = 1000; // in ohms
+C2 = 1e-9; // IN fARADS
+C3 = 1e-9; // IN fARADS
+
+//calculation:
+//the bridge is balanced
+R1 = R4/((R3/R2) + (C2/C3))
+//frequency, f
+f = 1/(2*%pi*((C2*C3*R2*R3)^0.5))
+
+printf("\n\n Result \n\n")
+printf("\n (a)Resistance R1 = %.0f ohm",R1)
+printf("\n (b)frequency, f is %.2E Hz ",f)
\ No newline at end of file diff --git a/608/CH27/EX27.03/27_03.sce b/608/CH27/EX27.03/27_03.sce new file mode 100755 index 000000000..9e0a19be8 --- /dev/null +++ b/608/CH27/EX27.03/27_03.sce @@ -0,0 +1,29 @@ +//Problem 27.03: A Schering bridge network is as shown in Figure 27.7. Given C2 = 0.2 μF, R4 = 200 ohm, R3 = 600 ohm, C3 = 4000 pF and the supply frequency is 1.5 kHz, determine, when the bridge is balanced, (a) the value of resistance Rx, (b) the value of capacitance Cx, (c) the phase angle of the unknown arm, (d) the power factor of the unknown arm and (e) its loss angle.
+
+//initializing the variables:
+R3 = 600; // in ohms
+R4 = 200; // in ohms
+C2 = 0.2e-6; // IN fARADS
+C3 = 4000e-12; // IN fARADS
+f = 1500; //in Hz
+
+//calculation:
+//the bridge is balanced
+//Resistance, Rx
+Rx = R4*C3/C2
+//Capacitance, Cx
+Cx = C2*R3/R4
+//Phase angle
+phi = atan(1/(2*%pi*f*Cx*Rx))
+phid = phi*180/%pi // in degrees
+//Power factor of capacitor
+Pc = cos(phi)
+//Loss angle,
+del = 90 - phid
+
+printf("\n\n Result \n\n")
+printf("\n (a)Resistance Rx = %.0f ohm ",Rx)
+printf("\n (b)capacitance, Cx is %.2E Farad ",Cx)
+printf("\n (c)phasor diagram = %.2f° ",phid)
+printf("\n (d)power factor is %.4f ",Pc)
+printf("\n (e)Loss angle = %.2f° ",del)
\ No newline at end of file diff --git a/608/CH28/EX28.01/28_01.sce b/608/CH28/EX28.01/28_01.sce new file mode 100755 index 000000000..ee2c3f6f2 --- /dev/null +++ b/608/CH28/EX28.01/28_01.sce @@ -0,0 +1,17 @@ +//Problem 28.01: A coil having a resistance of 10 ohm and an inductance of 75 mH is connected in series with a 40 μF capacitor across a 200 V a.c. supply. Determine at what frequency resonance occurs, and (b) the current flowing at resonance.
+
+//initializing the variables:
+R = 10; // in ohms
+C = 40e-6; // IN fARADS
+L = 0.075; // IN Henry
+V = 200; // in Volts
+
+//calculation:
+//Resonant frequency,
+fr = 1/(2*%pi*((L*C)^0.5))
+//Current at resonance, I
+I = V/R
+
+printf("\n\n Result \n\n")
+printf("\n (a)Resonant frequency = %.1f Hz ",fr)
+printf("\n (b)Current at resonance, I is %.0f A ",I)
\ No newline at end of file diff --git a/608/CH28/EX28.02/28_02.sce b/608/CH28/EX28.02/28_02.sce new file mode 100755 index 000000000..e4422ccdf --- /dev/null +++ b/608/CH28/EX28.02/28_02.sce @@ -0,0 +1,14 @@ +//Problem 28.02: An R–L–C series circuit is comprised of a coil of inductance 10 mH and resistance 8 ohm and a variable capacitor C. The supply frequency is 1 kHz. Determine the value of capacitor C for series resonance.
+
+//initializing the variables:
+R = 8; // in ohms
+L = 0.010; // IN Henry
+f = 1000; // in Hz
+
+//calculation:
+//At resonance
+//capacitance C
+C = 1/(L*(2*%pi*f)^2)
+
+printf("\n\n Result \n\n")
+printf("\n capacitance, C is %.2E F\n",C)
\ No newline at end of file diff --git a/608/CH28/EX28.03/28_03.sce b/608/CH28/EX28.03/28_03.sce new file mode 100755 index 000000000..1590663fd --- /dev/null +++ b/608/CH28/EX28.03/28_03.sce @@ -0,0 +1,17 @@ +//Problem 28.03: A coil having inductance L is connected in series with a variable capacitor C. The circuit possesses stray capacitance CS which is assumed to be constant and effectively in parallel with the variable capacitor C. When the capacitor is set to 1000 pF the resonant frequency of the circuit is 92.5 kHz, and when the capacitor is set to 500 pF the resonant frequency is 127.8 kHz Determine the values of (a) the stray capacitance CS, and (b) the coil inductance L.
+
+//initializing the variables:
+C1 = 1000e-12; // IN fARADS
+C2 = 500e-12; // IN fARADS
+fr1 = 92500; // in Hz
+fr2 = 127800; // in Hz
+
+//calculation:
+//For a series R–L–C circuit the resonant frequency fr is given by:
+//fr = 1/(2pi*(L*C)^2)
+Cs = ((C1 - C2)/((fr2/fr1)^2 - 1)) - C2
+L = 1/((C1 + Cs)*(2*%pi*fr1)^2)
+
+printf("\n\n Result \n\n")
+printf("\n (a)stray capacitance, Cs is %.2E F ",Cs)
+printf("\n (b)inductance, L is %.2E H ",L)
\ No newline at end of file diff --git a/608/CH28/EX28.04/28_04.sce b/608/CH28/EX28.04/28_04.sce new file mode 100755 index 000000000..211ededbf --- /dev/null +++ b/608/CH28/EX28.04/28_04.sce @@ -0,0 +1,34 @@ +//Problem 28.04: A series circuit comprises a 10 ohm resistance, a 5 μF capacitor and a variable inductance L. The supply voltage is 20/_0° volts at a frequency of 318.3 Hz. The inductance is adjusted until the p.d. across the 10 ohm resistance is a maximum. Determine for this condition (a) the value of inductance L, (b) the p.d. across each component and (c) the Q-factor.
+
+//initializing the variables:
+R = 10; // in ohms
+C = 5e-6; // IN fARADS
+rv = 20; //in volts
+thetav = 0; // in degrees
+f = 318.3; // in Hz
+
+//calculation:
+wr = 2*%pi*f
+//The maximum voltage across the resistance occurs at resonance when the current is a maximum. At resonance,L = 1/c*wr^2
+L = 1/(C*wr^2)
+//voltage
+V = rv*cos(thetav*%pi/180) + %i*rv*sin(thetav*%pi/180)
+//Current at resonance Ir
+Ir = V/R
+//p.d. across resistance, VR
+VR = Ir*R
+//inductive reactance, XL
+XL = wr*L
+//p.d. across inductance, VL
+VL = Ir*(%i*XL)
+//capacitive reactance, Xc
+Xc = 1/(wr*C)
+//p.d. across capacitor, Vc
+Vc = Ir*(-1*%i*Xc)
+//Q-factor at resonance, Qr
+Qr = imag(VL)/V
+
+printf("\n\n Result \n\n")
+printf("\n (a)inductance, L is %.2E H ",L)
+printf("\n (b)p.d. across resistance, VR is %.2f V, p.d. across inductance, VL %.0fi V and p.d. across capacitor, VC %.0fi V ",VR, imag(VL), imag(Vc))
+printf("\n (c)Q-factor at resonance, Qr is %.0f ",Qr)
\ No newline at end of file diff --git a/608/CH28/EX28.05/28_05.sce b/608/CH28/EX28.05/28_05.sce new file mode 100755 index 000000000..1d2f7dcda --- /dev/null +++ b/608/CH28/EX28.05/28_05.sce @@ -0,0 +1,25 @@ +//Problem 28.05: A series L–R–C circuit has a sinusoidal input voltage of maximum value 12 V. If inductance, L = 20 mH, resistance, R = 80 ohm, and capacitance, C = 400 nF, determine (a) the resonant frequency, (b) the value of the p.d. across the capacitor at the resonant frequency, (c) the frequency at which the p.d. across the capacitor is a maximum, and (d) the value of the maximum voltage across the capacitor.
+
+//initializing the variables:
+R = 80; // in ohms
+C = 0.4e-6; // IN fARADS
+L = 0.020; // IN Henry
+Vm = 12; //in volts
+
+//calculation:
+//Resonant frequency,
+fr = 1/(2*%pi*((L*C)^0.5))
+wr = 2*%pi*fr
+//Q = wr*L/R
+Q = wr*L/R
+Vc = Q*Vm
+//the frequency f at which VC is a maximum value,
+f = fr*(1 - (1/(2*Q*Q)))^0.5
+//the maximum value of the p.d. across the capacitor is given by:
+Vcm = Vc/(1 - (1/(2*Q))^2)^0.5
+
+printf("\n\n Result \n\n")
+printf("\n (a)The resonant frequency is %.1f Hz ",fr)
+printf("\n (b)the value of the p.d. across the capacitor at the resonant frequency %.2f V ",Vc)
+printf("\n (c)the frequency f at which Vc is a maximum value, is %.1f Hz ",f)
+printf("\n (d)the maximum value of the p.d. across the capacitor is %.1f V ",Vcm)
\ No newline at end of file diff --git a/608/CH28/EX28.06/28_06.sce b/608/CH28/EX28.06/28_06.sce new file mode 100755 index 000000000..33626afd8 --- /dev/null +++ b/608/CH28/EX28.06/28_06.sce @@ -0,0 +1,11 @@ +//Problem 28.06: An inductor of Q-factor 60 is connected in series with a capacitor having a Q-factor of 390. Determine the overall Q-factor of the circuit.
+
+//initializing the variables:
+QL = 60; // Q-factor
+Qc = 390; // Q-factor
+
+//calculation:
+QT = QL*Qc/(QL + Qc)
+
+printf("\n\n Result \n\n")
+printf("\n the overall Q-factor is %.0f \n",QT)
\ No newline at end of file diff --git a/608/CH28/EX28.07/28_07.sce b/608/CH28/EX28.07/28_07.sce new file mode 100755 index 000000000..5df2f7c54 --- /dev/null +++ b/608/CH28/EX28.07/28_07.sce @@ -0,0 +1,16 @@ +//Problem 28.07: A filter in the form of a series L–R–C circuit is designed to operate at a resonant frequency of 10 kHz. Included within the filter is a 10 mH inductance and 5 ohm resistance. Determine the bandwidth of the filter.
+
+//initializing the variables:
+R = 5; // in ohms
+L = 0.010; // IN Henry
+fr = 10000; // in Hz
+
+//calculation:
+wr = 2*%pi*fr
+//Q-factor at resonance is given by
+Qr = wr*L/R
+//Since Qr = fr/(f2 - f1),
+bw = fr/Qr
+
+printf("\n\n Result \n\n")
+printf("\n bandwidth of the filter is %.1f Hz\n",bw)
\ No newline at end of file diff --git a/608/CH28/EX28.08/28_08.sce b/608/CH28/EX28.08/28_08.sce new file mode 100755 index 000000000..fc8407a4b --- /dev/null +++ b/608/CH28/EX28.08/28_08.sce @@ -0,0 +1,33 @@ +//Problem 28.08: An R–L–C series circuit has a resonant frequency of 1.2 kHz and a Q-factor at resonance of 30. If the impedance of the circuit at resonance is 50 ohm determine the values of (a) the inductance, and (b) the capacitance. Find also (c) the bandwidth, (d) the lower and upper half-power frequencies and (e) the value of the circuit impedance at the half-power frequencies.
+
+//initializing the variables:
+Zr = 50; // in ohms
+fr = 1200; // in Hz
+Qr = 30; // Q-factor
+
+//calculation:
+//At resonance the circuit impedance, Z
+R = Zr
+wr = 2*%pi*fr
+//Q-factor at resonance is given by Qr = wr*L/R, then L is
+L = Qr*R/wr
+//At resonance r*L = 1/(wr*C)
+//capacitance, C
+C = 1/(L*wr*wr)
+//bandwidth,.(f2 − f1)
+bw = fr/Qr
+//upper half-power frequency, f2
+f2 = (bw + ((bw^2) + 4*(fr^2))^0.5)/2
+//lower half-power frequency, f1
+f1 = f2 - bw
+//At the half-power frequencies, current I
+//I = 0.707*Ir
+//Hence impedance
+Z = (2^0.5)*R
+
+printf("\n\n Result \n\n")
+printf("\n (a)inductance, L is %.3f H ",L)
+printf("\n (b)capacitance, C is %.2E F ",C)
+printf("\n (c)bandwidth is %.0f Hz ",bw)
+printf("\n (d)the upper half-power frequency, f2 is %.0f Hz and the lower half-power frequency, f1 is %.0f Hz ",f2,f1)
+printf("\n (e)impedance at the half-power frequencies is %.2f ohm ",Z)
\ No newline at end of file diff --git a/608/CH28/EX28.09/28_09.sce b/608/CH28/EX28.09/28_09.sce new file mode 100755 index 000000000..833f3247c --- /dev/null +++ b/608/CH28/EX28.09/28_09.sce @@ -0,0 +1,27 @@ +//Problem 28.09: A series R–L–C circuit is connected to a 0.2 V supply and the current is at its maximum value of 4 mA when the supply frequency is adjusted to 3 kHz. The Q-factor of the circuit under these conditions is 100. Determine the value of (a) the circuit resistance, (b) the circuit inductance, (c) the circuit capacitance, and (d) the voltage across the capacitor
+
+//initializing the variables:
+V = 0.2; // in Volts
+I = 0.004; // in Amperes
+fr = 3000; // in Hz
+Qr = 100; // Q-factor
+
+//calculation:
+wr = 2*%pi*fr
+//At resonance, impedance
+Z = V/I
+//At resonance the circuit impedance, Z
+R = Z
+//Q-factor at resonance is given by Qr = wr*L/R, then L is
+L = Qr*R/wr
+//At resonance r*L = 1/(wr*C)
+//capacitance, C
+C = 1/(L*wr*wr)
+//Q-factor at resonance in a series circuit represents the voltage magnification Qr = Vc/V, then Vc is
+Vc = Qr*V
+
+printf("\n\n Result \n\n")
+printf("\n (a)the circuit resistance is %.0f ohm ",R)
+printf("\n (b)inductance, L is %.3f H ",L)
+printf("\n (c)capacitance, C is %.2E F ",C)
+printf("\n (d)the voltage across the capacitor is %.0f V ",Vc)
\ No newline at end of file diff --git a/608/CH28/EX28.10/28_10.sce b/608/CH28/EX28.10/28_10.sce new file mode 100755 index 000000000..5900d486e --- /dev/null +++ b/608/CH28/EX28.10/28_10.sce @@ -0,0 +1,25 @@ +//Problem 28.10: A coil of inductance 351.8 mH and resistance 8.84 ohm is connected in series with a 20 μF capacitor. Determine (a) the resonant frequency, (b) the Q-factor at resonance, (c) the bandwidth, and (d) the lower and upper -3dB frequencies.
+
+//initializing the variables:
+R = 8.84; // in ohms
+L = 0.3518; // IN Henry
+C = 20e-6; // IN fARADS
+
+//calculation:
+//Resonant frequency,
+fr = 1/(2*%pi*((L*C)^0.5))
+wr = 2*%pi*fr
+//Q-factor at resonance, Q = wr*L/R
+Qr = wr*L/R
+//bandwidth,.(f2 − f1)
+bw = fr/Qr
+//the lower −3 dB frequency
+f1 = fr - bw/2
+//the upper −3 dB frequency
+f2 = fr + bw/2
+
+printf("\n\n Result \n\n")
+printf("\n (a)Resonant frequency, fr is %.0f Hz",fr)
+printf("\n (b)Q-factor at resonance is %.0f",Qr)
+printf("\n (c)Bandwidth is %.0f Hz ",bw)
+printf("\n (d)the lower -3dB frequency, f1 is %.0f Hz and the upper -3dB frequency, f2 is %.0f Hz ",f1,f2)
\ No newline at end of file diff --git a/608/CH28/EX28.11/28_11.sce b/608/CH28/EX28.11/28_11.sce new file mode 100755 index 000000000..706669966 --- /dev/null +++ b/608/CH28/EX28.11/28_11.sce @@ -0,0 +1,31 @@ +//Problem 28.11: In an L–R–C series network, the inductance, L = 8 mH, the capacitance, C = 0.3 μF, and the resistance, R = 15 ohm. Determine the current flowing in the circuit when the input voltage is 7.56/_0° V and the frequency is (a) the resonant frequency, (b) a frequency 3% above the resonant frequency. Find also (c) the impedance of the circuit when the frequency is 3% above the resonant frequency.
+
+//initializing the variables:
+R = 15; // in ohms
+L = 0.008; // IN Henry
+C = 0.3e-6; // IN fARADS
+rv = 7.56; //in volts
+thetav = 0; // in degrees
+x = 0.03;
+
+//calculation:
+//Resonant frequency,
+fr = 1/(2*%pi*((L*C)^0.5))
+wr = 2*%pi*fr
+//At resonance,
+Zr = R
+//voltage
+V = rv*cos(thetav*%pi/180) + %i*rv*sin(thetav*%pi/180)
+//Current at resonance
+Ir = V/Zr
+//Q-factor at resonance, Q = wr*L/R
+Qr = wr*L/R
+//If the frequency is 3% above fr, then
+del = x
+I = Ir/(1 + (2*del*Qr*%i))
+Z = V/I
+
+printf("\n\n Result \n\n")
+printf("\n (a)Current at resonance, Ir is %.2f A ",Ir)
+printf("\n (b)current flowing in the circuit when frequency 3 percent above the resonant frequency is %.4f + (%.4f)i A ",real(I), imag(I))
+printf("\n (c)impedance of the circuit when the frequency is 3 percent above the resonant frequency is %.0f + (%.2f)i A ",real(Z), imag(Z))
\ No newline at end of file diff --git a/608/CH29/EX29.01/29_01.sce b/608/CH29/EX29.01/29_01.sce new file mode 100755 index 000000000..dd18c145b --- /dev/null +++ b/608/CH29/EX29.01/29_01.sce @@ -0,0 +1,24 @@ +//Problem 29.01: A coil of inductance 5 mH and resistance 10 ohm is connected in parallel with a 250 nF capacitor across a 50 V variable-frequency supply. Determine (a) the resonant frequency, (b) the dynamic resistance, (c) the current at resonance, and (d) the circuit Q-factor at resonance.
+
+//initializing the variables:
+R = 10; // in ohms
+L = 0.005; // IN Henry
+C = 0.25e-6; // IN fARADS
+V = 50; //in volts
+
+//calculation:
+//Resonant frequency, for parallel
+fr = ((1/(L*C) - ((R^2)/(L^2)))^0.5)/(2*%pi)
+//dynamic resistance
+Rd = L/(C*R)
+//Current at resonance
+Ir = V/Rd
+wr = 2*%pi*fr
+//Q-factor at resonance, Q = wr*L/R
+Qr = wr*L/R
+
+printf("\n\n Result \n\n")
+printf("\n (a)Resonance frequency is %.0f Hz ",fr)
+printf("\n (b)dynamic resistance %.0f ohm ",Rd)
+printf("\n (c)Current at resonance, Ir is %.3f A ",Ir)
+printf("\n (d)Q-factor at resonance is %.1f ",Qr)
\ No newline at end of file diff --git a/608/CH29/EX29.02/29_02.sce b/608/CH29/EX29.02/29_02.sce new file mode 100755 index 000000000..9b5f9e7e0 --- /dev/null +++ b/608/CH29/EX29.02/29_02.sce @@ -0,0 +1,22 @@ +//Problem 29.02: In the parallel network of Figure 29.6, inductance, L = 100 mH and capacitance, C = 40 μF. Determine the resonant frequency for the network if (a) RL = 0 and (b) RL = 30 ohm.
+
+//initializing the variables:
+RL1 = 0; // in ohms
+RL2 = 30; // in ohms
+L = 0.100; // IN Henry
+C = 40e-6; // IN fARADS
+V = 50; //in volts
+
+//calculation:
+//for RL1
+//Resonant frequency,
+wr1 = (1/(L*C))^0.5
+fr1 = wr1/(2*%pi)
+//for RL2
+//Resonant frequency,
+wr2 = (1/(L*C) - ((RL2^2)/(L^2)))^0.5
+fr2 = wr2/(2*%pi)
+
+printf("\n\n Result \n\n")
+printf("\n (a)Resonance frequency at RL = 0 is %.1f Hz",fr1)
+printf("\n (b)Resonance frequency at RL = 30 ohm is %.1f Hz\n",fr2)
\ No newline at end of file diff --git a/608/CH29/EX29.03/29_03.sce b/608/CH29/EX29.03/29_03.sce new file mode 100755 index 000000000..90da727af --- /dev/null +++ b/608/CH29/EX29.03/29_03.sce @@ -0,0 +1,34 @@ +//Problem 29.03: A coil of inductance 120 mH and resistance 150 ohm is connected in parallel with a variable capacitor across a 20 V, 4 kHz supply. Determine for the condition when the supply current is a minimum, (a) the capacitance of the capacitor, (b) the dynamic resistance, (c) the supply current, (d) the Q-factor, (e) the band-width, (f) the upper and lower -3 dB frequencies, and (g) the value of the circuit impedance at the -3 dB frequencies.
+
+//initializing the variables:
+R = 150; // in ohms
+L = 0.120; // IN Henry
+V = 20; //in volts
+fr = 4000; // in Hz
+
+//calculation:
+//capacitance, C
+C = 1/(L*[(2*%pi*fr)^2 + ((R^2)/(L^2))])
+Rd = L/(C*R)
+//Current at resonance
+Ir = V/Rd
+wr = 2*%pi*fr
+//Q-factor at resonance, Q = wr*L/R
+Qr = wr*L/R
+//bandwidth,.(f2 − f1)
+bw = fr/Qr
+//upper half-power frequency, f2
+f2 = (bw + ((bw^2) + 4*(fr^2))^0.5)/2
+//lower half-power frequency, f1
+f1 = f2 - bw
+//impedance at the −3 dB frequencies
+Z = Rd/(2^0.5)
+
+printf("\n\n Result \n\n")
+printf("\n (a)the capacitance of the capacitor,C is %.2E F",C)
+printf("\n (b)dynamic resistance %.2E ohm ",Rd)
+printf("\n (c)Current at resonance, Ir is %.3E A ",Ir)
+printf("\n (d)Q-factor at resonance is %.2f ",Qr)
+printf("\n (e)bandwidth is %.0f Hz ",bw)
+printf("\n (f)the upper half-power frequency, f2 is %.0f Hz and the lower half-power frequency, f1 is %.0f Hz ",f2,f1)
+printf("\n (g)impedance at the −3 dB frequencies is %.3E ohm",Z)
\ No newline at end of file diff --git a/608/CH29/EX29.04/29_04.sce b/608/CH29/EX29.04/29_04.sce new file mode 100755 index 000000000..e00a0d3a1 --- /dev/null +++ b/608/CH29/EX29.04/29_04.sce @@ -0,0 +1,14 @@ +//Problem 29.03: A two-branch parallel network is shown in Figure 29.8. Determine the resonant frequency of the network.
+
+//initializing the variables:
+RL = 5; // in ohms
+L = 0.002; // IN Henry
+C = 25e-6; // IN fARADS
+Rc = 3; // in ohms
+
+//calculation:
+//Resonant frequency, for parallel
+fr = (1/(2*%pi*((L*C)^0.5)))*((RL^2 - (L/C))/(Rc^2 - (L/C)))^0.5
+
+printf("\n\n Result \n\n")
+printf("\n resonant frequency, fr is %.2f Hz",fr)
\ No newline at end of file diff --git a/608/CH29/EX29.05/29_05.sce b/608/CH29/EX29.05/29_05.sce new file mode 100755 index 000000000..d6ac22778 --- /dev/null +++ b/608/CH29/EX29.05/29_05.sce @@ -0,0 +1,18 @@ +//Problem 29.05: Determine for the parallel network shown in Figure 29.9 the values of inductance L for which the network is resonant at a frequency of 1 kHz.
+
+//initializing the variables:
+RL = 3; // in ohms
+fr = 1000; // in Hz
+Xc = 10; // IN ohms
+Rc = 4; // in ohms
+
+//calculation:
+XL1 = (((Rc^2 + Xc^2)/Xc) + ((((Rc^2 + Xc^2)/Xc)^2) - 4*(RL^2))^0.5)/2
+XL2 = (((Rc^2 + Xc^2)/Xc) - ((((Rc^2 + Xc^2)/Xc)^2) - 4*(RL^2))^0.5)/2
+wr = 2*%pi*fr
+//inductance
+L1 = XL1/wr
+L2 = XL2/wr
+
+printf("\n\n Result \n\n")
+printf("\n inductance is either %.2E H or %.2E H",L1, L2)
\ No newline at end of file diff --git a/608/CH29/EX29.06/29_06.sce b/608/CH29/EX29.06/29_06.sce new file mode 100755 index 000000000..9f3b1ad34 --- /dev/null +++ b/608/CH29/EX29.06/29_06.sce @@ -0,0 +1,11 @@ +//Problem 29.06: A capacitor having a Q-factor of 300 is connected in parallel with a coil having a Q-factor of 60. Determine the overall Q-factor of the parallel combination.
+
+//initializing the variables:
+QL = 60; // Q-factor
+Qc = 300; // Q-factor
+
+//calculation:
+QT = QL*Qc/(QL + Qc)
+
+printf("\n\n Result \n\n")
+printf("\n the overall Q-factor is %.0f \n",QT)
\ No newline at end of file diff --git a/608/CH29/EX29.07/29_07.sce b/608/CH29/EX29.07/29_07.sce new file mode 100755 index 000000000..8c02ee8f3 --- /dev/null +++ b/608/CH29/EX29.07/29_07.sce @@ -0,0 +1,22 @@ +//Problem 29.07: In an LR–C network, the capacitance is 10.61 nF, the bandwidth is 500 Hz and the resonant frequency is 150 kHz. Determine for the circuit (a) the Q-factor, (b) the dynamic resistance, and (c) the magnitude of the impedance when the supply frequency is 0.4% greater than the tuned frequency.
+
+//initializing the variables:
+C = 10.61E-9; // in Farad
+bw = 500; // in Hz
+fr = 150000; // in Hz
+x = 0.004
+
+//calculation:
+//Q-factor
+Q = fr/bw
+wr = 2*%pi*fr
+//dynamic resistance, RD
+Rd = L*Q/(C*wr*L)
+del = x
+Z = Rd/(1 + (2*del*Q*%i))
+Zmag = (real(Z)^2 + imag(Z)^2)^0.5
+
+printf("\n\n Result \n\n")
+printf("\n (a)Q-factor %.2f",Q)
+printf("\n (b)dynamic resistance %.2E ohm",Rd)
+printf("\n (c)magnitude of the impedance %.2E ohm",Zmag)
\ No newline at end of file diff --git a/608/CH3/EX3.01/3_01.sce b/608/CH3/EX3.01/3_01.sce new file mode 100755 index 000000000..f756e01e8 --- /dev/null +++ b/608/CH3/EX3.01/3_01.sce @@ -0,0 +1,15 @@ +//Problem 3.01: The resistance of a 5 m length of wire is 600 ohms. Determine (a) the resistance of an 8 m length of the same wire, and (b) the length of the same wire when the resistance is 420 ohms.
+
+//initializing the variables:
+R = 600; // in ohms
+L = 5; // in meter
+L1 = 8; // in meter
+R2 = 420; // in ohms
+
+//calculation:
+R1 = R*L1/L
+L2 = R2*L/R
+
+printf("\n\nResult\n\n")
+printf("\n(a)Resistance %.0f Ohms",R1)
+printf("\n(b)Length: %.1f meters(m)\n",L2)
\ No newline at end of file diff --git a/608/CH3/EX3.02/3_02.sce b/608/CH3/EX3.02/3_02.sce new file mode 100755 index 000000000..927c3434e --- /dev/null +++ b/608/CH3/EX3.02/3_02.sce @@ -0,0 +1,15 @@ +//Problem 3.02: A piece of wire of cross-sectional area 2 mm2 has a resistance of 300 ohms. Find (a) the resistance of a wire of the same length and material if the cross-sectional area is 5 mm2, (b) the cross-sectional area of a wire of the same length and material of resistance 750 ohms.
+
+//initializing the variables:
+R = 300; // in ohms
+A = 2; // in mm2
+A1 = 5; // in mm2
+R2 = 750; // in ohms
+
+//calculation:
+R1 = R*A/A1
+A2 = R*A/R2
+
+printf("\n\nResult\n\n")
+printf("\n(a)Resistance %.0f Ohms",R1)
+printf("\n(b)C.S.A: %.1f mm2\n",A2)
\ No newline at end of file diff --git a/608/CH3/EX3.03/3_03.sce b/608/CH3/EX3.03/3_03.sce new file mode 100755 index 000000000..028ba8c1f --- /dev/null +++ b/608/CH3/EX3.03/3_03.sce @@ -0,0 +1,14 @@ +//Problem 3.03: A wire of length 8 m and cross-sectional area 3 mm2 has a resistance of 0.16 ohms. If the wire is drawn out until its crosssectional area is 1 mm2, determine the resistance of the wire.
+
+//initializing the variables:
+R = 0.16; // in ohms
+A = 3; // in mm2
+L = 8; // in m
+A1 = 1; // in mm2
+
+//calculation:
+L1 = L*3
+R1 = R*A*L1/(A1*L)
+
+printf("\n\nResult\n\n")
+printf("\nResistance %.2f Ohms\n",R1)
diff --git a/608/CH3/EX3.04/3_04.sce b/608/CH3/EX3.04/3_04.sce new file mode 100755 index 000000000..ba6bd1ed8 --- /dev/null +++ b/608/CH3/EX3.04/3_04.sce @@ -0,0 +1,12 @@ +//Problem 3.04: Calculate the resistance of a 2 km length of aluminium overhead power cable if the cross-sectional area of the cable is 100 mm2. Take the resistivity of aluminium to be 0.03E-6 ohm m.
+
+//initializing the variables:
+A = 100E-6; // in m2
+L = 2000; // in m
+p = 0.03E-6; // in ohm m
+
+//calculation:
+R = p*L/A
+
+printf("\n\nResult\n\n")
+printf("\nResistance %.1f Ohms\n",R)
diff --git a/608/CH3/EX3.05/3_05.sce b/608/CH3/EX3.05/3_05.sce new file mode 100755 index 000000000..764985d56 --- /dev/null +++ b/608/CH3/EX3.05/3_05.sce @@ -0,0 +1,12 @@ +//Problem 3.05: Calculate the cross-sectional area, in mm2, of a piece of copper wire, 40 m in length and having a resistance of 0.25 ohms. Take the resistivity of copper as 0.02E-6ohm m.
+
+//initializing the variables:
+R = 0.25; // in ohms
+L = 40; // in m
+p = 0.02E-6; // in ohm m
+
+//calculation:
+A = p*L*1E6/R
+
+printf("\n\nResult\n\n")
+printf("\nC.S.A %.1f Ohms\n",A)
diff --git a/608/CH3/EX3.06/3_06.sce b/608/CH3/EX3.06/3_06.sce new file mode 100755 index 000000000..cc12aea3b --- /dev/null +++ b/608/CH3/EX3.06/3_06.sce @@ -0,0 +1,12 @@ +//Problem 3.06: The resistance of 1.5 km of wire of cross-sectional area 0.17 mm2 is 150 ohms. Determine the resistivity of the wire.
+
+//initializing the variables:
+R = 150; // in ohms
+L = 1500; // in m
+A = 0.17E-6; // in m2
+
+//calculation:
+p = R*A/L
+
+printf("\n\nResult\n\n")
+printf("\nresistivity %.3E Ohm m\n",p)
\ No newline at end of file diff --git a/608/CH3/EX3.07/3_07.sce b/608/CH3/EX3.07/3_07.sce new file mode 100755 index 000000000..ac0a3c3de --- /dev/null +++ b/608/CH3/EX3.07/3_07.sce @@ -0,0 +1,14 @@ +//Problem 3.07: Determine the resistance of 1200 m of copper cable having a diameter of 12 mm if the resistivity of copper is 1.7E-8 ohm m.
+
+//initializing the variables:
+d = 0.012; // in m
+L = 1200; // in m
+p = 1.7E-8; // in ohm m
+pi = 3.14;
+
+//calculation:
+A = pi*d*d/4
+R = p*L/A
+
+printf("\n\nResult\n\n")
+printf("\nresistance %.3f Ohm\n",R)
\ No newline at end of file diff --git a/608/CH3/EX3.08/3_08.sce b/608/CH3/EX3.08/3_08.sce new file mode 100755 index 000000000..a38d5e3eb --- /dev/null +++ b/608/CH3/EX3.08/3_08.sce @@ -0,0 +1,14 @@ +//Problem 3.08: A coil of copper wire has a resistance of 100 ohms when its temperature is 0°C. Determine its resistance at 70°C if the temperature coefficient of resistance of copper at 0°C is 0.0043/°C
+
+//initializing the variables:
+R0 = 100; // in ohms
+T0 = 0; // in °C
+T1 = 70; // in °C
+a0 = 0.0043; // in per°C
+pi = 3.14;
+
+//calculation:
+R70 = R0*[1 + (a0*T1)]
+
+printf("\n\nResult\n\n")
+printf("\nresistance %.1f Ohm\n",R70)
\ No newline at end of file diff --git a/608/CH3/EX3.09/3_09.sce b/608/CH3/EX3.09/3_09.sce new file mode 100755 index 000000000..6be6ce8a8 --- /dev/null +++ b/608/CH3/EX3.09/3_09.sce @@ -0,0 +1,14 @@ +//Problem 3.09: An aluminium cable has a resistance of 27 ohms at a temperature of 35°C. Determine its resistance at 0°C. Take the temperature coefficient of resistance at 0°C to be 0.0038/°C
+
+//initializing the variables:
+R1 = 27; // in ohms
+T0 = 0; // in °C
+T1 = 35; // in °C
+a0 = 0.0038; // in per°C
+pi = 3.14;
+
+//calculation:
+R0 = R1/[1 + (a0*T1)]
+
+printf("\n\nResult\n\n")
+printf("\nresistance %.2f Ohm\n",R0)
\ No newline at end of file diff --git a/608/CH3/EX3.10/3_10.sce b/608/CH3/EX3.10/3_10.sce new file mode 100755 index 000000000..1a7eb6c38 --- /dev/null +++ b/608/CH3/EX3.10/3_10.sce @@ -0,0 +1,14 @@ +//Problem 3.10: A carbon resistor has a resistance of 1 kohms at 0°C. Determine its resistance at 80°C. Assume that the temperature coefficient of resistance for carbon at 0°C is 0.0005/°C
+
+//initializing the variables:
+R0 = 1000; // in ohms
+T0 = 0; // in °C
+T1 = 80; // in °C
+a0 = -0.0005; // in per°C
+pi = 3.14;
+
+//calculation:
+R80 = R0*[1 + (a0*T1)]
+
+printf("\n\nResult\n\n")
+printf("\nresistance %.0f Ohm\n",R80)
\ No newline at end of file diff --git a/608/CH3/EX3.11/3_11.sce b/608/CH3/EX3.11/3_11.sce new file mode 100755 index 000000000..6d5575e3a --- /dev/null +++ b/608/CH3/EX3.11/3_11.sce @@ -0,0 +1,14 @@ +//Problem 3.11: A coil of copper wire has a resistance of 10 ohms at 20°C. If the temperature coefficient of resistance of copper at 20°C is 0.004/°C determine the resistance of the coil when the temperature rises to 100°C.
+
+//initializing the variables:
+R20 = 10; // in ohms
+T0 = 20; // in °C
+T1 = 100; // in °C
+a20 = 0.004; // in per°C
+pi = 3.14;
+
+//calculation:
+R100 = R20*[1 + (a20)*(T1 - T0)]
+
+printf("\n\nResult\n\n")
+printf("\nresistance %.1f Ohm\n",R100)
\ No newline at end of file diff --git a/608/CH3/EX3.12/3_12.sce b/608/CH3/EX3.12/3_12.sce new file mode 100755 index 000000000..c59ced9b7 --- /dev/null +++ b/608/CH3/EX3.12/3_12.sce @@ -0,0 +1,14 @@ +//Problem 3.12: The resistance of a coil of aluminium wire at 18°C is 200 ohms. The temperature of the wire is increased and the resistance rises to 240 ohms. If the temperature coefficient of resistance of aluminium is 0.0039/°C at 18°C determine the temperature to which the coil has risen.
+
+//initializing the variables:
+R18 = 200; // in ohms
+R1 = 240; // in ohms
+T0 = 18; // in °C
+a18 = 0.0039; // in per°C
+pi = 3.14;
+
+//calculation:
+T1 = (((R1/R18)-1)/a18) + T0
+
+printf("\n\nResult\n\n")
+printf("\nTemperature %.2f °C\n",T1)
\ No newline at end of file diff --git a/608/CH3/EX3.13/3_13.sce b/608/CH3/EX3.13/3_13.sce new file mode 100755 index 000000000..f6c7fc3c2 --- /dev/null +++ b/608/CH3/EX3.13/3_13.sce @@ -0,0 +1,14 @@ +//Problem 3.13: Some copper wire has a resistance of 200 ohms at 20°C. A current is passed through the wire and the temperature rises to 90°C. Determine the resistance of the wire at 90°C, correct to thenearest ohm, assuming that the temperature coefficient of resistance is 0.004/°C at 0°C.
+
+//initializing the variables:
+R20 = 200; // in ohms
+T0 = 20; // in °C
+T1 = 90; // in °C
+a0 = 0.004; // in per°C
+pi = 3.14;
+
+//calculation:
+R90 = R20*[1 + (a0*T1)]/[1 + (a0*T0)]
+
+printf("\n\nResult\n\n")
+printf("\nResistance %.0f ohms\n",R90)
\ No newline at end of file diff --git a/608/CH30/EX30.01/30_01.sce b/608/CH30/EX30.01/30_01.sce new file mode 100755 index 000000000..8d98d0a0e --- /dev/null +++ b/608/CH30/EX30.01/30_01.sce @@ -0,0 +1,27 @@ +//Problem 30.01: Use Kirchhoff’s laws to find the current flowing in each branch of the network shown in Figure 30.3.
+
+//initializing the variables:
+rv1 = 100; // in volts
+rv2 = 50; // in volts
+thetav1 = 0; // in degrees
+thetav2 = 90; // in degrees
+R1 = 25; // in ohm
+R2 = 20; // in ohm
+R3 = 10; // in ohm
+
+//calculation:
+//voltage
+V1 = rv1*cos(thetav1*%pi/180) + %i*rv1*sin(thetav1*%pi/180)
+V2 = rv2*cos(thetav2*%pi/180) + %i*rv2*sin(thetav2*%pi/180)
+//The branch currents and their directions are labelled as shown in Figure 30.4
+//Two loops are chosen. loop ABEF, and loop BCDE
+//using kirchoff rule in 3 loops
+//two eqns obtained
+//(R1 + R2)*I1 + R2*I2 = V1
+//R2*I1 + (R2 + R3)*I2 = V2
+I1 = (3*V1 - 2*V2)/(3*(R1 + R2) - 2*(R2))
+I2 = (V2 - R2*I1)/(R2 + R3)
+I = I1 + I2
+
+printf("\n\n Result \n\n")
+printf("\n current, I1 is %.2f + (%.2f)i A, current, I2 is %.2f + (%.2f)i A and total current, I is %.2f + (%.2f)i A",real(I1), imag(I1),real(I2), imag(I2),real(I), imag(I))
\ No newline at end of file diff --git a/608/CH30/EX30.02/30_02.sce b/608/CH30/EX30.02/30_02.sce new file mode 100755 index 000000000..d48f214fa --- /dev/null +++ b/608/CH30/EX30.02/30_02.sce @@ -0,0 +1,39 @@ +//Problem 30.02: Determine the current flowing in the 2 ohm resistor of the circuit shown in Figure 30.5 using Kirchhoff’s laws. Find also the power dissipated in the 3 ohm resistance.
+
+//initializing the variables:
+V = 8; // in volts
+R1 = 1; // in ohm
+R2 = 2; // in ohm
+R3 = 3; // in ohm
+R4 = 4; // in ohm
+R5 = 5; // in ohm
+R6 = 6; // in ohm
+
+//calculation:
+//Currents and their directions are assigned as shown in Figure 30.6.
+//Three loops are chosen since three unknown currents are required. The choice of loop directions is arbitrary. loop ABCDE, and loop EDGF and loop DCHG
+//using kirchoff rule in 3 loops
+//three eqns obtained
+//R5*I1 + (R6 + R4)*I2 - R4*I3 = V
+//-1*R1*I1 + (R6 + R1)*I2 + R2*I3 = 0
+// R3*I1 - (R3 + R4)*I2 + (R2 + R3 + R4)*I3 = 0
+//using determinants
+d1 = [V (R6 + R4) -1*R4; 0 (R6 + R1) R2; 0 (-1*(R3 + R4)) (R2 + R3 + R4)]
+D1 = det(d1)
+d2 = [R5 V -1*R4; -1*R1 0 R2; R3 0 (R2 + R3 + R4)]
+D2 = det(d2)
+d3 = [R5 (R6 + R4) V; -1*R1 (R6 + R1) 0; R3 (-1*(R3 + R4)) 0]
+D3 = det(d3)
+d = [R5 (R6 + R4) -1*R4; -1*R1 (R6 + R1) R2; R3 (-1*(R3 + R4)) (R2 + R3 + R4)]
+D = det(d)
+I1 = D1/D
+I2 = D2/D
+I3 = D3/D
+//Current in the 2 ohm resistance
+I = I1 - I2 + I3
+//power dissipated in the 3 ohm resistance
+P3 = R3*I^2
+
+printf("\n\n Result \n\n")
+printf("\n (a)current through 2 ohm resistor is %.2f A",I2)
+printf("\n (b)power dissipated in the 3 ohm resistor is %.2f W",P3)
\ No newline at end of file diff --git a/608/CH30/EX30.03/30_03.sce b/608/CH30/EX30.03/30_03.sce new file mode 100755 index 000000000..970bb40a9 --- /dev/null +++ b/608/CH30/EX30.03/30_03.sce @@ -0,0 +1,22 @@ +//Problem 30.03: For the a.c. network shown in Figure 30.7, determine the current flowing in each branch using Kirchhoff’s laws.
+
+//initializing the variables:
+E1 = 5 + %i*0; // in volts
+E2 = 2 + %i*4; // in volts
+Z1 = 3 + %i*4; // in ohm
+Z2 = 2 - %i*5; // in ohm
+Z3 = 6 + %i*8; // in ohm
+
+//calculation:
+//Currents I1 and I2 with their directions are shown in Figure 30.8.
+//Two loops are chosen with their directions both clockwise.loop ABEF and loop BCDE,
+//using kirchoff rule in 3 loops
+//two eqns obtained
+//(Z1 + Z3)*I1 - Z3*I2 = E1
+//-1*Z3*I1 + (Z2 + Z3)*I2 = E2
+I1 = ((Z2 + Z3)*E1 + Z3*E2)/((Z2 + Z3)*(Z1 + Z3) - Z3*Z3)
+I2 = -1*(E1 - (Z1 + Z3)*I1)/Z3
+I3 = I1 - I2
+
+printf("\n\n Result \n\n")
+printf("\n current, I1 is %.2f + (%.2f)i A, current, I2 is %.2f + (%.2f)i A and current in Z3, I3 is %.3f + (%.3f)i A",real(I1), imag(I1),real(I2), imag(I2),real(I3), imag(I3))
\ No newline at end of file diff --git a/608/CH30/EX30.04/30_04.sce b/608/CH30/EX30.04/30_04.sce new file mode 100755 index 000000000..51570c9a2 --- /dev/null +++ b/608/CH30/EX30.04/30_04.sce @@ -0,0 +1,44 @@ +//Problem 30.04: For the network shown in Figure 30.9, use Kirchhoff’s laws to determine the magnitude of the current in the (4 + i3)ohm impedance.
+
+//initializing the variables:
+rv1 = 10; // in volts
+rv2 = 12; // in volts
+rv3 = 15; // in volts
+thetav1 = 0; // in degrees
+thetav2 = 0; // in degrees
+thetav3 = 0; // in degrees
+R1 = 4; // in ohm
+R2 = -1*5*%i; // in ohm
+R3 = 8; // in ohm
+R4 = 4; // in ohm
+R5 = %i*3; // in ohm
+
+//calculation:
+//voltages
+V1 = rv1*cos(thetav1*%pi/180) + %i*rv1*sin(thetav1*%pi/180)
+V2 = rv2*cos(thetav2*%pi/180) + %i*rv2*sin(thetav2*%pi/180)
+V3 = rv3*cos(thetav3*%pi/180) + %i*rv3*sin(thetav3*%pi/180)
+//Currents I1, I2 and I3 with their directions are shown in Figure 30.10.
+//Three loops are chosen. The choice of loop directions is arbitrary. loop ABGH, and loopBCFG and loop CDEF
+Z4 = R4 + R5
+//using kirchoff rule in 3 loops
+//three eqns obtained
+//R1*I1 + R2*I2 = V1 + V2
+//-1*R3*I1 + (R3 + R2)*I2 + R3*I3 = V2 + V3
+// -1*R3*I1 + R3*I2 + (R3 + Z4)*I3 = V3
+//using determinants
+d1 = [(V1 + V2) R2 0; (V2 + V3) (R3 + R2) R3; V3 R3 (R3 + Z4)]
+D1 = det(d1)
+d2 = [R1 (V1 + V2) 0; -1*R3 (V2 + V3) R3; -1*R3 V3 (R3 + Z4)]
+D2 = det(d2)
+d3 = [R1 R2 (V1 + V2); -1*R3 (R3 + R2) (V2 + V3); -1*R3 R3 V3]
+D3 = det(d3)
+d = [R1 R2 0; -1*R3 (R3 + R2) R3; -1*R3 R3 (R3 + Z4)]
+D = det(d)
+I1 = D1/D
+I2 = D2/D
+I3 = D3/D
+I3mag = (real(I3)^2 + imag(I3)^2)^0.5
+
+printf("\n\n Result \n\n")
+printf("\n magnitude of the current through (4 + i3)ohm impedance is %.2f A",I3mag)
\ No newline at end of file diff --git a/608/CH31/EX31.01/31_01.sce b/608/CH31/EX31.01/31_01.sce new file mode 100755 index 000000000..86228e4ec --- /dev/null +++ b/608/CH31/EX31.01/31_01.sce @@ -0,0 +1,36 @@ +//Problem 31.01: Use mesh-current analysis to determine the current flowing in (a) the 5 ohm resistance, and (b) the 1ohm resistance of the d.c. circuit shown in Figure 31.2.
+
+//initializing the variables:
+V1 = 4; // in volts
+V2 = 5; // in volts
+R1 = 3; // in ohm
+R2 = 5; // in ohm
+R3 = 4; // in ohm
+R4 = 1; // in ohm
+R5 = 6; // in ohm
+R6 = 8; // in ohm
+
+//calculation:
+//The mesh currents I1, I2 and I3 are shown in Figure 31.2. Using Kirchhoff’s voltage law in 3 loops
+//three eqns obtained
+//(R1 + R2)*I1 - R2*I2 = V1
+//-1*R2*I1 + (R2 + R3 + R4 + R5)*I2 - R4*I3 = 0
+// -1*R4*I2 + (R4 + R6)*I3 = -1*V2
+//using determinants
+d1 = [V1 -1*R2 0; 0 (R2 + R3 + R4 + R5) -1*R4; -1*V2 -1*R4 (R4 + R6)]
+D1 = det(d1)
+d2 = [(R1 + R2) V1 0; -1*R2 0 -1*R4; 0 -1*V2 (R4 + R6)]
+D2 = det(d2)
+d3 = [(R1 + R2) -1*R2 V1; -1*R2 (R2 + R3 + R4 + R5) 0; 0 -1*R4 -1*V2]
+D3 = det(d3)
+d = [(R1 + R2) -1*R2 0; -1*R2 (R2 + R3 + R4 + R5) -1*R4; 0 -1*R4 (R4 + R6)]
+D = det(d)
+I1 = D1/D
+I2 = D2/D
+I3 = D3/D
+IR2 = I1 - I2
+IR4 = I2 - I3
+
+printf("\n\n Result \n\n")
+printf("\n (a)current in the 5 ohm resistance is %.2f A",IR2)
+printf("\n (b)current in the 1 ohm resistance is %.2f A",IR4)
\ No newline at end of file diff --git a/608/CH31/EX31.02/31_02.sce b/608/CH31/EX31.02/31_02.sce new file mode 100755 index 000000000..16f18b47f --- /dev/null +++ b/608/CH31/EX31.02/31_02.sce @@ -0,0 +1,40 @@ +//Problem 31.02: For the a.c. network shown in Figure 31.3 determine, using mesh-current analysis, (a) the mesh currents I1 and I2 (b) the current flowing in the capacitor, and (c) the active power delivered by the 100/_0° V voltage source.
+
+//initializing the variables:
+rv = 100; // in volts
+thetav = 0; // in degrees
+R1 = 5; // in ohm
+R2 = -1*4*%i; // in ohm
+R3 = 4; // in ohm
+R4 = %i*3; // in ohm
+
+//calculation:
+//voltages
+V = rv*cos(thetav*%pi/180) + %i*rv*sin(thetav*%pi/180)
+//Currents I1, I2 with their directions are shown in Figure 31.03.
+//Two loops are chosen. The choice of loop directions is arbitrary.
+//using kirchoff rule in 2 loops
+//two eqns obtained
+//(R1 + R2)*I1 - R2*I2 = V
+//-1*R2*I1 + (R3 + R2 + R4)*I2 = 0
+//using determinants
+d1 = [V -1*R2; 0 (R3 + R2 + R4)]
+D1 = det(d1)
+d2 = [(R1 + R2) V; -1*R2 0]
+D2 = det(d2)
+d = [(R1 + R2) -1*R2; -1*R2 (R3 + R2 + R4)]
+D = det(d)
+I1 = D1/D
+I2 = D2/D
+I1mag = (real(I1)^2 + imag(I1)^2)^0.5
+//Current flowing in capacitor
+Ic = I1 - I2
+//Source power P
+phi = atan(imag(I1)/real(I1))
+P = V*I1mag*cos(phi)
+Icmag = (real(Ic)^2 + imag(Ic)^2)^0.5
+
+printf("\n\n Result \n\n")
+printf("\n (a)current, I1 is %.2f + (%.2f)i A, current, I2 is %.2f + (%.2f)i A",real(I1), imag(I1),real(I2), imag(I2))
+printf("\n (b)current in the capacitor is %.2f A",Icmag)
+printf("\n (c)Source power P is %.2f W",P)
\ No newline at end of file diff --git a/608/CH31/EX31.03/31_03.sce b/608/CH31/EX31.03/31_03.sce new file mode 100755 index 000000000..19cc488b6 --- /dev/null +++ b/608/CH31/EX31.03/31_03.sce @@ -0,0 +1,38 @@ +//Problem 31.03: A balanced star-connected 3-phase load is shown in Figure 31.4. Determine the value of the line currents IR, IY and IB using mesh-current analysis.
+
+//initializing the variables:
+rv1 = 415; // in volts
+rv2 = 415; // in volts
+thetav1 = 120; // in degrees
+thetav2 = 0; // in degrees
+R = 3 + %i*4; // in ohm
+
+//calculation:
+//voltages
+V1 = rv1*cos(thetav1*%pi/180) + %i*rv1*sin(thetav1*%pi/180)
+V2 = rv2*cos(thetav2*%pi/180) + %i*rv2*sin(thetav2*%pi/180)
+//Two mesh currents I1 and I2 are chosen as shown in Figure 31.4.
+//Two loops are chosen. The choice of loop directions is arbitrary.
+//using kirchoff rule in 2 loops
+//two eqns obtained
+//2*R*I1 - R*I2 = V1
+//-1*R*I1 + 2*R*I2 = V2
+//using determinants
+d1 = [V1 -1*R; V2 2*R]
+D1 = det(d1)
+d2 = [2*R V1; -1*R V2]
+D2 = det(d2)
+d = [2*R -1*R; -1*R 2*R]
+D = det(d)
+I1 = D1/D
+I2 = D2/D
+I1mag = (real(I1)^2 + imag(I1)^2)^0.5
+//line current IR
+IR = I1
+//line current IB
+IB = -1*I2
+//line current IY
+IY = I2 - I1
+
+printf("\n\n Result \n\n")
+printf("\n current, IR is %.2f + (%.2f)i A, current, IB is %.2f + (%.2f)i A and current, IY is %.2f + (%.2f)i A",real(IR), imag(IR),real(IB), imag(IB),real(IY), imag(IY))
\ No newline at end of file diff --git a/608/CH31/EX31.04/31_04.sce b/608/CH31/EX31.04/31_04.sce new file mode 100755 index 000000000..a5ad7b13c --- /dev/null +++ b/608/CH31/EX31.04/31_04.sce @@ -0,0 +1,21 @@ +//Problem 31.04: For the network shown in Figure 31.8, determine the voltage VAB, by using nodal analysis.
+
+//initializing the variables:
+ri = 20; // in amperes
+thetai = 0; // in degrees
+R1 = 10; // in ohm
+R2 = %i*3; // in ohm
+R3 = 4; // in ohm
+R4 = 16; // in ohm
+
+//calculation:
+//current
+I = ri*cos(thetai*%pi/180) + %i*ri*sin(thetai*%pi/180)
+//Figure 31.8 contains two principal nodes (at 1 and B) and thus only one nodal equation is required. B is taken as the reference node and the equation for node 1 is obtained as follows. Applying Kirchhoff’s current law to node 1 gives:
+//IX + IY = I
+V1 = I/((1/R4) +(1/(R2 +R3)))
+IY = V1/(R2 + R3)
+VAB = IY*R3
+
+printf("\n\n Result \n\n")
+printf("\n voltage VAB is %.2f + (%.2f)i V",real(VAB), imag(VAB))
\ No newline at end of file diff --git a/608/CH31/EX31.05/31_05.sce b/608/CH31/EX31.05/31_05.sce new file mode 100755 index 000000000..7a612c1d8 --- /dev/null +++ b/608/CH31/EX31.05/31_05.sce @@ -0,0 +1,22 @@ +//Problem 31.05: Determine the value of voltage VXY shown in the circuit of Figure 31.9.
+
+//initializing the variables:
+rv1 = 8; // in volts
+rv2 = 8; // in volts
+thetav1 = 0; // in degrees
+thetav2 = 90; // in degrees
+R1 = 5; // in ohm
+R2 = %i*6; // in ohm
+R3 = 4; // in ohm
+R4 = 3; // in ohm
+
+//calculation:
+//voltages
+V1 = rv1*cos(thetav1*%pi/180) + %i*rv1*sin(thetav1*%pi/180)
+V2 = rv2*cos(thetav2*%pi/180) + %i*rv2*sin(thetav2*%pi/180)
+//The circuit contains no principal nodes. However, if point Y is chosen as the reference node then an equation may be written for node X assuming that current leaves point X by both branches
+VX = [(V1/(R1 + R3) + V2/(R2 + R4))/(1/(R1 + R3) + 1/(R2 + R4))]
+VXY = VX
+
+printf("\n\n Result \n\n")
+printf("\n voltage VXY is %.2f + (%.2f)i V",real(VXY), imag(VXY))
\ No newline at end of file diff --git a/608/CH31/EX31.06/31_06.sce b/608/CH31/EX31.06/31_06.sce new file mode 100755 index 000000000..9c6f82966 --- /dev/null +++ b/608/CH31/EX31.06/31_06.sce @@ -0,0 +1,24 @@ +//Problem 31.06: Use nodal analysis to determine the current flowing in each branch of the network shown in Figure 31.10.
+
+//initializing the variables:
+rv1 = 100; // in volts
+rv2 = 50; // in volts
+thetav1 = 0; // in degrees
+thetav2 = 90; // in degrees
+R1 = 25; // in ohm
+R2 = 20; // in ohm
+R3 = 10; // in ohm
+
+//calculation:
+//voltages
+V1 = rv1*cos(thetav1*%pi/180) + %i*rv1*sin(thetav1*%pi/180)
+V2 = rv2*cos(thetav2*%pi/180) + %i*rv2*sin(thetav2*%pi/180)
+//There are only two principal nodes in Figure 31.10 so only one nodal equation is required. Node 2 is taken as the reference node.
+//The equation at node 1 is I1 + I2 + I3 = 0
+Vn1 = [(V1/R1 + V2/R3)/(1/R1 + 1/R2 + 1/R3)]
+I1 = (Vn1 - V1)/R1
+I2 = Vn1/R2
+I3 = (Vn1 - V2)/R3
+
+printf("\n\n Result \n\n")
+printf("\n current, I1 is %.2f + (%.2f)i A, current, I2 is %.2f + (%.2f)i A and current, I3 is %.2f + (%.2f)i A",real(I1), imag(I1),real(I2), imag(I2),real(I3), imag(I3))
\ No newline at end of file diff --git a/608/CH31/EX31.07/31_07.sce b/608/CH31/EX31.07/31_07.sce new file mode 100755 index 000000000..cd396bd9e --- /dev/null +++ b/608/CH31/EX31.07/31_07.sce @@ -0,0 +1,44 @@ +//Problem 31.07: In the network of Figure 31.11 use nodal analysis to determine (a) the voltage at nodes 1 and 2, (b) the current in the j4 ohm inductance, (c) the current in the 5 ohm resistance, and (d) the magnitude of the active power dissipated in the 2.5 ohm resistance.
+
+//initializing the variables:
+rv1 = 25; // in volts
+rv2 = 25; // in volts
+thetav1 = 0; // in degrees
+thetav2 = 90; // in degrees
+R1 = 2; // in ohm
+R2 = -1*%i*4; // in ohm
+R3 = 5; // in ohm
+R4 = %i*4; // in ohm
+R5 = 2.5; // in ohm
+
+//calculation:
+//voltages
+V1 = rv1*cos(thetav1*%pi/180) + %i*rv1*sin(thetav1*%pi/180)
+V2 = rv2*cos(thetav2*%pi/180) + %i*rv2*sin(thetav2*%pi/180)
+//The equation at node 1
+//Vn1*(1/R1 + 1/R2 + 1/R3) - Vn2/R3 = V1/R1
+//The equation at node 2
+//Vn1*(-1/R3) + Vn2*(1/R4 + 1/R5 + 1/R3) = V2/R5
+//using determinants
+d1 = [V1/R1 -1/R3; V2/R5 (1/R4 + 1/R5 + 1/R3)]
+D1 = det(d1)
+d2 = [(1/R1 + 1/R2 + 1/R3) V1/R1; -1/R3 V2/R5]
+D2 = det(d2)
+d = [(1/R1 + 1/R2 + 1/R3) -1/R3; -1/R3 (1/R4 + 1/R5 + 1/R3)]
+D = det(d)
+Vn1 = D1/D
+Vn2 = D2/D
+//current in the j4 ohm inductance is given by:
+I4 = Vn2/R4
+//current in the 5 ohm resistance is given by:
+I3 = (Vn1 - Vn2)/R3
+//active power dissipated in the 2.5 ohm resistor is given by
+P5 = R5*((Vn2 - V2)/R5)^2
+//magnitude of the active power dissipated
+P5mag = (real(P5)^2 + imag(P5)^2)^0.5
+
+printf("\n\n Result \n\n")
+printf("\n (a) the voltage at nodes 1 and 2 is %.2f + (%.2f)i V and %.2f + (%.2f)i V",real(Vn1), imag(Vn1),real(Vn2), imag(Vn2))
+printf("\n (b)the current in the j4 ohm inductance is %.2f + (%.2f)i A",real(I4), imag(I4))
+printf("\n (c)the current in the 5 ohm resistance is %.2f + (%.2f)i A",real(I3), imag(I3))
+printf("\n (d) magnitude of the active power dissipated in the 2.5 ohm resistance is %.2f W",P5mag)
\ No newline at end of file diff --git a/608/CH31/EX31.08/31_08.sce b/608/CH31/EX31.08/31_08.sce new file mode 100755 index 000000000..a71af8f54 --- /dev/null +++ b/608/CH31/EX31.08/31_08.sce @@ -0,0 +1,36 @@ +//Problem 31.08: In the network shown in Figure 31.12 determine the voltage VXY using nodal analysis
+
+//initializing the variables:
+ri = 25; // in amperes
+thetai = 0; // in degrees
+R1 = 4; // in ohm
+R2 = %i*3; // in ohm
+R3 = 5; // in ohm
+R4 = %i*10; // in ohm
+R5 = %i*20; // in ohm
+
+//calculation:
+//current
+I = ri*cos(thetai*%pi/180) + %i*ri*sin(thetai*%pi/180)
+//Node 3 is taken as the reference node.
+//At node 1,
+//V1*(1/(R1 + R2) + 1/R3) - V2/R3 = I
+//The equation at node 2
+//V1*(-1/R3) + V2*(1/R4 + 1/R5 + 1/R3) = 0
+//using determinants
+d1 = [I -1/R3; 0 (1/R4 + 1/R5 + 1/R3)]
+D1 = det(d1)
+d2 = [(1/(R1 + R2) + 1/R3) I; -1/R3 0]
+D2 = det(d2)
+d = [(1/(R1 + R2) + 1/R3) -1/R3; -1/R3 (1/R4 + 1/R5 + 1/R3)]
+D = det(d)
+V1 = D1/D
+V2 = D2/D
+//the voltage between point X and node 3 is
+VX = V1*R2/(R1 + R2)
+//Thus the voltage
+VY = V2
+VXY = VX - VY
+
+printf("\n\n Result \n\n")
+printf("\n voltage VXY is %.2f + (%.2f)i V",real(VXY), imag(VXY))
\ No newline at end of file diff --git a/608/CH31/EX31.09/31_09.sce b/608/CH31/EX31.09/31_09.sce new file mode 100755 index 000000000..3d0f14dcd --- /dev/null +++ b/608/CH31/EX31.09/31_09.sce @@ -0,0 +1,41 @@ +//Problem 31.09: Use nodal analysis to determine the voltages at nodes 2 and 3 in Figure 31.13 and hence determine the current flowing in the 2 ohm resistor and the power dissipated in the 3 ohm resistor.
+
+//initializing the variables:
+V = 8; // in volts
+R1 = 1; // in ohm
+R2 = 2; // in ohm
+R3 = 3; // in ohm
+R4 = 4; // in ohm
+R5 = 5; // in ohm
+R6 = 6; // in ohm
+
+//calculation:
+//In Figure 31.13, the reference node is shown at point A.
+//At node 1,
+//V1*(1/R1 + 1/R6 + 1/R5) - V2/R1 - V3/R5 = V/R5
+//The equation at node 2
+//V1*(-1/R1) + V2*(1/R2 + 1/R1 + 1/R3) - V3/R3 = 0
+//At node 3
+// - V1/R5 - V2/R3 + V3*(1/R4 + 1/R3 + 1/R5) = -1*V/R5
+//using determinants
+d1 = [V/R5 -1/R1 -1/R5; 0 (1/R2 + 1/R1 + 1/R3) -1/R3; -1*V/R5 -1/R3 (1/R4 + 1/R3 + 1/R5)]
+D1 = det(d1)
+d2 = [(1/R1 + 1/R6 + 1/R5) V/R5 -1/R5; -1/R1 0 -1/R3; -1/R5 -1*V/R5 (1/R4 + 1/R3 + 1/R5)]
+D2 = det(d2)
+d3 = [(1/R1 + 1/R6 + 1/R5) -1/R1 V/R5; -1/R1 (1/R2 + 1/R1 + 1/R3) 0; -1/R5 -1/R3 -1*V/R5]
+D3 = det(d3)
+d = [(1/R1 + 1/R6 + 1/R5) -1/R1 -1/R5; -1/R1 (1/R2 + 1/R1 + 1/R3) -1/R3; -1/R5 -1/R3 (1/R4 + 1/R3 + 1/R5)]
+D = det(d)
+Vn1 = D1/D
+Vn2 = D2/D
+Vn3 = D3/D
+//the current in the 2 ohm resistor
+I2 = Vn2/R2
+//power dissipated in the 3 ohm resistance
+P3 = R3*((Vn2 - Vn3)/R3)^2
+
+printf("\n\n Result \n\n")
+printf("\n voltage at node 2 is %.2f V",Vn2)
+printf("\n voltage at node 3 is %.2f V",Vn3)
+printf("\n (a)current through 2 ohm resistor is %.2f A",I2)
+printf("\n (b)power dissipated in the 3 ohm resistor is %.2f W",P3)
\ No newline at end of file diff --git a/608/CH32/EX32.01/32_01.sce b/608/CH32/EX32.01/32_01.sce new file mode 100755 index 000000000..de9fd9215 --- /dev/null +++ b/608/CH32/EX32.01/32_01.sce @@ -0,0 +1,38 @@ +//Problem 32.01:A.c. sources of 100/_0° V and internal resistance 25 ohm and 50/_90° V and internal resistance 10 ohm, are connected in parallel across a 20 ohm load. Determine using the superposition theorem, the current in the 20 ohm load and the current in each voltage source
+
+//initializing the variables:
+rv1 = 100; // in volts
+rv2 = 50; // in volts
+thetav1 = 0; // in degrees
+thetav2 = 90; // in degrees
+r1 = 25; // in ohm
+R = 20; // in ohm
+r2 = 10; // in ohm
+
+//calculation:
+//voltage
+V1 = rv1*cos(thetav1*%pi/180) + %i*rv1*sin(thetav1*%pi/180)
+V2 = rv2*cos(thetav2*%pi/180) + %i*rv2*sin(thetav2*%pi/180)
+//The circuit diagram is shown in Figure 32.7. Following the above procedure:
+//The network is redrawn with the 50/_90° V source removed as shown in Figure 32.8
+//Currents I1, I2 and I3 are labelled as shown in Figure 32.8.
+I1 = V1/(r1 + r2*R/(R + r2))
+I2 = (r2/(r2 + R))*I1
+I3 = (R/(r2 + R))*I1
+//The network is redrawn with the 100/_0° V source removed as shown in Figure 32.9
+//Currents I4, I5 and I6 are labelled as shown in Figure 32.9.
+I4 = V2/(r2 + r1*R/(r1 + R))
+I5 = (r1/(r1 + R))*I4
+I6 = (R/(r1 + R))*I4
+//Figure 32.10 shows Figure 32.9 superimposed on Figure 32.8, giving the currents shown.
+//Current in the 20 ohm load,
+I20 = I2 + I5
+//Current in the 100/_0° V source
+IV1 = I1 - I6
+//Current in the 50/_90° V source
+IV2 = I4 - I3
+
+printf("\n\n Result \n\n")
+printf("\n (a)current in the 20 ohm load is %.3f + (%.3f)i A",real(I20), imag(I20))
+printf("\n (b)Current in the 100/_0° V source is %.3f + (%.3f)i A",real(IV1), imag(IV1))
+printf("\n (b)Current in the 50/_90° V source is %.3f + (%.3f)i A",real(IV2), imag(IV2))
\ No newline at end of file diff --git a/608/CH32/EX32.02/32_02.sce b/608/CH32/EX32.02/32_02.sce new file mode 100755 index 000000000..23238da4e --- /dev/null +++ b/608/CH32/EX32.02/32_02.sce @@ -0,0 +1,30 @@ +//Problem 32.02:Use the superposition theorem to determine the current in the 4 ohm resistor of the network shown in Figure 32.11.
+
+//initializing the variables:
+V1 = 12; // in volts
+V2 = 20; // in volts
+R1 = 5; // in ohm
+R2 = 4; // in ohm
+R3 = 2.5; // in ohm
+R4 = 6; // in ohm
+R5 = 2; // in ohm
+
+//calculation:
+//Removing the 20 V source gives the network shown in Figure 32.12.
+//Currents I1 and I2 are shown labelled in Figure 32.12
+Re1 = (R4*R5/(R4 + R5)) + R3
+Re2 = Re1*R2/(Re1 + R2) + R1
+I1 = V1/Re2
+I2 = (R2/(Re1 + R2))*I1
+//Removing the 12 V source from the original network gives the network shown in Figure 32.14.
+//Currents I3, I4 and I5 are shown labelled in Figure 32.14.
+Re3 = (R1*R2/(R1 + R2)) + R3
+Re4 = Re3*R4/(Re3 + R4) + R5
+I3 = V2/Re4
+I4 = (R4/(Re3 + R4))*I3
+I5 = (R1/(R1 + R2))*I4
+//Superimposing Figure 32.14 on Figure 32.12 shows that the current flowing in the 4 ohm resistor is given by
+Ir4 = I5 - I2
+
+printf("\n\n Result \n\n")
+printf("\ncurrent in the 4 ohm resistor of the network is %.3f A",Ir4)
\ No newline at end of file diff --git a/608/CH32/EX32.03/32_03.sce b/608/CH32/EX32.03/32_03.sce new file mode 100755 index 000000000..b4998c6bd --- /dev/null +++ b/608/CH32/EX32.03/32_03.sce @@ -0,0 +1,34 @@ +//Problem 32.03: Use the superposition theorem to obtain the current flowing in the (4 + i3) ohm impedance of Figure 32.16.
+
+//initializing the variables:
+rv1 = 30; // in volts
+rv2 = 30; // in volts
+thetav1 = 45; // in degrees
+thetav2 = -45; // in degrees
+R1 = 4; // in ohm
+R2 = 4; // in ohm
+R3 = %i*3; // in ohm
+R4 = -1*%i*10; // in ohm
+
+//calculation:
+//voltage
+V1 = rv1*cos(thetav1*%pi/180) + %i*rv1*sin(thetav1*%pi/180)
+V2 = rv2*cos(thetav2*%pi/180) + %i*rv2*sin(thetav2*%pi/180)
+//The network is redrawn with V2 removed, as shown in Figure 32.17.
+//Current I1 and I2 are shown in Figure 32.17. From Figure 32.17,
+Re1 = R4*(R2 + R3)/(R4 + R3 + R2)
+Re2 = Re1 + R1
+//current
+I1 = V1/Re2
+I2 = (R4/(R2 + R3 + R4))*I1
+//The original network is redrawn with V1 removed, as shown in Figure 32.18
+//Currents I3 and I4 are shown in Figure 32.18. From Figure 32.18,
+Re3 = R1*(R2 + R3)/(R1 + R3 + R2)
+Re4 = Re3 + R4
+I3 = V2/Re4
+I4 = (R1/(R2 + R3 + R1))*I3
+//If the network of Figure 32.18 is superimposed on the network of Figure 32.17, it can be seen that the current in the (4+i3) ohm impedance is given by
+Ir4i3 = I2 - I4
+
+printf("\n\n Result \n\n")
+printf("\ncurrent in the (4 + i3) ohm impedance of the network is %.3f + (%.3f)i A",real(Ir4i3), imag(Ir4i3))
\ No newline at end of file diff --git a/608/CH32/EX32.04/32_04.sce b/608/CH32/EX32.04/32_04.sce new file mode 100755 index 000000000..e927b3583 --- /dev/null +++ b/608/CH32/EX32.04/32_04.sce @@ -0,0 +1,46 @@ +//Problem 32.04: For the a.c. network shown in Figure 32.19 determine, using the superposition theorem, (a) the current in each branch, (b) the magnitude of the voltage across the(6 + i8) ohm impedance, and (c) the total active power delivered to the network.
+
+//initializing the variables:
+E1 = 5 + %i*0; // in volts
+E2 = 2 + %i*4; // in volts
+Z1 = 3 + %i*4; // in ohm
+Z2 = 2 - %i*5; // in ohm
+Z3 = 6 + %i*8; // in ohm
+
+//calculation:
+//The original network is redrawn with E2 removed, as shown in Figure 32.20.
+//Currents I1, I2 and I3 are labelled as shown in Figure 32.20.
+Ze1 = Z3*Z2/(Z3 + Z2)
+Ze2 = Ze1 + Z1
+//current
+I1 = E1/Ze2
+I2 = (Z2/(Z3 + Z2))*I1
+I3 = (Z3/(Z3 + Z2))*I1
+//The original network is redrawn with E1 removed, as shown in Figure 32.22
+//Currents I4, I5 and I6 are shown labelled in Figure 32.22 with I4 flowing away from the positive terminal of the E2 source.
+Ze3 = Z3*Z1/(Z3 + Z1)
+Ze4 = Ze3 + Z2
+I4 = E2/Ze4
+I5 = (Z1/(Z3 + Z1))*I4
+I6 = (Z3/(Z3 + Z1))*I4
+//If the network of Figure 32.18 is superimposed on the network of Figure 32.17, it can be seen that the current in the (4+i3) ohm impedance is given by
+i1 = I1 + I6
+i2 = I3 + I4
+i3 = I2 - I5
+//magnitude
+i1mag = (real(i1)^2 + imag(i1)^2)^0.5
+i2mag = (real(i2)^2 + imag(i2)^2)^0.5
+E1mag = (real(E1)^2 + imag(E1)^2)^0.5
+E2mag = (real(E2)^2 + imag(E2)^2)^0.5
+//phase
+phi1 = atan(imag(i1)/real(i1))
+phi2 = atan(imag(i2)/real(i2))
+//voltage across the(6 + i8) ohm impedance
+V6i8 = i3*Z3
+V6i8m = (real(V6i8)^2 + imag(V6i8)^2)^0.5
+//power
+P = (E1mag*i1mag*cos(phi1)) + (E2mag*i2mag*cos(phi2 - atan(imag(E2)/real(E2))))
+
+printf("\n\n Result \n\n")
+printf("\n(b)current in the (6 + i8) ohm resistor of the network is %.3f V",V6i8m)
+printf("\n(c)the total active power delivered to the network is %.3f W",P)
\ No newline at end of file diff --git a/608/CH32/EX32.05/32_05.sce b/608/CH32/EX32.05/32_05.sce new file mode 100755 index 000000000..db921720f --- /dev/null +++ b/608/CH32/EX32.05/32_05.sce @@ -0,0 +1,70 @@ +//Problem 32.05: Use the superposition theorem to determine, for the network shown in Figure 32.25, (a) the magnitude of the current flowing in the capacitor, (b) the p.d. across the 5 ohm resistance, (c) the active power dissipated in the 20 ohm resistance and (d) the total active power taken from the supply.
+
+//initializing the variables:
+rv1 = 50; // in volts
+rv2 = 30; // in volts
+thetav1 = 0; // in degrees
+thetav2 = 90; // in degrees
+R1 = 20; // in ohm
+R2 = 5; // in ohm
+R3 = -1*%i*3; // in ohm
+R4 = 8; // in ohm
+R5 = 8; // in ohm
+
+//calculation:
+//voltage
+V1 = rv1*cos(thetav1*%pi/180) + %i*rv1*sin(thetav1*%pi/180)
+V2 = rv2*cos(thetav2*%pi/180) + %i*rv2*sin(thetav2*%pi/180)
+//The network is redrawn with the V2 source removed, as shown in Figure 32.26.
+//Currents I1 to I5 are shown labelled in Figure 32.26.
+//current
+Re1 = R4*R5/(R5 + R4) + R3
+Re2 = Re1*R2/(R2 + Re1)
+I1 = V1/(Re2 + R1)
+I2 = (Re1/(R2 + Re1))*I1
+I3 = (R2/(Re1 + R2))*I1
+I4 = (R4/(R4 + R5))*I3
+I5 = I3 - I4
+//The original network is redrawn with the V1 source removed, as shown in Figure 32.27.
+//Currents I6 to I10 are shown labelled in Figure 32.27
+Re3 = R1*R2/(R1 + R2)
+Re4 = Re3 + R3
+Re5 = Re4*R4/(Re4 + R4)
+Re6 = Re5 + R5
+I6 = V2/Re6
+I7 = (Re4/(Re4 + R4))*I6
+I8 = (R4/(Re4 + R4))*I6
+I9 = (R1/(R1 + R2))*I8
+I10 = (R2/(R1 + R2))*I8
+//current flowing in the capacitor is given by
+Ic = I3 - I8
+//magnitude of the current in the capacitor
+Icmag = (real(Ic)^2 + imag(Ic)^2)^0.5
+//
+i1 = I2 + I9
+i1mag = (real(i1)^2 + imag(i1)^2)^0.5
+//magnitude of the p.d. across the 5 ohm resistance is given by
+Vr5m = i1mag*R2
+//Active power dissipated in the 20 ohm resistance is given by
+i2 = I1 - I10
+i2mag = (real(i2)^2 + imag(i2)^2)^0.5
+phii2 = atan(imag(i2)/real(i2))
+Pr20 = R1*(i2mag)^2
+//Active power developed by the V1
+P1 = rv1*i2mag*cos(phii2)
+//Active power developed by V2 source
+i3 = I6 - I5
+i3mag = (real(i3)^2 + imag(i3)^2)^0.5
+phii3 = atan(imag(i3)/real(i3))
+if ((imag(i3)>0) & (real(i3)<0)) then
+ phii3 = phii3 + %pi
+end
+P2 = rv2*i3mag*cos(phii3 - (thetav2*%pi/180))
+//Total power developed
+P = P1 + P2
+
+printf("\n\n Result \n\n")
+printf("\n(a)the magnitude of the current flowing in the capacitor is %.2f A",Icmag)
+printf("\n(b) the p.d. across the 5 ohm resistance is %.3f V",Vr5m)
+printf("\n(c)the active power dissipated in the 20 ohm resistance is %.0f W",Pr20)
+printf("\n(d)the total active power taken from the supply is %.1f W",P)
\ No newline at end of file diff --git a/608/CH33/EX33.01/33_01.sce b/608/CH33/EX33.01/33_01.sce new file mode 100755 index 000000000..eb83d2bb4 --- /dev/null +++ b/608/CH33/EX33.01/33_01.sce @@ -0,0 +1,35 @@ +//Problem 33.01: For the circuit shown in Figure 33.12, use Th´evenin’s theorem to determine (a) the current flowing in the capacitor, and (b) the p.d. across the 150 kohm resistor.
+
+//initializing the variables:
+rv = 200; // in volts
+thetav = 0; // in degrees
+R1 = 5000; // in ohm
+R2 = 20000; // in ohm
+R3 = -1*%i*120000; // in ohm
+R4 = 150000; // in ohm
+
+//calculation:
+//voltage
+V = rv*cos(thetav*%pi/180) + %i*rv*sin(thetav*%pi/180)
+//Initially the (150-i120)kohm impedance is removed from the circuit as shown in Figure 33.13.
+//Note that, to find the current in the capacitor, only the capacitor need have been initially removed from the circuit. However, removing each of the components from the branch through which the current is required will often result in a simpler solution.
+//From Figure 33.13,
+//current, I1
+I1 = V/(R1 + R2)
+//The open-circuit e.m.f. E is equal to the p.d. across the 20 kohm resistor, i.e.
+E = I1*R2
+//Removing the V1 source gives the network shown in Figure 33.14.
+//The impedance, z, ‘looking in’ at the open-circuited terminals is given by
+z = R1*R2/(R1 + R2)
+//The Th´evenin equivalent circuit is shown in Figure 33.15, where current iL is given by
+ZL = R3 + R4
+IL = E/(ZL + z)
+ILmag = (real(IL)^2 + imag(IL)^2)^0.5
+//current flowing in the capacitor
+Ic = ILmag
+//P.d. across the 150 kohm resistor,
+Vr150 = ILmag*R4
+
+printf("\n\n Result \n\n")
+printf("\n(a)the current flowing in the capacitor is %.1E A",Ic)
+printf("\n(b) the p.d. across the 150 ohm resistance is %.0f V",Vr150)
\ No newline at end of file diff --git a/608/CH33/EX33.02/33_02.sce b/608/CH33/EX33.02/33_02.sce new file mode 100755 index 000000000..7742fc3cc --- /dev/null +++ b/608/CH33/EX33.02/33_02.sce @@ -0,0 +1,28 @@ +//Problem 33.02: Determine, for the network shown in Figure 33.16, the value of current I. Each of the voltage sources has a frequency of 2 kHz.
+
+//initializing the variables:
+V1 = 20; // in volts
+V2 = 10; // in volts
+R1 = 2; // in ohm
+R2 = 1.5; // in ohm
+L = 235E-6; // in Henry
+R4 = 3; // in ohm
+f = 2000; // in Hz
+
+//calculation:
+//The impedance through which current I is flowing is initially removed from the network, as shown in Figure 33.17.
+//From Figure 33.17,
+//current, I1
+I1 = (V1 - V2)/(R1 + R4)
+//the open circuit e.m.f. E
+E = V1 - I1*R1
+//When the sources of e.m.f. are removed from the circuit, the impedance, z, ‘looking in’ at the break is given by
+z = R1*R4/(R1 + R4)
+//The Th´evenin equivalent circuit is shown in Figure 33.18, where inductive reactance,
+XL = 2*%pi*f*L
+R3 = %i*XL
+//Hence current
+I = E/(R2 + R3 + z)
+
+printf("\n\n Result \n\n")
+printf("\n the current I is %.2f + (%.2f)i A",real(I), imag(I))
\ No newline at end of file diff --git a/608/CH33/EX33.03/33_03.sce b/608/CH33/EX33.03/33_03.sce new file mode 100755 index 000000000..e0004c896 --- /dev/null +++ b/608/CH33/EX33.03/33_03.sce @@ -0,0 +1,30 @@ +//Problem 33.03: Use Th´evenin’s theorem to determine the power dissipated in the 48 ohm resistor of the network shown in Figure 33.19
+
+//initializing the variables:
+rv = 50; // in volts
+thetav = 0; // in degrees
+R1 = -1*%i*400; // in ohm
+R2 = 300; // in ohm
+R3 = %i*144; // in ohm
+R4 = 48; // in ohm
+
+//calculation:
+//voltage
+V = rv*cos(thetav*%pi/180) + %i*rv*sin(thetav*%pi/180)
+//The R3 and R4 impedance is initially removed from the network as shown in Figure 33.20.
+//From Figure 33.20,
+//current, I
+i = V/(R1 + R2)
+//the open circuit e.m.f. E
+E = i*R2
+//When the V is removed from the circuit, the impedance, z, ‘looking in’ at the break is given by
+z = R1*R2/(R1 + R2)
+//The Th´evenin equivalent circuit is shown in Figure 33.21 connected to R# and R4,
+//Hence current
+I = E/(R4 + R3 + z)
+Imag = (real(I)^2 + imag(I)^2)^0.5
+//the power dissipated in the 48 ohm resistor
+Pr48 = R4*Imag^2
+
+printf("\n\n Result \n\n")
+printf("\n the power dissipated in the 48 ohm resistor is %.2f W",Pr48)
\ No newline at end of file diff --git a/608/CH33/EX33.04/33_04.sce b/608/CH33/EX33.04/33_04.sce new file mode 100755 index 000000000..a6209a296 --- /dev/null +++ b/608/CH33/EX33.04/33_04.sce @@ -0,0 +1,29 @@ +//Problem 33.04:For the network shown in Figure 33.22, use Th´evenin’s theorem to determine the current flowing in the 80 ohm resistor.
+
+//initializing the variables:
+V = 100; // in volts
+R1 = 5; // in ohm
+R2 = 20; // in ohm
+R3 = 46; // in ohm
+R4 = 50; // in ohm
+R5 = 15; // in ohm
+R6 = 60; // in ohm
+R7 = 16; // in ohm
+R8 = 80; // in ohm
+
+//calculation:
+//One method of analysing a multi-branch network as shown in Figure 33.22 is to use Th´evenin’s theorem on one part of the network at a time. For example, the part of the circuit to the left of AA may be reduced to a Th´evenin equivalent circuit.
+//From Figure 33.23,
+E1 = (R2/(R1 + R2))*V
+z1 = R1*R2/(R1 + R2)
+//Thus the network of Figure 33.22 reduces to that of Figure 33.24. The part of the network shown in Figure 33.24 to the left of BB may be reduced to a Th´evenin equivalent circuit, where
+E2 = (R4/(R3 + R4 + z1))*E1
+z2 = R4*(z1 + R3)/(R4 + z1 + R3)
+//Thus the original network reduces to that shown in Figure 33.25. The part of the network shown in Figure 33.25 to the left of CC may be reduced to a Th´evenin equivalent circuit, where
+E3 = (R6/(R5 + R6 + z2))*E2
+z3 = R6*(z2 + R5)/(R5 + z2 + R6)
+//Thus the original network reduces to that of Figure 33.26, from which the current in the 80 ohm resistor is given by
+I = E3/(z3 + R7 + R8)
+
+printf("\n\n Result \n\n")
+printf("\n the current flowing in the 80 ohm resistor is %.2f A",I)
\ No newline at end of file diff --git a/608/CH33/EX33.05/33_05.sce b/608/CH33/EX33.05/33_05.sce new file mode 100755 index 000000000..8d8e86489 --- /dev/null +++ b/608/CH33/EX33.05/33_05.sce @@ -0,0 +1,30 @@ +//Problem 33.05:Determine the Th´evenin equivalent circuit with respect to terminals AB of the circuit shown in Figure 33.27. Hence determine (a) the magnitude of the current flowing in a (3.75 + i11) ohm impedance connected across terminals AB, and (b) the magnitude of the p.d. across the( 3.75 + i11)ohm impedance.
+
+//initializing the variables:
+rv = 24; // in volts
+thetav = 0; // in degrees
+R1 = -1*%i*3; // in ohm
+R2 = 4; // in ohm
+R3 = %i*3; // in ohm
+
+//calculation:
+//voltage
+V = rv*cos(thetav*%pi/180) + %i*rv*sin(thetav*%pi/180)
+//Current I1 shown in Figure 33.27 is given by
+I1 = V/(R1 + R2 + R3)
+//The Th´evenin equivalent voltage, i.e., the open-circuit voltage across terminals AB, is given by
+E = I1*(R2 + R3)
+//When the voltage source is removed, the impedance z ‘looking in’ at AB is given by
+z = (R2 + R3)*R1/(R1 + R2 + R3)
+//Thus the Th´evenin equivalent circuit is as shown in Figure 33.28.
+//when (3.75 + i11) ohm impedance connected across terminals AB, the current I flowing in the impedance is given by
+R = 3.75 + %i*11; // in ohms
+I = E/(R + z)
+Imag = (real(I)^2 + imag(I)^2)^0.5
+//the p.d. across the( 3.75 + i11)ohm impedance.
+VR = I*R
+VRmag = (real(VR)^2 + imag(VR)^2)^0.5
+
+printf("\n\n Result \n\n")
+printf("\n (a) the current I flowing in the (3.75 + i11) impedance is given by is %.0f A",Imag)
+printf("\n (b) the magnitude of the p.d. across the impedance is %.1f V",VRmag)
\ No newline at end of file diff --git a/608/CH33/EX33.06/33_06.sce b/608/CH33/EX33.06/33_06.sce new file mode 100755 index 000000000..8e3938d9b --- /dev/null +++ b/608/CH33/EX33.06/33_06.sce @@ -0,0 +1,31 @@ +//Problem 33.06: Use Th´evenin’s theorem to determine the current flowing in the capacitor of the network shown in Figure 33.29.
+
+//initializing the variables:
+rv = 16.55; // in volts
+thetav = -22.62; // in degrees
+R1 = 4; // in ohm
+R2 = %i*2; // in ohm
+R3 = %i*6; // in ohm
+R4 = 3; // in ohm
+R5 = 5; // in ohm
+R6 = -1*%i*8; // in ohm
+
+//calculation:
+//voltage
+V = rv*cos(thetav*%pi/180) + %i*rv*sin(thetav*%pi/180)
+//The capacitor is removed from branch AB, as shown in Figure 33.30.
+//Impedance, Z
+Z1 = R3 + R4 + R5
+Z = R1 + (Z1*R2/(R2 + Z1))
+I1 = V/Z
+I2 = (R2/(R2 +Z1))*I1
+//The open-circuit voltage, E
+E = I2*R5
+//If the voltage source is removed from Figure 33.30, the impedance, z, ‘looking in’ at AB is given by
+z = R5*((R1*R2/(R1 + R2)) + R3 + R4)/(R5 + ((R1*R2/(R1 + R2)) + R3 + R4))
+//The Th´evenin equivalent circuit is shown in Figure 33.31, where the current flowing in the capacitor, I, is given by
+I = E/(z + R6)
+Imag = (real(I)^2 + imag(I)^2)^0.5
+phiid = (atan(imag(I)/real(I)))*180/%pi
+printf("\n\n Result \n\n")
+printf("\n the current flowing in the capacitor of the network is %.2f/_%.2f° A",Imag,phiid)
\ No newline at end of file diff --git a/608/CH33/EX33.07/33_07.sce b/608/CH33/EX33.07/33_07.sce new file mode 100755 index 000000000..e9eb93988 --- /dev/null +++ b/608/CH33/EX33.07/33_07.sce @@ -0,0 +1,34 @@ +//Problem 33.07: For the network shown in Figure 33.32, derive the Th´evenin equivalent circuit with respect to terminals PQ, and hence determine the power dissipated by a 2 ohm resistor connected across PQ.
+
+//initializing the variables:
+rv1 = 5; // in volts
+rv2 = 10; // in volts
+thetav1 = 45; // in degrees
+thetav2 = 0; // in degrees
+R1 = 8; // in ohm
+R2 = 5; // in ohm
+R3 = %i*3; // in ohm
+R4 = 4; // in ohm
+
+//calculation:
+//voltage
+V1 = rv1*cos(thetav1*%pi/180) + %i*rv1*sin(thetav1*%pi/180)
+V2 = rv2*cos(thetav2*%pi/180) + %i*rv2*sin(thetav2*%pi/180)
+//Current I1 shown in Figure 33.32 is given by
+I1 = V2/(R2 + R3 + R4)
+//Hence the voltage drop across the 5 ohm resistor is given by VX is in the direction shown in Figure 33.32,
+Vx = I1*R2
+//The open-circuit voltage E across PQ is the phasor sum of V1, Vx and V2, as shown in Figure 33.33.
+E = V2 - V1 - Vx
+//The impedance, z, ‘looking in’ at terminals PQ with the voltage sources removed is given by
+z = R1 + R2*(R3 + R4)/(R2 + R3 + R4)
+//The Th´evenin equivalent circuit is shown in Figure 33.34 with the 2 ohm resistance connected across terminals PQ.
+//The current flowing in the 2 ohm resistance is given by
+R = 2; // in ohms
+I = E/(z + R)
+Imag = (real(I)^2 + imag(I)^2)^0.5
+//power P dissipated in the 2 ohm resistor is given by
+Pr2 = R*Imag^2
+
+printf("\n\n Result \n\n")
+printf("\n power P dissipated in the 2 ohm resistor is %.4f W",Pr2)
\ No newline at end of file diff --git a/608/CH33/EX33.08/33_08.sce b/608/CH33/EX33.08/33_08.sce new file mode 100755 index 000000000..0669f2cb5 --- /dev/null +++ b/608/CH33/EX33.08/33_08.sce @@ -0,0 +1,41 @@ +//Problem 33.08: For the a.c. bridge network shown in Figure 33.35, determine the current flowing in the capacitor, and its direction, by using Th´evenin’s theorem. Assume the 306/_0° V source to have negligible internal impedance.
+
+//initializing the variables:
+rv = 30; // in volts
+thetav = 0; // in degrees
+R1 = 15; // in ohm
+R2 = 40; // in ohm
+R3 = %i*20; // in ohm
+R4 = 20; // in ohm
+R5 = %i*5; // in ohm
+R6 = 5; // in ohm
+R7 = -1*%i*25; // in ohm
+
+//calculation:
+//voltage
+V = rv*cos(thetav*%pi/180) + %i*rv*sin(thetav*%pi/180)
+//The R7 is initially removed from the network, as shown in Figure 33.36
+Z1 = R1
+Z2 = R2
+Z3 = R3 + R4
+Z4 = R5 + R6
+//P.d. between A and C,
+Vac = (Z1/(Z1 + Z4))*V
+//P.d. between B and C,
+Vbc = (Z2/(Z2 + Z3))*V
+//Assuming that point A is at a higher potential than point B, then the p.d. between A and B is
+Vab = Vac - Vbc
+//the open-circuit voltage across AB is given by
+E = Vab
+//Point C is at a potential of V . Between C and A is a volt drop of Vac. Hence the voltage at point A is
+Va = V - Vac
+//Between points C and B is a voltage drop of Vbc. Hence the voltage at point B
+Vb = V - Vbc
+//Replacing the V source with a short-circuit (i.e., zero internal impedance) gives the network shown in Figure 33.37(a). The network is shown redrawn in Figure 33.37(b) and simplified in Figure 33.37(c). Hence the impedance, z, ‘looking in’ at terminals AB is given by
+z = Z1*Z4/(Z1 + Z4) + Z2*Z3/(Z2 + Z3)
+//The Th´evenin equivalent circuit is shown in Figure 33.38, where current I is given by
+I = E/(z + R7)
+Imag = (real(I)^2 + imag(I)^2)^0.5
+
+printf("\n\n Result \n\n")
+printf("\n the current flowing in the capacitor is %.3f A in direction from B to A.",Imag)
\ No newline at end of file diff --git a/608/CH33/EX33.09/33_09.sce b/608/CH33/EX33.09/33_09.sce new file mode 100755 index 000000000..5c0f8ac60 --- /dev/null +++ b/608/CH33/EX33.09/33_09.sce @@ -0,0 +1,22 @@ +//Problem 33.09: Use Norton’s theorem to determine the value of current I in the circuit shown in Figure 33.47.
+
+//initializing the variables:
+V = 5; // in volts
+R1 = 2; // in ohm
+R2 = 3; // in ohm
+R3 = -1*%i*3; // in ohm
+R4 = 2.8; // in ohm
+
+//calculation:
+//The branch containing the R4 is short-circuited, as shown in Figure 33.48.
+//The R2 in parallel with a short-circuit is the same as R2 in parallel with 0 ohm giving an equivalent impedance of
+Z1 = R2*0/(R3 + 0)
+//Hence the network reduces to that shown in Figure 33.49, where
+Isc = V/R1
+//If the Voltage source is removed from the network the input impedance, z, ‘looking-in’ at a break made in AB of Figure 33.48 gives
+z = R1*R2/(R1 + R2)
+//The Norton equivalent network is shown in Figure 33.51, where current I is given by
+I = (z/(z + R4 + R3))*Isc
+
+printf("\n\n Result \n\n")
+printf("\n the current flowing in the capacitor is %.2f + (%.2f)i A",real(I), imag(I))
\ No newline at end of file diff --git a/608/CH33/EX33.10/33_10.sce b/608/CH33/EX33.10/33_10.sce new file mode 100755 index 000000000..4f942091c --- /dev/null +++ b/608/CH33/EX33.10/33_10.sce @@ -0,0 +1,23 @@ +//Problem 33.10: For the circuit shown in Figure 33.52 determine the current flowing in the inductive branch by using Norton’s theorem.
+
+//initializing the variables:
+V1 = 20; // in volts
+V2 = 10; // in volts
+R1 = 2; // in ohm
+R2 = 1.5; // in ohm
+R3 = %i*2.95; // in ohm
+R4 = 3; // in ohm
+
+//calculation:
+//The inductive branch is initially short-circuited, as shown in Figure 33.53.
+//From Figure 33.53,
+I1 = V1/R1
+I2 = V2/R4
+Isc = I1 + I2
+//If the voltage sources are removed, the impedance, z, ‘looking in’ at a break made in AB is given by
+z = R1*R4/(R1 + R4)
+//The Norton equivalent network is shown in Figure 33.54, where current I is given by
+I = (z/(z + R2 + R3))*Isc
+
+printf("\n\n Result \n\n")
+printf("\n the current flowing in the inductive branch is %.2f + (%.2f)i A",real(I), imag(I))
\ No newline at end of file diff --git a/608/CH33/EX33.11/33_11.sce b/608/CH33/EX33.11/33_11.sce new file mode 100755 index 000000000..1f1050fb3 --- /dev/null +++ b/608/CH33/EX33.11/33_11.sce @@ -0,0 +1,22 @@ +//Problem 33.11: Use Norton’s theorem to determine the magnitude of the p.d. across the 1 ohm resistance of the network shown in Figure 33.55.
+
+//initializing the variables:
+V = 10; // in volts
+R1 = 4; // in ohm
+R2 = 4; // in ohm
+R3 = -1*%i*2; // in ohm
+R4 = 1; // in ohm
+
+//calculation:
+//The branch containing the R4 is initially short-circuited, as shown in Figure 33.56.
+//R2 in parallel with R3 in parallel with 0 ohm (i.e., the short-circuit) is equivalent 0 ohm giving the equivalent circuit of Figure 33.57. Hence Isc
+Isc = V/R1
+//The voltage source is removed from the network of Figure 33.55, as shown in Figure 33.58, and the impedance z, ‘looking in’ at a break made in AB is given by
+z = 1/(1/R1 + 1/R2 + 1/R3)
+//The Norton equivalent network is shown in Figure 33.59, from which current I is given by
+I = (z/(z + R4))*Isc
+Imag = (real(I)^2 + imag(I)^2)^0.5
+Vr1 = Imag*R4
+
+printf("\n\n Result \n\n")
+printf("\n the magnitude of the p.d. across the 1 ohm resistor is %.2f V", Vr1)
\ No newline at end of file diff --git a/608/CH33/EX33.12/33_12.sce b/608/CH33/EX33.12/33_12.sce new file mode 100755 index 000000000..fbeff8302 --- /dev/null +++ b/608/CH33/EX33.12/33_12.sce @@ -0,0 +1,31 @@ +//Problem 33.12:For the network shown in Figure 33.60, obtain the Norton equivalent network at terminals AB. Hence determine the power dissipated in a 5 ohm resistor connected between A and B.
+
+//initializing the variables:
+rv = 20; // in volts
+thetav = 0; // in degrees
+R1 = 2; // in ohm
+R2 = 4; // in ohm
+R3 = %i*3; // in ohm
+R4 = -1*%i*3; // in ohm
+
+//calculation:
+//voltage
+V = rv*cos(thetav*%pi/180) + %i*rv*sin(thetav*%pi/180)
+//Terminals AB are initially short-circuited, as shown in Figure 33.61.
+//The circuit impedance Z presented to the voltage source is given by
+Z = R1 + R4*(R2 + R3)/(R2 + R3 + R4)
+//Thus current I in Figure 33.61 is given by
+I = V/Z
+Isc = ((R2 + R3)/(R2 + R3 + R4))*I
+//Removing the voltage source of Figure 33.60 gives the network Figure 33.62 of Figure 33.62. Impedance, z, ‘looking in’ at terminals AB is given by
+z = R4 + R1*(R2 + R3)/(R2 + R3 + R1)
+//The Norton equivalent network is shown in Figure 33.63.
+R = 5; // in ohms
+//Current IL
+IL = (z/(z + R))*Isc
+ILmag = (real(IL)^2 + imag(IL)^2)^0.5
+//the power dissipated in the 5 ohm resistor is
+Pr5 = R*ILmag^2
+
+printf("\n\n Result \n\n")
+printf("\n the power dissipated in the 5 ohm resistor is %.2f W", Pr5)
\ No newline at end of file diff --git a/608/CH33/EX33.13/33_13.sce b/608/CH33/EX33.13/33_13.sce new file mode 100755 index 000000000..4980e3f92 --- /dev/null +++ b/608/CH33/EX33.13/33_13.sce @@ -0,0 +1,42 @@ +//Problem 33.13:Derive the Norton equivalent network with respect to terminals PQ for the network shown in Figure 33.64 and hence determine the magnitude of the current flowing in a 2 ohm resistor connected across PQ.
+
+//initializing the variables:
+rv1 = 5; // in volts
+rv2 = 10; // in volts
+thetav1 = 45; // in degrees
+thetav2 = 0; // in degrees
+R1 = 8; // in ohm
+R2 = 5; // in ohm
+R3 = %i*3; // in ohm
+R4 = 4; // in ohm
+
+//calculation:
+//voltage
+V1 = rv1*cos(thetav1*%pi/180) + %i*rv1*sin(thetav1*%pi/180)
+V2 = rv2*cos(thetav2*%pi/180) + %i*rv2*sin(thetav2*%pi/180)
+//Terminals PQ are initially short-circuited, as shown in Figure 33.65.
+//Currents I1 and I2 are shown labelled. Kirchhoff’s laws are used.
+//For loop ABCD, and moving anticlockwise,
+//I1*(R2 + R3 + R4) + I2*(R3 + R4) = V2
+//For loop DPQC, and moving clockwise,
+//R2*I1 - R1*I2 = V2 - V1
+//Solving Equations by using determinants gives
+d1 = [V2 (R3 + R4); (V2 - V1) -1*R1]
+D1 = det(d1)
+d2 = [(R2 + R3 + R4) V2; R2 (V2 - V1)]
+D2 = det(d2)
+d = [(R2 + R3 + R4) (R3 + R4); R2 -1*R1]
+D = det(d)
+I1 = D1/D
+I2 = D2/D
+//the short-circuit current Isc
+Isc = I2
+//The impedance, z, ‘looking in’ at a break made between P and Q is given by
+z = R1 + R2*(R3 + R4)/(R2 + R3 + R4)
+//The Norton equivalent circuit is shown in Figure 33.66, where current I is given by
+R = 2; //in ohm
+I = (z/(z + R))*Isc
+Imag = (real(I)^2 + imag(I)^2)^0.5
+
+printf("\n\n Result \n\n")
+printf("\n the magnitude of the current flowing 5 ohm resistor is %.2f A", Imag)
\ No newline at end of file diff --git a/608/CH33/EX33.15/33_15.sce b/608/CH33/EX33.15/33_15.sce new file mode 100755 index 000000000..c6cc546c0 --- /dev/null +++ b/608/CH33/EX33.15/33_15.sce @@ -0,0 +1,32 @@ +//Problem 33.15: (a) Convert the circuit to the left of terminals AB in Figure 33.72 to an equivalent Th´evenin circuit by initially converting to a Norton equivalent circuit. (b) Determine the magnitude of the current flowing in the (1.8+i4) ohm impedance connected between terminals A and B of Figure 33.72.
+
+//initializing the variables:
+E1 = 12; // in volts
+E2 = 24; // in volts
+Z1 = 3; // in ohm
+Z2 = 2; // in ohm
+R1 = %i*4; // in ohm
+R2 = 1.8; // in ohm
+
+//calculation:
+Z3 = R1 + R2
+//For the branch containing the E1 source, conversion to a Norton equivalent network gives
+Isc1 = E1/Z1
+//For the branch containing the E2 source, conversion to a Norton equivalent circuit gives
+Isc2 = E2/Z2
+//Thus Figure 33.73 shows a network equivalent to Figure 33.72. From Figure 33.73, the total short-circuit current
+Isc = Isc1 + Isc2
+//the total impedance is given by
+z = Z1*Z2/(Z1 + Z2)
+//Thus Figure 33.73 simplifies to Figure 33.74.
+//The open-circuit voltage across AB of Figure 33.74, E
+E = Isc*z
+//the impedance ‘looking in’ at AB,is z
+//the Th´evenin equivalent circuit is as shown in Figure 33.75.
+R = 1.8 + %i*4; // in ohm
+//when R impedance is connected to terminals AB of Figure 33.75, the current I flowing is given by
+I = E/(z + R)
+Imag = (real(I)^2 + imag(I)^2)^0.5
+
+printf("\n\n Result \n\n")
+printf("\n the magnitude of the current flowing (1.8 + i4) ohm resistor is %.2f A", Imag)
\ No newline at end of file diff --git a/608/CH33/EX33.16/33_16.sce b/608/CH33/EX33.16/33_16.sce new file mode 100755 index 000000000..2165d8bfb --- /dev/null +++ b/608/CH33/EX33.16/33_16.sce @@ -0,0 +1,44 @@ +//Problem 33.16: Determine, by successive conversions between Th´evenin’s and Norton’s equivalent networks, a Th´evenin equivalent circuit for terminals AB of Figure 33.76. Hence determine the magnitude of the current flowing in the capacitive branch connected to terminals AB.
+
+//initializing the variables:
+V1 = 5; // in volts
+V2 = 10; // in volts
+i = 0.001; // in Amperes
+R1 = 1000; // in ohm
+R2 = 4000; // in ohm
+R3 = 2000; // in ohm
+R4 = 200; // in ohm
+R5 = -1*%i*4000; // in ohm
+
+//calculation:
+//For the branch containing the V1 source, conversion to a Norton equivalent network gives
+Isc1 = V1/R1
+z1 = R1
+//For the branch containing the V2 source, conversion to a Norton equivalent circuit gives
+Isc2 = V2/R2
+z2 = R2
+//Thus the circuit of Figure 33.76 converts to that of Figure 33.77.
+//The above two Norton equivalent networks shown in Figure 33.77 may be combined, since the total short-circuit current is
+Isc = Isc1 + Isc2
+//the total impedance is given by
+Z1 = z1*z2/(z1 + z2)
+//Both of the Norton equivalent networks shown in Figure 33.78 may be converted to Th´evenin equivalent circuits. Open-circuit voltage across CD is
+Ecd = Isc*Z1
+//the impedance ‘looking in’ at CD is Z1
+//Open-circuit voltage across EF
+Eef = i*R3
+//the impedance ‘looking in’ Figure 33.79 at EF
+Z2 = R3
+//Thus Figure 33.78 converts to Figure 33.79.
+//Combining the two Th´evenin circuits gives e.m.f.
+E = Ecd - Eef
+//impedance z
+z = Z1 + Z2
+//the Th´evenin equivalent circuit for terminals AB of Figure 33.76 is as shown in Figure 33.80.
+Z3 = R4 + R5
+//If an impedance Z3 is connected across terminals AB, then the current I flowing is given by
+I = E/(z + Z3)
+Imag = (real(I)^2 + imag(I)^2)^0.5
+
+printf("\n\n Result \n\n")
+printf("\n the current in the capacitive branch is %.2E A", Imag)
\ No newline at end of file diff --git a/608/CH33/EX33.17/33_17.sce b/608/CH33/EX33.17/33_17.sce new file mode 100755 index 000000000..e8de21846 --- /dev/null +++ b/608/CH33/EX33.17/33_17.sce @@ -0,0 +1,26 @@ +//Problem 33.17: (a) Determine an equivalent Th´evenin circuit for terminals AB of the network shown in Figure 33.81. (b) Calculate the power dissipated in a (600 - i800)ohm impedance connected between A and B of Figure 33.81.
+
+//initializing the variables:
+V = 5; // in volts
+i = 0.004; // in Amperes
+R1 = 2000; // in ohm
+R2 = %i*1000; // in ohm
+
+//calculation:
+//Converting the Th´evenin circuit to a Norton network gives
+Isc1 = V/R2
+//Thus Figure 33.81 converts to that shown in Figure 33.82. The two Norton equivalent networks may be combined, giving
+Isc = Isc1 + i
+z = R1*R2/(R1 + R2)
+//This results in the equivalent network shown in Figure 33.83. Converting to an equivalent Th´evenin circuit gives open circuit e.m.f. across AB,
+E = Isc*z
+//Thus the The´venin equivalent circuit is as shown in Figure 33.84.
+R = 600 - %i*800; // in ohms
+//When a R impedance is connected across AB, the current I flowing is given by
+I = E/(z + R)
+Imag = (real(I)^2 + imag(I)^2)^0.5
+//the power dissipated in the R resistor is
+PR = R*Imag^2
+
+printf("\n\n Result \n\n")
+printf("\n the power dissipated in the (600 - i800) ohm resistor is %.2E W", PR)
\ No newline at end of file diff --git a/608/CH34/EX34.02/34_02.sce b/608/CH34/EX34.02/34_02.sce new file mode 100755 index 000000000..9e7c827c8 --- /dev/null +++ b/608/CH34/EX34.02/34_02.sce @@ -0,0 +1,34 @@ +//Problem 34.02: For the network shown in Figure 34.7, determine (a) the equivalent circuit impedance across terminals AB, (b) supply current I and (c) the power dissipated in the 10 ohm resistor.
+
+//initializing the variables:
+rv = 40; // in volts
+thetav = 0; // in degrees
+ZA = %i*10; // in ohm
+ZB = %i*15; // in ohm
+ZC = %i*25; // in ohm
+ZD = -1*%i*8; // in ohm
+ZE = 10; // in ohm
+
+//calculation:
+//voltage
+V = rv*cos(thetav*%pi/180) + %i*rv*sin(thetav*%pi/180)
+//The network of Figure 34.7 is redrawn, as in Figure 34.8, showing more clearly the part of the network 1, 2, 3 forming a delta connection This may he transformed into a star connection as shown in Figure 34.9.
+Z1 = ZA*ZB/(ZA + ZB + ZC)
+Z2 = ZC*ZB/(ZA + ZB + ZC)
+Z3 = ZA*ZC/(ZA + ZB + ZC)
+//The equivalent network is shown in Figure 34.10 and is further simplified in Figure 34.11
+//(ZE + Z3) in parallel with (Z1 + ZD) gives an equivalent impedance of
+z = (ZE + Z3)*(Z1 + ZD)/(Z1 + ZD + ZE + Z3)
+//Hence the total circuit equivalent impedance across terminals AB is given by
+Zab = z + Z2
+//Supply current I
+I = V/Zab
+I1 = ((Z1 + ZD)/(Z1 + ZD + ZE + Z3))*I
+I1mag = (real(I1)^2 + imag(I1)^2)^0.5
+//Power P dissipated in the 10 ohm resistance of Figure 34.7 is given by
+Pr10 = ZE*I1mag^2
+
+printf("\n\n Result \n\n")
+printf("\n (a)the equivalent circuit impedance across terminals AB is %.2f + (%.2f)i ohm",real(Zab), imag(Zab))
+printf("\n (b)supply current I is %.2f + (%.2f)i A",real(I), imag(I))
+printf("\n (c)power P dissipated in the 10 ohm resistor is %.2f W",Pr10)
\ No newline at end of file diff --git a/608/CH34/EX34.03/34_03.sce b/608/CH34/EX34.03/34_03.sce new file mode 100755 index 000000000..d3a0581f9 --- /dev/null +++ b/608/CH34/EX34.03/34_03.sce @@ -0,0 +1,36 @@ +//Problem 34.03: Determine, for the bridge network shown in Figure 34.12, (a) the value of the single equivalent resistance that replaces the network between terminals A and B, (b) the current supplied by the 52 V source, and (c) the current flowing in the 8 ohm resistance.
+
+//initializing the variables:
+V = 52; // in volts
+ZA = 8; // in ohm
+ZB = 16; // in ohm
+ZC = 40; // in ohm
+ZD = 1; // in ohm
+ZE = 4; // in ohm
+
+//calculation:
+//In Figure 34.12, no resistances are directly in parallel or directly in series with each other. However, ACD and BCD are both delta connections and either may be converted into an equivalent star connection. The delta network BCD is redrawn in Figure 34.13(a) and is transformed into an equivalent star connection as shown in Figure 34.13(b), where
+Z1 = ZA*ZB/(ZA + ZB + ZC)
+Z2 = ZC*ZB/(ZA + ZB + ZC)
+Z3 = ZA*ZC/(ZA + ZB + ZC)
+//The network of Figure 34.12 may thus be redrawn as shown in Figure 34.14. The Z1 and ZE are in series with each other, as are the ZD and Z3 resistors. Hence the equivalent network is as shown in Figure 34.15. The total equivalent resistance across terminals A and B is given by
+Zab = (Z1 + ZE)*(ZD + Z3)/(Z1 + ZE + ZD + Z3) + Z2
+//Current supplied by the source, i.e., current I in Figure 34.15, is given by
+I = V/Zab
+//From Figure 34.15, current I1
+I1 = ((ZD + Z3)/(Z1 + ZE + ZD + Z3))*I
+//current I2
+I2 = I - I1
+//From Figure 34.14, p.d. across AC,
+Vac = I1*ZE
+//p.d. across AD
+Vad = I2*ZD
+//Hence p.d. between C and D is given
+Vcd = Vac - Vad
+//current in the 8 ohm resistance
+Ir8 = Vcd/ZA
+
+printf("\n\n Result \n\n")
+printf("\n (a)the equivalent circuit impedance across terminals AB is %.2f ohm",Zab)
+printf("\n (b)the current supplied by the 52 V source is %.2f A",I)
+printf("\n (c)the current flowing in the 8 ohm resistance is %.2f A",Ir8)
\ No newline at end of file diff --git a/608/CH34/EX34.05/34_05.sce b/608/CH34/EX34.05/34_05.sce new file mode 100755 index 000000000..2a598ffa8 --- /dev/null +++ b/608/CH34/EX34.05/34_05.sce @@ -0,0 +1,33 @@ +//Problem 34.05: For the network shown in Figure 34.20, determine (a) the current flowing in the (0+i10) ohm impedance, and (b) the power dissipated in the (20 + i0) ohm impedance.
+
+//initializing the variables:
+rv = 120; // in volts
+thetav = 0; // in degrees
+ZA = 25 - %i*5; // in ohm
+ZB = 15 + %i*10; // in ohm
+ZC = 20 - %i*30; // in ohm
+ZD = 20 + %i*0; // in ohm
+ZE = 0 + %i*10; // in ohm
+ZF = 2.5 - %i*5; // in ohm
+
+//calculation:
+//voltage
+V = rv*cos(thetav*%pi/180) + %i*rv*sin(thetav*%pi/180)
+//The network may initially be simplified by transforming the delta PQR to its equivalent star connection as represented by impedances Z1, Z2 and Z3 in Figure 34.21. From equation (34.7),
+Z1 = ZA*ZB/(ZA + ZB + ZC)
+Z2 = ZC*ZB/(ZA + ZB + ZC)
+Z3 = ZA*ZC/(ZA + ZB + ZC)
+//The network is shown redrawn in Figure 34.22 and further simplified in Figure 34.23, from which,
+Zab = ((Z3 + ZE)*(ZD + Z2)/(Z2 + ZE + ZD + Z3)) + (Z1 + ZF)
+//Current I1
+I1 = V/Zab
+//current I2
+I2 = ((ZE + Z3)/(Z2 + ZE + ZD + Z3))*I1
+//current I3
+I3 = I1 - I2
+//The power P dissipated in the ZD impedance of Figure 34.20 is given by
+Pzd = ZD*I2^2
+
+printf("\n\n Result \n\n")
+printf("\n (a)the current flowing in the (0+i10) ohm impedance is %.2f A",I3)
+printf("\n (b) the power dissipated in the (20 + i0) ohm impedance is %.2f W",Pzd)
\ No newline at end of file diff --git a/608/CH35/EX35.01/35_01.sce b/608/CH35/EX35.01/35_01.sce new file mode 100755 index 000000000..afdd653e8 --- /dev/null +++ b/608/CH35/EX35.01/35_01.sce @@ -0,0 +1,23 @@ +//Problem 35.01: For the circuit shown in Figure 35.2 the load impedance Z is a pure resistance. Determine (a) the value of R for maximum power to be transferred from the source to the load, and (b) the value of the maximum power delivered to R.
+
+//initializing the variables:
+rv = 120; // in volts
+thetav = 0; // in degrees
+Z = 15 + %i*20; // in ohm
+
+//calculation:
+//voltage
+V = rv*cos(thetav*%pi/180) + %i*rv*sin(thetav*%pi/180)
+//maximum power transfer occurs when R = mod(Z)
+R = (real(Z)^2 + imag(Z)^2)^0.5
+//the total circuit impedance
+ZT = Z + R
+//Current I flowing in the load is given by
+I = V/ZT
+Imag = (real(I)^2 + imag(I)^2)^0.5
+//maximum power delivered
+P = R*Imag^2
+
+printf("\n\n Result \n\n")
+printf("\n (a)maximum power transfer occurs when R is %.0f ohm",R)
+printf("\n (b) maximum power delivered is %.0f W",P)
\ No newline at end of file diff --git a/608/CH35/EX35.02/35_02.sce b/608/CH35/EX35.02/35_02.sce new file mode 100755 index 000000000..ab38df2cb --- /dev/null +++ b/608/CH35/EX35.02/35_02.sce @@ -0,0 +1,23 @@ +//Problem 35.02: If the load impedance Z in Figure 35.2 of problem 35.01 consists of variable resistance R and variable reactance X, determine (a) the value of Z that results in maximum power transfer, and (b) the value of the maximum power.
+
+//initializing the variables:
+rv = 120; // in volts
+thetav = 0; // in degrees
+Z = 15 + %i*20; // in ohm
+
+//calculation:
+//voltage
+V = rv*cos(thetav*%pi/180) + %i*rv*sin(thetav*%pi/180)
+//maximum power transfer occurs when X = -1*imag(Z) and R = real(Z)
+z = real(Z) - %i*imag(Z)
+//Total circuit impedance at maximum power transfer condition,
+ZT = Z + z
+//Current I flowing in the load is given by
+I = V/ZT
+Imag = (real(I)^2 + imag(I)^2)^0.5
+//maximum power delivered
+P = real(Z)*I^2
+
+printf("\n\n Result \n\n")
+printf("\n (a)maximum power transfer occurs when Z is %.0f + (%.0f)i ohm",real(z), imag(z))
+printf("\n (b) maximum power delivered is %.0f W",P)
\ No newline at end of file diff --git a/608/CH35/EX35.03/35_03.sce b/608/CH35/EX35.03/35_03.sce new file mode 100755 index 000000000..faf0183a8 --- /dev/null +++ b/608/CH35/EX35.03/35_03.sce @@ -0,0 +1,29 @@ +//Problem 35.03: For the network shown in Figure 35.3, determine (a) the value of the load resistance R required for maximum power transfer, and (b) the value of the maximum power transferred.
+
+//initializing the variables:
+rv = 200; // in volts
+thetav = 0; // in degrees
+R1 = 100; // in ohm
+C = 1E-6; // in farad
+f = 1000; // in Hz
+
+//calculation:
+//voltage
+V = rv*cos(thetav*%pi/180) + %i*rv*sin(thetav*%pi/180)
+//Capacitive reactance, Xc
+Xc = 1/(2*%pi*f*C)
+//Hence source impedance,
+z = R1*(%i*Xc)/(R1 + %i*Xc)
+//maximum power transfer is achieved when R = mod(z)
+R = (real(z)^2 + imag(z)^2)^0.5
+//Total circuit impedance at maximum power transfer condition,
+ZT = z + R
+//Current I flowing in the load is given by
+I = V/ZT
+Imag = (real(I)^2 + imag(I)^2)^0.5
+//maximum power transferred,
+P = R*Imag^2
+
+printf("\n\n Result \n\n")
+printf("\n (a)maximum power transfer occurs when R is %.2f ohm",R)
+printf("\n (b) maximum power delivered is %.0f W",P)
\ No newline at end of file diff --git a/608/CH35/EX35.04/35_04.sce b/608/CH35/EX35.04/35_04.sce new file mode 100755 index 000000000..d32490165 --- /dev/null +++ b/608/CH35/EX35.04/35_04.sce @@ -0,0 +1,27 @@ +//Problem 35.04: In the network shown in Figure 35.4 the load consists of a fixed capacitive reactance of 7 ohm and a variable resistance R. Determine (a) the value of R for which the power transferred to the load is a maximum, and (b) the value of the maximum power.
+
+//initializing the variables:
+rv = 60; // in volts
+thetav = 0; // in degrees
+R1 = 4; // in ohm
+XL = 10; // in ohm
+Xc = 7; // in ohm
+R2 = %i*XL; // in ohm
+R3 = -1*%i*Xc; // in ohm
+
+//calculation:
+//voltage
+V = rv*cos(thetav*%pi/180) + %i*rv*sin(thetav*%pi/180)
+//maximum power transfer is achieved when
+R = (R1^2 + (XL - Xc)^2)^0.5
+//Hence source impedance,
+ZT = R1 + R2 + R3 + R
+//Current I flowing in the load is given by
+I = V/ZT
+Imag = (real(I)^2 + imag(I)^2)^0.5
+//maximum power transferred,
+P = R*Imag^2
+
+printf("\n\n Result \n\n")
+printf("\n (a)maximum power transfer occurs when R is %.2f ohm",R)
+printf("\n (b) maximum power delivered is %.0f W",P)
\ No newline at end of file diff --git a/608/CH35/EX35.05/35_05.sce b/608/CH35/EX35.05/35_05.sce new file mode 100755 index 000000000..b40f34f2d --- /dev/null +++ b/608/CH35/EX35.05/35_05.sce @@ -0,0 +1,23 @@ +//Problem 35.05: Determine the value of the load resistance R shown in Figure 35.5 that gives maximum power dissipation and calculate the value of this power.
+
+//initializing the variables:
+V = 20; // in volts
+R1 = 5; // in ohm
+R2 = 15; // in ohm
+
+//calculation:
+//R is removed from the network as shown in Figure 35.6
+//P.d. across AB, E
+E = (R2/(R1 + R2))*V
+//Impedance ‘looking-in’ at terminals AB with the source removed is given by
+r = R1*R2/(R1 + R2)
+//The equivalent Th´evenin circuit supplying terminals AB is shown in Figure 35.7. From condition (2), for maximum power transfer
+R = r
+//Current I flowing in the load is given by
+I = E/(R + r)
+//maximum power transferred,
+P = R*I^2
+
+printf("\n\n Result \n\n")
+printf("\n (a)maximum power transfer occurs when R is %.2f ohm",R)
+printf("\n (b) maximum power delivered is %.0f W",P)
\ No newline at end of file diff --git a/608/CH35/EX35.06/35_06.sce b/608/CH35/EX35.06/35_06.sce new file mode 100755 index 000000000..7a6dbbdda --- /dev/null +++ b/608/CH35/EX35.06/35_06.sce @@ -0,0 +1,30 @@ +//Problem 35.06: Determine, for the network shown in Figure 35.8, (a) the values of R and X that will result in maximum power being transferred across terminals AB, and (b) the value of the maximum power.
+
+//initializing the variables:
+rv = 100; // in volts
+thetav = 30; // in degrees
+R1 = 5; // in ohm
+R2 = 5; // in ohm
+R3 = %i*10; // in ohm
+
+//calculation:
+//voltage
+V = rv*cos(thetav*%pi/180) + %i*rv*sin(thetav*%pi/180)
+//Resistance R and reactance X are removed from the network as shown in Figure 35.9
+//P.d. across AB,
+E = ((R2 + R3)/(R1 + R2 + R3))*V
+//With the source removed the impedance, z, ‘looking in’ at terminals AB is given by:
+z = (R2 + R3)*R1/(R1 + R2 + R3)
+//The equivalent Th´evenin circuit is shown in Figure 35.10. From condition 3, maximum power transfer is achieved when X = -1*imag(z) and R = real(z)
+X = -1*imag(z)
+R = real(z)
+Z = R + %i*X
+//Current I flowing in the load is given by
+I = E/(z + Z)
+Imag = (real(I)^2 + imag(I)^2)^0.5
+//maximum power transferred,
+P = R*Imag^2
+
+printf("\n\n Result \n\n")
+printf("\n (a)maximum power transfer occurs when R is %.2f ohm and X is %.2f ohm",R, X)
+printf("\n (b) maximum power delivered is %.0f W",P)
\ No newline at end of file diff --git a/608/CH35/EX35.07/35_07.sce b/608/CH35/EX35.07/35_07.sce new file mode 100755 index 000000000..7e57609a1 --- /dev/null +++ b/608/CH35/EX35.07/35_07.sce @@ -0,0 +1,13 @@ +//Problem 35.07: Determine the optimum value of load resistance for maximum power transfer if the load is connected to an amplifier of output resistance 448 ohm through a transformer with a turns ratio of 8:1.
+
+//initializing the variables:
+Ro = 448; // in ohm
+tr = 8; // turn ratio N1/N2
+
+//calculation:
+//The equivalent input resistance r of the transformer must be Ro for maximum power transfer.
+r = Ro
+RL = r*(1/tr)^2
+
+printf("\n\n Result \n\n")
+printf("\n the optimum value of load resistance is %.0f ohm",RL)
\ No newline at end of file diff --git a/608/CH35/EX35.08/35_08.sce b/608/CH35/EX35.08/35_08.sce new file mode 100755 index 000000000..027923f80 --- /dev/null +++ b/608/CH35/EX35.08/35_08.sce @@ -0,0 +1,14 @@ +//Problem 35.08: A generator has an output impedance of (450 + i60) ohm. Determine the turns ratio of an ideal transformer necessary to match the generator to a load of (40 + i19) ohm for maximum transfer of power.
+
+//initializing the variables:
+Zo = 450 + %i*60; // in ohm
+ZL = 40 + %i*19; // in ohm
+
+//calculation:
+//transformer turns ratio tr = (N1/N2)
+Zomag = (real(Zo)^2 + imag(Zo)^2)^0.5
+ZLmag = (real(ZL)^2 + imag(ZL)^2)^0.5
+tr = (Zomag/ZLmag)^0.5
+
+printf("\n\n Result \n\n")
+printf("\n the transformer turns ratio is %.2f",tr)
\ No newline at end of file diff --git a/608/CH35/EX35.09/35_09.sce b/608/CH35/EX35.09/35_09.sce new file mode 100755 index 000000000..c7a967aba --- /dev/null +++ b/608/CH35/EX35.09/35_09.sce @@ -0,0 +1,27 @@ +//Problem 35.09: A single-phase, 240 V/1920 V ideal transformer is supplied from a 240 V source through a cable of resistance 5 ohm. If the load across the secondary winding is 1.60 kohm determine (a) the primary current flowing, and (b) the power dissipated in the load resistance.
+
+//initializing the variables:
+V1 = 240; // in volts
+V2 = 1920; // in volts
+R1 = 5; // in ohms
+R2 = 1600; // in ohms
+
+//calculation:
+//The network is shown in Figure 35.12.
+//turn ratio N1/N2 = V1/V2
+tr = V1/V2
+//Equivalent input resistance of the transformer,
+RL = R2
+r = RL*tr^2
+//Total input resistance,
+Rin = R1 + r
+//primary current, I1
+I1 = V1/Rin
+//For an ideal transformer V1/V2 = I2/I1
+I2 = I1*(V1/V2)
+//Power dissipated in the load resistance
+P = RL*I2^2
+
+printf("\n\n Result \n\n")
+printf("\n (a) primary current flowing is %.0f A",I1)
+printf("\n (b) Power dissipated in the load resistance is %.0fW",P)
\ No newline at end of file diff --git a/608/CH35/EX35.10/35_10.sce b/608/CH35/EX35.10/35_10.sce new file mode 100755 index 000000000..90a7b3aed --- /dev/null +++ b/608/CH35/EX35.10/35_10.sce @@ -0,0 +1,28 @@ +//Problem 35.10: An ac. source of 30/_0° V and internal resistance 20 kohm is matched to a load by a 20:1 ideal transformer. Determine for maximum power transfer (a) the value of the load resistance, and (b) the power dissipated in the load.
+
+//initializing the variables:
+rv = 30; // in volts
+thetav = 0; // in degrees
+r = 20000; // in ohms
+tr = 20; // turn ratio
+
+//calculation:
+//voltage
+V = rv*cos(thetav*%pi/180) + %i*rv*sin(thetav*%pi/180)
+//The network diagram is shown in Figure 35.13.
+//For maximum power transfer, r1 must be equal to
+r1 = r
+//load resistance RL
+RL = r1/tr^2
+//The total input resistance when the source is connected to the matching transformer is
+RT = r + r1
+//Primary current
+I1 = V/RT
+//N1/N2 = I2/I1
+I2 = I1*tr
+//Power dissipated in load resistance RL is given by
+P = RL*I2^2
+
+printf("\n\n Result \n\n")
+printf("\n (a)the value of the load resistance is %.0f ohm",RL)
+printf("\n (b) Power dissipated in the load resistance is %.2E W",P)
\ No newline at end of file diff --git a/608/CH36/EX36.03/36_03.sce b/608/CH36/EX36.03/36_03.sce new file mode 100755 index 000000000..a034401b5 --- /dev/null +++ b/608/CH36/EX36.03/36_03.sce @@ -0,0 +1,13 @@ +//Problem 36.03: Determine the rms value of the current waveform represented by i = 100sinwt + 20sin(3wt + pi/6) + 10sin(5wt + 2*pi/3) mA
+
+//initializing the variables:
+A1 = 0.100; // in amperes
+A3 = 0.020; // in amperes
+A5 = 0.010; // in amperes
+
+//calculation:
+//the rms value of current is given by
+Irms = ((A1^2 + A3^2 + A5^2)/2)^0.5
+
+printf("\n\n Result \n\n")
+printf("\n the rms value of current is %.5f A",Irms)
\ No newline at end of file diff --git a/608/CH36/EX36.04/36_04.sce b/608/CH36/EX36.04/36_04.sce new file mode 100755 index 000000000..77090619c --- /dev/null +++ b/608/CH36/EX36.04/36_04.sce @@ -0,0 +1,25 @@ +//Problem 36.04: A complex voltage is represented by
+// v = 10sinwt + 3sin(3wt) + 2sin(5wt) Volts
+//Determine for the voltage, (a) the rms value, (b) the mean value and (c) the form factor.
+
+//initializing the variables:
+A1 = 10; // in volts
+A3 = 3; // in volts
+A5 = 2; // in volts
+
+//calculation:
+//the rms value of voltage is given by
+Vrms = ((A1^2 + A3^2 + A5^2)/2)^0.5
+//the mean value of voltage is given by
+//x = wt
+function [Y]=f(x)
+ Y = (10*sin(x) + 3*sin(3*x) + 2*sin(5*x));
+endfunction
+Vav = (1/%pi)*(integrate('f', 'x', 0, %pi))
+//form factor is given by
+ff = Vrms/Vav
+
+printf("\n\n Result \n\n")
+printf("\n (a)the rms value of voltage is %.2f V",Vrms)
+printf("\n (b)the mean value of voltage is %.2f V",Vav)
+printf("\n (c)form factor is %.3f ",ff)
\ No newline at end of file diff --git a/608/CH36/EX36.06/36_06.sce b/608/CH36/EX36.06/36_06.sce new file mode 100755 index 000000000..3808cbfe6 --- /dev/null +++ b/608/CH36/EX36.06/36_06.sce @@ -0,0 +1,17 @@ +//Problem 36.06: Determine the average power in a 20 # resistance if the current i flowing through it is of the form
+// i = 12sinwt + 5sin(3wt) + 2sin(5wt) A
+
+//initializing the variables:
+A1 = 12; // in amperes
+A3 = 5; // in amperes
+A5 = 2; // in amperes
+R = 20; // in ohms
+
+//calculation:
+//rms current
+Irms = ((A1^2 + A3^2 + A5^2)/2)^0.5
+//average power
+P = R*Irms^2
+
+printf("\n\n Result \n\n")
+printf("\n average power %.0f W",P)
\ No newline at end of file diff --git a/608/CH36/EX36.07/36_07.sce b/608/CH36/EX36.07/36_07.sce new file mode 100755 index 000000000..f2c037672 --- /dev/null +++ b/608/CH36/EX36.07/36_07.sce @@ -0,0 +1,41 @@ +//Problem 36.07: A complex voltage v given by
+// v = 60sinwt + 15sin(3wt + pi/4) + 10sin(5wt - pi/2) Volts
+//is applied to a circuit and the resulting current i is given by
+// i = 2sin(wt - pi/6) + 0.30sin(3wt - pi/12) + 0.1sin(5wt - 8pi/9) A
+//Determine (a) the total active power supplied to the circuit, and (b) the overall power factor.
+
+//initializing the variables:
+Ia1 = 2; // in amperes
+Ia3 = 0.3; // in amperes
+Ia5 = 0.1; // in amperes
+Va1 = 60; // in volts
+Va3 = 15; // in volts
+Va5 = 10; // in volts
+Phii1 = -1*%pi/6; // in radians
+Phii3 = -1*%pi/12; // in radians
+Phii5 = -8*%pi/9; // in radians
+Phiv1 = 0; // in radians
+Phiv3 = %pi/4; // in radians
+Phiv5 = -1*%pi/2; // in radians
+
+
+//calculation:
+//rms values;
+I1 = Ia1/(2^0.5); // in amperes
+I3 = Ia3/(2^0.5); // in amperes
+I5 = Ia5/(2^0.5); // in amperes
+V1 = Va1/(2^0.5); // in volts
+V3 = Va3/(2^0.5); // in volts
+V5 = Va5/(2^0.5); // in volts
+//total power supplied,
+P = V1*I1*cos(Phiv1 - Phii1) + V3*I3*cos(Phiv3 - Phii3) + V5*I5*cos(Phiv5 - Phii5)
+//rms current
+Irms = ((I1^2 + I3^2 + I5^2))^0.5
+//rms voltage
+Vrms = ((V1^2 + V3^2 + V5^2))^0.5
+//overall power factor
+pf = P/(Vrms*Irms)
+
+printf("\n\n Result \n\n")
+printf("\n(a)the total active power supplied to the circuit %.2f W",P)
+printf("\n(b)overall power factor %.3f",pf)
\ No newline at end of file diff --git a/608/CH36/EX36.09/36_09.sce b/608/CH36/EX36.09/36_09.sce new file mode 100755 index 000000000..1eb397a7f --- /dev/null +++ b/608/CH36/EX36.09/36_09.sce @@ -0,0 +1,52 @@ +//Problem 36.09: A supply voltage v given by
+// v = 240sin314t + 40sin942t + 30sin1570t Volts
+//is applied to a circuit comprising a resistance of 12 ohm connected in series with a coil of inductance 9.55 mH. Determine (a) an expression to represent the instantaneous value of the current, (b) the rms voltage, (c) the rms current, (d) the power dissipated, and (e) the overall power factor.
+
+//initializing the variables:
+V1m = 240; // in volts
+V3m = 40; // in volts
+V5m = 30; // in volts
+w1 = 314; // fundamental
+R = 12; // in ohm
+L = 0.00955; // in Henry
+
+//calculation:
+//fundamental or first harmonic
+//inductive reactance,
+XL1 = w1*L
+//impedance at the fundamental frequency,
+Z1 = R + %i*XL1
+//Maximum current at fundamental frequency
+I1m = V1m/Z1
+I1mag = (real(I1m)^2 + imag(I1m)^2)^0.5
+phii1 = atan(imag(I1m)/real(I1m))
+//Third harmonic
+XL3 = 3*XL1
+//impedance at the third harmonic frequency,
+Z3 = R + %i*XL3
+//Maximum current at third harmonic frequency
+I3m = V3m/Z3
+I3mag = (real(I3m)^2 + imag(I3m)^2)^0.5
+phii3 = atan(imag(I3m)/real(I3m))
+//fifth harmonic
+XL5 = 5*XL1
+//impedance at the third harmonic frequency,
+Z5 = R + %i*XL5
+//Maximum current at third harmonic frequency
+I5m = V5m/Z5
+I5mag = (real(I5m)^2 + imag(I5m)^2)^0.5
+phii5 = atan(imag(I5m)/real(I5m))
+//rms voltage
+Vrms = ((V1m^2 + V3m^2 + V5m^2)/2)^0.5
+//rms current
+Irms = ((I1mag^2 + I3mag^2 + I5mag^2)/2)^0.5
+//power dissipated
+P = R*Irms^2
+//overall power factor
+pf = P/(Vrms*Irms)
+
+printf("\n\n Result \n\n")
+printf("\n(b)the rms value of current is %.2f A",Irms)
+printf("\n(c)the rms value of voltage is %.2f V",Vrms)
+printf("\n(d)the total power dissipated %.0f W",P)
+printf("\n(e)overall power factor %.3f",pf)
\ No newline at end of file diff --git a/608/CH36/EX36.10/36_10.sce b/608/CH36/EX36.10/36_10.sce new file mode 100755 index 000000000..d6b415379 --- /dev/null +++ b/608/CH36/EX36.10/36_10.sce @@ -0,0 +1,54 @@ +//Problem 36.10: An e.m.f. is represented by
+// e = 50 + 200sinwt + 40sin(2wt - pi/2) + 5sin(4wt + pi/4) Volts
+//the fundamental frequency being 50 Hz. The e.m.f. is applied across a circuit comprising a 100 μF capacitor connected in series with a 50 ohm resistor. Obtain an expression for the current flowing and hence determine the rms value of current.
+
+//initializing the variables:
+Vom = 50; // in volts
+V1m = 200; // in volts
+V2m = 40; // in volts
+V4m = 5; // in volts
+f = 50; // in Hz
+R = 50; // in ohm
+C = 100E-6; // in farad
+phiv1 = 0; // in rad
+phiv2 = -1*%pi/2; // in rad
+phiv4 = %pi/4; // in rad
+
+//calculation:
+//voltage
+V1 = V1m*cos(phiv1) + %i*V1m*sin(phiv1)
+V2 = V2m*cos(phiv2) + %i*V2m*sin(phiv2)
+V4 = V4m*cos(phiv4) + %i*V4m*sin(phiv4)
+//Inductance has no effect on a steady current. Hence the d.c. component of the current, i0, is given by
+Iom = 0
+//fundamental or first harmonic
+w1 = 2*%pi*f
+//inductive reactance,
+Xc1 = 1/(w1*C)
+//impedance at the fundamental frequency,
+Z1 = R + %i*Xc1
+//Maximum current at fundamental frequency
+I1m = V1/Z1
+I1mag = (real(I1m)^2 + imag(I1m)^2)^0.5
+phii1 = atan(imag(I1m)/real(I1m))
+//second harmonic
+Xc2 = Xc1/2
+//impedance at the third harmonic frequency,
+Z2 = R + %i*Xc2
+//Maximum current at third harmonic frequency
+I2m = V2/Z2
+I2mag = (real(I2m)^2 + imag(I2m)^2)^0.5
+phii2 = atan(imag(I2m)/real(I2m))
+//fourth harmonic
+Xc4 = Xc1/4
+//impedance at the third harmonic frequency,
+Z4 = R + %i*Xc4
+//Maximum current at third harmonic frequency
+I4m = V4/Z4
+I4mag = (real(I4m)^2 + imag(I4m)^2)^0.5
+phii4 = atan(imag(I4m)/real(I4m))
+//rms current
+Irms = (Iom^2 + (I1mag^2 + I2mag^2 + I4mag^2)/2)^0.5
+
+printf("\n\n Result \n\n")
+printf("\n(b)the rms value of current is %.2f A",Irms)
\ No newline at end of file diff --git a/608/CH36/EX36.11/36_11.sce b/608/CH36/EX36.11/36_11.sce new file mode 100755 index 000000000..dc09d93f1 --- /dev/null +++ b/608/CH36/EX36.11/36_11.sce @@ -0,0 +1,56 @@ +//Problem 36.11: A supply voltage v given by
+// v = 25 + 100sinwt + 40sin(3wt + pi/6) + 20sin(5wt + pi/12) Volts
+//where w = 10000 rad/s. The voltage is applied to a series circuit comprising a 5.0 ohm resistance and a 500 μH inductance. Determine (a) an expression to represent the current flowing in the circuit, (b) the rms value of current, correct to two decimal places, and (c) the power dissipated in the circuit, correct to three significant figures.
+
+//initializing the variables:
+Vom = 25; // in volts
+V1m = 100; // in volts
+V3m = 40; // in volts
+V5m = 20; // in volts
+w1 = 10000; // fundamental
+R = 5; // in ohm
+L = 500E-6; // in Henry
+phiv1 = 0; // in rad
+phiv3 = %pi/6; // in rad
+phiv5 = %pi/12; // in rad
+
+//calculation:
+//voltage
+V1 = V1m*cos(phiv1) + %i*V1m*sin(phiv1)
+V3 = V3m*cos(phiv3) + %i*V3m*sin(phiv3)
+V5 = V5m*cos(phiv5) + %i*V5m*sin(phiv5)
+//Inductance has no effect on a steady current. Hence the d.c. component of the current, i0, is given by
+Iom = Vom/R
+//fundamental or first harmonic
+//inductive reactance,
+XL1 = w1*L
+//impedance at the fundamental frequency,
+Z1 = R + %i*XL1
+//Maximum current at fundamental frequency
+I1m = V1/Z1
+I1mag = (real(I1m)^2 + imag(I1m)^2)^0.5
+phii1 = atan(imag(I1m)/real(I1m))
+//Third harmonic
+XL3 = 3*XL1
+//impedance at the third harmonic frequency,
+Z3 = R + %i*XL3
+//Maximum current at third harmonic frequency
+I3m = V3/Z3
+I3mag = (real(I3m)^2 + imag(I3m)^2)^0.5
+phii3 = atan(imag(I3m)/real(I3m))
+//fifth harmonic
+XL5 = 5*XL1
+//impedance at the third harmonic frequency,
+Z5 = R + %i*XL5
+//Maximum current at third harmonic frequency
+I5m = V5/Z5
+I5mag = (real(I5m)^2 + imag(I5m)^2)^0.5
+phii5 = atan(imag(I5m)/real(I5m))
+//rms current
+Irms = (Iom^2 + (I1mag^2 + I3mag^2 + I5mag^2)/2)^0.5
+//power dissipated
+P = R*Irms^2
+
+printf("\n\n Result \n\n")
+printf("\n(b)the rms value of current is %.2f A",Irms)
+printf("\n(c)the total power dissipated %.0f W",P)
\ No newline at end of file diff --git a/608/CH36/EX36.12/36_12.sce b/608/CH36/EX36.12/36_12.sce new file mode 100755 index 000000000..881823137 --- /dev/null +++ b/608/CH36/EX36.12/36_12.sce @@ -0,0 +1,37 @@ +//Problem 36.12: The voltage applied to a particular circuit comprising two components connected in series is given by
+// v = 30 + 40sinwt + 25sin(2wt) + 15sin(4wt) Volts
+//and the resulting current is given by
+// v = 0.743sin(wt + 1.19) + 0.78sin(2wt + 0.896) + 0.636sin(4wt + 0.559) A
+//Determine (a) the average power supplied, (b) the type of components present, and (c) the values of the components.
+
+//initializing the variables:
+Vom = 30; // in volts
+V1m = 40; // in volts
+V2m = 25; // in volts
+V4m = 15; // in volts
+Iom = 0; // in amperes
+I1m = 0.743; // in Amperes
+I2m = 0.781; // in Amperes
+I4m = 0.636; // in Amperes
+phii1 = 1.190; // in rad
+phii2 = 0.896; // in rad
+phii4 = 0.559; // in rad
+w = 1000; // in rad
+
+//calculation:
+//the average power P is given by
+P = Vom*Iom + (0.707*V1m)*(0.707*I1m)*cos(phii1) + + (0.707*V2m)*(0.707*I2m)*cos(phii2) + (0.707*V4m)*(0.707*I4m)*cos(phii4)
+//rms current
+Irms = (Iom^2 + (I1m^2 + I2m^2 + I4m^2)/2)^0.5
+//resistance R
+R = P/(Irms^2)
+//impedance
+Z1 = V1m/I1m
+//Xc1
+Xc1 = (Z1^2 - R^2)^0.5
+//capacitance
+C = 1/(w*Xc1)
+
+printf("\n\n Result \n\n")
+printf("\n(a)the average power P is %.2f W",P)
+printf("\n(c)the resistance R %.0f ohm and capacitance %.2E F",R,C)
\ No newline at end of file diff --git a/608/CH36/EX36.13/36_13.sce b/608/CH36/EX36.13/36_13.sce new file mode 100755 index 000000000..144c51540 --- /dev/null +++ b/608/CH36/EX36.13/36_13.sce @@ -0,0 +1,40 @@ +//Problem 36.13: In the circuit shown in Figure 36.17 the supply voltage v is given by v = 300sin314t + 120sin(942t + 0.698) Volts.Determine (a) an expression for the supply current, i, (b) the percentage harmonic content of the supply current, (c) the total power dissipated, (d) an expression for the p.d. shown as v1, and (e) an expression for current ic.
+
+//initializing the variables:
+V1m = 300; // in volts
+V3m = 120; // in volts
+phiv1 = 0; // in rad
+phiv2 = 0.698; // in rad
+w1 = 314; // in rad
+C = 2.123E-6; // in farads
+R1 = 560; // in ohms
+R2 = 2000; // in Ohm
+
+//calculation:
+//voltage
+V1 = V1m*cos(phiv1) + %i*V1m*sin(phiv1)
+V3 = V3m*cos(phiv3) + %i*V3m*sin(phiv3)
+//capacitive reactance,
+Xc1 = 1/(w1*C)
+//impedance at the fundamental frequency,
+Z1 = R1 + %i*Xc1*R2/(R2 + %i*Xc1)
+//Maximum current at fundamental frequency
+I1m = V1/Z1
+I1mag = (real(I1m)^2 + imag(I1m)^2)^0.5
+phii1 = atan(imag(I1m)/real(I1m))
+//Third harmonic
+Xc3 = Xc1/3
+//impedance at the third harmonic frequency,
+Z3 = R1 + %i*Xc3*R2/(R2 + %i*Xc3)
+//Maximum current at third harmonic frequency
+I3m = V3/Z3
+I3mag = (real(I3m)^2 + imag(I3m)^2)^0.5
+phii3 = atan(imag(I3m)/real(I3m))
+//Percentage harmonic content of the supply current is given by
+percent = I3mag*100/I1mag
+//total active power
+P = (0.707*V1m)*(0.707*I1mag)*cos(phiv1 - phii1) + (0.707*V3m)*(0.707*I3m)*cos(phiv3 - phii3)
+
+printf("\n\n Result \n\n")
+printf("\n(b)Percentage harmonic content of the supply current is %.0f percent",percent)
+printf("\n(c)total active power is %.2f W",P)
\ No newline at end of file diff --git a/608/CH36/EX36.14/36_14.sce b/608/CH36/EX36.14/36_14.sce new file mode 100755 index 000000000..9c6d0e828 --- /dev/null +++ b/608/CH36/EX36.14/36_14.sce @@ -0,0 +1,31 @@ +//Problem 36.14: A voltage waveform having a fundamental of maximum value 400 V and a third harmonic of maximum value 10 V is applied to the circuit shown in Figure 36.18. Determine (a) the fundamental frequency for resonance with the third harmonic, and (b) the maximum value of the fundamental and third harmonic components of current.
+
+//initializing the variables:
+V1m = 400; // in volts
+V3m = 10; // in volts
+C = 0.2E-6; // in farads
+R = 2; // in ohms
+L = 0.5; // in Henry
+
+//calculation:
+//Resonance with the third harmonic means that
+w = (1/(9*L*C))^0.5
+//fundamental frequency, f
+f = w/(2*%pi)
+//At the fundamental frequency,
+//impedance Z1
+Z1 = R + %i*(w*L - 1/(w*C))
+Z1mag = (real(Z1)^2 + imag(Z1)^2)^0.5
+phiZ1 = atan(imag(Z1)/real(Z1))
+//Maximum value of current at the fundamental frequency,
+I1m = V1m/Z1mag
+//At the third harmonic frequency,
+Z3 = R + %i*(3*w*L - 1/(3*w*C))
+Z3mag = (real(Z3)^2 + imag(Z3)^2)^0.5
+phiZ3 = atan(imag(Z3)/real(Z3))
+//Maximum value of current at the third harmonic frequency,
+I3m = V3m/Z3
+
+printf("\n\n Result \n\n")
+printf("\n(a)fundamental frequency for resonance with the third harmonic is %.2f Hz",f)
+printf("\n(b)Maximum value of current at the fundamental frequency is %.3f A and at the third harmonic frequency %.2f A",I1m, I3m)
\ No newline at end of file diff --git a/608/CH36/EX36.15/36_15.sce b/608/CH36/EX36.15/36_15.sce new file mode 100755 index 000000000..9be820846 --- /dev/null +++ b/608/CH36/EX36.15/36_15.sce @@ -0,0 +1,40 @@ +//Problem 36.15: A voltage wave has an amplitude of 800 V at the fundamental frequency of 50 Hz and its nth harmonic has an amplitude 1.5% of the fundamental. The voltage is applied to a series circuit containing resistance 5 ohm, inductance 0.369 H and capacitance 0.122 μF. Resonance occurs at the nth harmonic. Determine (a) the value of n, (b) the maximum value of current at the nth harmonic, (c) the p.d. across the capacitor at the nth harmonic and (d) the maximum value of the fundamental current.
+
+//initializing the variables:
+V1m = 800; // in volts
+f = 50; // in Hz
+x = 0.015;
+C = 0.122E-6; // in farads
+R = 5; // in ohms
+L = 0.369; // in Henry
+
+//calculation:
+//voltage at nth harmonic
+Vnm = x*V1m
+w = 2*%pi*f
+//For resonance at the nth harmonic nwL = 1/nwC
+n = 1/(w*(L*C)^0.5)
+//At resonance, impedance
+Zn = R
+//the maximum value of current at the nth harmonic
+Inm = Vnm/Zn
+//capacitive reactance, at nth harmonic
+Xcn = 1/(n*w*C)
+//the p.d. across the capacitor at the nth harmonic
+Vcn = Inm*Xcn
+//At the fundamental frequency, inductive reactance,
+XL1 = w*L
+//capacitive reactance
+Xc1 = 1/(w*C)
+//Impedance at the fundamental frequency,
+Z1 = R + %i*(XL1 - Xc1)
+Z1mag = (real(Z1)^2 + imag(Z1)^2)^0.5
+phiZ1 = atan(imag(Z1)/real(Z1))
+//Maximum value of current at the fundamental frequency,
+I1m = V1m/Z1mag
+
+printf("\n\n Result \n\n")
+printf("\n(a)n = %.0f",n)
+printf("\n(b)the maximum value of current at the nth harmonic %.2f A",Inm)
+printf("\n(c)the p.d. across the capacitor at the nth harmonic is %.2f",Vcn)
+printf("\n(d)the maximum value of the fundamental current. %.2f A",I1m)
\ No newline at end of file diff --git a/608/CH38/EX38.01/38_01.sce b/608/CH38/EX38.01/38_01.sce new file mode 100755 index 000000000..83b1561df --- /dev/null +++ b/608/CH38/EX38.01/38_01.sce @@ -0,0 +1,17 @@ +//Problem 38.01: The area of a hysteresis loop obtained from a ferromagnetic specimen is 12.5 cm2. The scales used were: horizontal axis 1 cm = 500 A/m; vertical axis 1 cm = 0.2 T. Determine (a) the hysteresis loss per m3 per cycle, and (b) the hysteresis loss per m3 at a frequency of 50 Hz.
+
+//initializing the variables:
+A = 12.5; // in cm2
+x = 500; // horizontal axis 1 cm = 500 A/m
+y = 0.2; // vertical axis 1 cm = 0.2 T
+f = 50; // in Hz
+
+//calculation:
+//hysteresis loss per cycle
+HL = A*x*y
+//At 50 Hz frequency, hysteresis loss
+HLf = HL*f
+
+printf("\n\n Result \n\n")
+printf("\n(a)hysteresis loss per cycle is = %.0f J/m3",HL)
+printf("\n(b)At 50 Hz frequency, hysteresis loss %.0f W/m3",HLf)
\ No newline at end of file diff --git a/608/CH38/EX38.02/38_02.sce b/608/CH38/EX38.02/38_02.sce new file mode 100755 index 000000000..c8fd3e077 --- /dev/null +++ b/608/CH38/EX38.02/38_02.sce @@ -0,0 +1,19 @@ +//Problem 38.02: If in problem 38.01, the maximum flux density is 1.5 T at a frequency of 50 Hz, determine the hysteresis loss per m3 for a maximum flux density of 1.1 T and frequency of 25 Hz. Assume the Steinmetz index to be 1.6
+
+//initializing the variables:
+n = 1.6; // the Steinmetz index
+f1 = 50; // in Hz
+f2 = 25; // in Hz
+Bm1 = 1.5; // in Tesla
+Bm2 = 1.1; // in Tesla
+Ph1 = 62500; // in W/m3
+v = 1;
+
+//calculation:
+//hysteresis loss Ph = kh*v*f*(Bm)^n
+kh = Ph1/(v*f1*(Bm1)^n)
+//When f = 25 Hz and Bm = 1.1 T,
+Ph2 = kh*v*f2*(Bm2)^n
+
+printf("\n\n Result \n\n")
+printf("\n hysteresis loss When f = 25 Hz and Bm = 1.1 T, is = %.0f W/m3",Ph2)
\ No newline at end of file diff --git a/608/CH38/EX38.03/38_03.sce b/608/CH38/EX38.03/38_03.sce new file mode 100755 index 000000000..2db1c3360 --- /dev/null +++ b/608/CH38/EX38.03/38_03.sce @@ -0,0 +1,23 @@ +//Problem 38.03: A ferromagnetic ring has a uniform cross-sectional area of 2000 mm2 and a mean circumference of 1000 mm. A hysteresis loop obtained for the specimen is plotted to scales of 10 mm = 0.1 T and 10 mm = 400 A/m and is found to have an area of 104 mm2. Determine the hysteresis loss at a frequency of 80 Hz.
+
+//initializing the variables:
+csa = 0.002; // in m2
+l = 1; // in m
+a = 400/0.01; // 10 mm = 400 A/m
+b = 0.1/0.01; // 10 mm = 0.1 T
+A = 0.01; // in m2
+f = 80; // in Hz
+
+//calculation:
+//hysteresis loss per cycle
+HL = A*a*b
+//At a frequency of 80 Hz,
+//hysteresis loss
+HLf = HL*f
+//Volume of ring
+v = csa*l
+//hysteresis loss
+Ph = HLf*v
+
+printf("\n\n Result \n\n")
+printf("\n the hysteresis loss at a frequency of 80 Hz is %.0f W",Ph)
\ No newline at end of file diff --git a/608/CH38/EX38.04/38_04.sce b/608/CH38/EX38.04/38_04.sce new file mode 100755 index 000000000..cc4738d35 --- /dev/null +++ b/608/CH38/EX38.04/38_04.sce @@ -0,0 +1,23 @@ +//Problem 38.04: The cross-sectional area of a transformer limb is 80 cm2 and the volume of the transformer core is 5000 cm3. The maximum value of the core flux is 10 mWb at a frequency of 50 Hz. Taking the Steinmetz constant as 1.7, the hysteresis loss is found to be 100 W. Determine the value of the hysteresis loss when the maximum core flux is 8 mWb and the frequency is 50 Hz.
+
+//initializing the variables:
+Phi1 = 0.01; // in Wb
+Phi2 = 0.008; // in Wb
+csa = 0.008; // in m2
+v = 0.005; // in m3
+f = 50; // in Hz
+n = 1.7; // the Steinmetz constant
+Ph1 = 100; // in Watt
+
+//calculation:
+//maximum flux density
+Bm1 = Phi1/csa
+//hysteresis loss Ph1 = kh*v*f*(Bm1)^n
+kh = Ph1/(v*f*(Bm1)^n)
+//When the maximum core flux is 8 mWb,
+Bm2 = Phi2/csa
+//hysteresis loss, Ph2
+Ph2 = kh*v*f*(Bm2)^n
+
+printf("\n\n Result \n\n")
+printf("\nthe value of the hysteresis loss when the maximum core flux is 8 mWb and the frequency is 50 Hz is %.1f W",Ph2)
\ No newline at end of file diff --git a/608/CH38/EX38.05/38_05.sce b/608/CH38/EX38.05/38_05.sce new file mode 100755 index 000000000..74a5525c9 --- /dev/null +++ b/608/CH38/EX38.05/38_05.sce @@ -0,0 +1,16 @@ +//Problem 38.05: The eddy current loss in a particular magnetic circuit is 10 W/m3. If the frequency of operation is reduced from 50 Hz to 30 Hz with the flux density remaining unchanged, determine the new value of eddy current loss per cubic metre.
+
+//initializing the variables:
+Pe1 = 10; // in W/m3
+f1 = 50; // in Hz
+f2 = 30; // in Hz
+
+//calculation:
+//When the eddy current loss is 10 W/m3, frequency f is 50 Hz.
+//constant k
+k = Pe1/(f1^2)
+//When the frequency is 30 Hz, eddy current loss,
+Pe2 = k*(f2^2)
+
+printf("\n\n Result \n\n")
+printf("\neddy current loss per cubic metre is %.1f W/m3",Pe2)
\ No newline at end of file diff --git a/608/CH38/EX38.06/38_06.sce b/608/CH38/EX38.06/38_06.sce new file mode 100755 index 000000000..e0cdac3ae --- /dev/null +++ b/608/CH38/EX38.06/38_06.sce @@ -0,0 +1,19 @@ +//Problem 38.06: The core of a transformer operating at 50 Hz has an eddy current loss of 100 W/m3 and the core laminations have a thickness of 0.50 mm. The core is redesigned so as to operate with the same eddy current loss but at a different voltage and at a frequency of 250 Hz. Assuming that at the new voltage the maximum flux density is one-third of its original value and the resistivity of the core remains unaltered, determine the necessary new thickness of the laminations.
+
+//initializing the variables:
+Pe = 100; // in W/m3
+f1 = 50; // in Hz
+t1 = 0.0005; // in m
+x = 1/3;
+f2 = 250; // in Hz
+
+//calculation:
+//Pe = ke*(Bm1*f1*t1)^2
+//Hence, at 50 Hz frequency
+ke = Pe/(Bm1*f1*t1)^2
+//At 250 Hz frequency
+Bm2 = x*Bm1
+t2 = ((Pe/ke)^0.5)/(Bm2*f2)
+
+printf("\n\n Result \n\n")
+printf("\nlamination thickness is %.2Em",t2)
\ No newline at end of file diff --git a/608/CH38/EX38.07/38_07.sce b/608/CH38/EX38.07/38_07.sce new file mode 100755 index 000000000..3cdd0c6d0 --- /dev/null +++ b/608/CH38/EX38.07/38_07.sce @@ -0,0 +1,33 @@ +//Problem 38.07: The core of an inductor has a hysteresis loss of 40 W and an eddy current loss of 20 W when operating at 50 Hz frequency. (a) Determine the values of the losses if the frequency is increased to 60 Hz. (b) What will be the total core loss if the frequency is 50 Hz and the lamination are made one-half of their original thickness? Assume that the flux density remains unchanged in each case
+
+//initializing the variables:
+Ph1 = 40; // in W
+Pe1 = 20; // in W
+f1 = 50; // in Hz
+x = 1/2;
+f2 = 60; // in Hz
+t1 = 1;
+//calculation:
+//hysteresis loss Ph = kh*v*f*(Bm)^n = k1*f
+//Thus when the hysteresis is 40 W and the frequency 50 Hz,
+k1 = Ph1/f1
+//If the frequency is increased to 60 Hz,
+Ph2 = k1*f2
+//eddy current loss, Pe = ke*(Bm1*f1*t1)^2 = k2*f^2
+//since the flux density and lamination thickness are constant.
+//When the eddy current loss is 20 W the frequency is 50 Hz. Thus
+k2 = Pe1/(f1^2)
+//If the frequency is increased to 60 Hz,
+Pe2 = k2*(f2^2)
+//hysteresis loss Ph = kh*v*f*(Bm)^n, is independent of the thickness of the laminations. Thus, if the thickness of the laminations is halved, the hysteresis loss remains at
+Phb2 = Ph1
+//eddy current loss, Pe = ke*(Bm1*f1*t1)^2 = k2*t^3
+k3 = Pe1/(t1^3)
+t2 = 0.5*t1
+Peb2 = k3*t2^3
+//total core loss when the thickness of the laminations is halved is given by
+TL = Phb2 + Peb2
+
+printf("\n\n Result \n\n")
+printf("\n(a)If the frequency is increased to 60 Hz,hysteresis loss is %.0f W and eddy current loss %.1f W",Ph2, Pe2)
+printf("\n(b)the total core loss when the thickness of the laminations is halved %.1f W",TL)
\ No newline at end of file diff --git a/608/CH38/EX38.08/38_08.sce b/608/CH38/EX38.08/38_08.sce new file mode 100755 index 000000000..93ee60357 --- /dev/null +++ b/608/CH38/EX38.08/38_08.sce @@ -0,0 +1,27 @@ +//Problem 38.08: When a transformer is connected to a 500 V, 50 Hz supply, the hysteresis and eddy current losses are 400 Wand 150 W respectively. The applied voltage is increased to 1 kV and the frequency to 100 Hz. Assuming the Steinmetz index to be 1.6, determine the new total core loss.
+
+//initializing the variables:
+V1 = 500; // in Volts
+V2 = 1000; // in Volts
+Ph1 = 400; // in W
+Pe1 = 150; // in W
+f1 = 50; // in Hz
+n = 1.6; // Steinmetz index
+f2 = 100; // in Hz
+
+//calculation:
+//hysteresis loss Ph = k1*f*(E/f)^n
+//At 500 V and 50 Hz
+k1 = Ph1/(f1*(V1/f1)^1.6)
+//At 1000 V and 100 Hz,
+Ph2 = k1*f2*(V2/f2)^1.6
+//eddy current loss, Pe = k2*E^2
+//At 500 V,
+k2 = Pe1/(V1^2)
+//At 1000 V,
+Pe2 = k2*(V2^2)
+//the new total core loss
+TL = Ph2 + Pe2
+
+printf("\n\n Result \n\n")
+printf("\n the new total core loss %.0f W",TL)
\ No newline at end of file diff --git a/608/CH38/EX38.10/38_10.sce b/608/CH38/EX38.10/38_10.sce new file mode 100755 index 000000000..9bd40d557 --- /dev/null +++ b/608/CH38/EX38.10/38_10.sce @@ -0,0 +1,33 @@ +//Problem 38.10: The core of a synchrogenerator has total losses of 400 W at 50 Hz and 498W at 60 Hz, the flux density being constant for the two tests. (a) Determine the hysteresis and eddy current losses at 50 Hz (b) If the flux density is increased by 25% and the lamination thickness is increased by 40%, determine the hysteresis and eddy current losses at 50 Hz. Assume the Steinmetz index to be 1.7.
+
+//initializing the variables:
+TL1 = 400; // in Watt
+TL2 = 498; // in Watt
+x = 0.25;
+y = 0.4;
+f1 = 50; // in Hz
+n = 1.7; // Steinmetz index
+f2 = 60; // in Hz
+
+//calculation:
+//if volume v and the maximum flux density are constant
+//hysteresis loss Ph = kh*v*f*(Bm)^n = k1*f
+//(if the maximum flux density and the lamination thickness are constant)
+//eddy current loss, Pe = ke*(Bm1*f1*t1)^2 = k2*f^2
+//At 50 Hz frequency, TL1 = k1*f1 + k2*f1^2
+//At 60 Hz frequency, TL2 = k1*f2 + k2*f2^2
+//Solving equations gives the values of k1 and k2.
+k2 = (5*TL2 - 6*TL1)/(5*(f2^2) - 6*(f1^2))
+k1 = (TL1 - k2*f1^2)/f1
+//hysteresis loss Ph = k1*f
+Ph1 = k1*f1
+//eddy current loss
+Pe1 = k2*f1^2
+//Since at 50 Hz the flux density is increased by 25%, the new hysteresis loss is
+Ph2 = Ph1*(1 + x)^1.7
+//Since at 50 Hz the flux density is increased by 25%, and the lamination thickness is increased by 40%, the new eddy current loss is
+Pe2 = Pe1*((1 + x)^2)*(1 + y)^3
+
+printf("\n\n Result \n\n")
+printf("\n (a)the hysteresis and eddy current losses at 50 Hz are %.0f W and %.0f W resp.",Ph1, Pe1)
+printf("\n (b)the hysteresis and eddy current losses at 50 Hz after increement are %.1f W and %.1f W resp.",Ph2, Pe2)
\ No newline at end of file diff --git a/608/CH39/EX39.01/39_01.sce b/608/CH39/EX39.01/39_01.sce new file mode 100755 index 000000000..0e889df3d --- /dev/null +++ b/608/CH39/EX39.01/39_01.sce @@ -0,0 +1,24 @@ +//Problem 39.01: The equivalent series circuit for a particular capacitor consists of a 1.5 ohm resistance in series with a 400 pF capacitor. Determine for the capacitor, at a frequency of 8 MHz, (a) the loss angle, (b) the power factor, (c) the Q-factor, and (d) the dissipation factor.
+
+//initializing the variables:
+Rs = 1.5; // in ohms
+Cs = 400E-12; // in Farads
+f = 8E6; // in Hz
+
+//calculation:
+//for a series equivalent circuit,
+//tan(del) = Rs*w*Cs
+//loss angle,
+del = atan(Rs*Cs*(2*%pi*f))
+//power factor
+pf = cos(del)
+//the Q-factor
+Q = 1/tan(del)
+//dissipation factor,
+D = 1/Q
+
+printf("\n\n Result \n\n")
+printf("\n (a)loss angle %.3f rad.",del)
+printf("\n (b)power factor %.3f rad.",del)
+printf("\n (c)Q-factor is %.2f ",Q)
+printf("\n (d)dissipation factor %.3f rad.",D)
\ No newline at end of file diff --git a/608/CH39/EX39.02/39_02.sce b/608/CH39/EX39.02/39_02.sce new file mode 100755 index 000000000..8f7e1d88b --- /dev/null +++ b/608/CH39/EX39.02/39_02.sce @@ -0,0 +1,17 @@ +//Problem 39.02: A capacitor has a loss angle of 0.025 rad, and when it is connected across a 5 kV, 50 Hz supply, the power loss is 20 .W Determine the component values of the equivalent parallel circuit.
+
+//initializing the variables:
+del = 0.025; // in rad.
+V = 5000; // in Volts
+PL = 20; // power loss
+f = 50; // in Hz
+
+//calculation:
+//power loss = w*C*V^2*tan(del)
+Cp = PL/(2*%pi*f*V*V*tan(del))
+//for a parallel equivalent circuit,
+//tan(del) = 1/(Rp*w*Cp)
+Rp = 1/(2*%pi*f*Cp*tan(del))
+
+printf("\n\n Result \n\n")
+printf("\n capacitance C %.2E F and parallel resistance %.2E ohm.",Cp, Rp)
\ No newline at end of file diff --git a/608/CH39/EX39.03/39_03.sce b/608/CH39/EX39.03/39_03.sce new file mode 100755 index 000000000..b5d906d17 --- /dev/null +++ b/608/CH39/EX39.03/39_03.sce @@ -0,0 +1,25 @@ +//Problem 39.03: A 2000 pF capacitor has an alternating voltage of 20 V connected across it at a frequency of 10 kHz. If the power dissipated in the dielectric is 500 μW, determine (a) the loss angle, (b) the equivalent series loss resistance, and (c) the equivalent parallel loss resistance.
+
+//initializing the variables:
+P = 500E-6; // in Watt
+C = 2000E-12; // in Farads
+V = 20; // in Volts
+f = 10000; // in Hz
+
+//calculation:
+//power loss = w*C*V^2*tan(del)
+//loss angle
+del = atan(P/(2*%pi*f*V*V*C))
+//for an equivalent series circuit,
+//tan(del) = (Rs*w*Cs)
+Cs = C
+Rs = (tan(del))/(2*%pi*f*Cp)
+//for an equivalent parallel circuit
+//tan(del) = 1/(Rp*w*Cp)
+Cp = C
+Rp = 1/(2*%pi*f*Cp*tan(del))
+
+printf("\n\n Result \n\n")
+printf("\n (a)loss angle %.6f rad.",del)
+printf("\n (b)series resistance %.2f ohm.",Rs)
+printf("\n (c)parallel resistance %.2E ohm.",Rp)
\ No newline at end of file diff --git a/608/CH4/EX4.01/4_01.sce b/608/CH4/EX4.01/4_01.sce new file mode 100755 index 000000000..885a45d3a --- /dev/null +++ b/608/CH4/EX4.01/4_01.sce @@ -0,0 +1,18 @@ +//Problem 4.01: Eight cells, each with an internal resistance of 0.2 ohms and an e.m.f. of 2.2 V are connected (a) in series, (b) in parallel. Determine the e.m.f. and the internal resistance of the batteries so formed.
+
+//initializing the variables:
+R = 0.2; // in ohms
+n = 8; // no. of cells
+e = 2.2; // in volts
+
+//calculation:
+es = n*e
+ep = e
+Rs = n*R
+Rp = R/n
+
+printf("\n\nResult\n\n")
+printf("\n(a)Resistance %.1f ohms",Rs)
+printf("\n(a)e.m.f %.1f Volts(V)",es)
+printf("\n(b)Resistance %.3f ohms",Rp)
+printf("\n(b)e.m.f %.1f Volts(V)",ep)
\ No newline at end of file diff --git a/608/CH4/EX4.02/4_02.sce b/608/CH4/EX4.02/4_02.sce new file mode 100755 index 000000000..31d30c617 --- /dev/null +++ b/608/CH4/EX4.02/4_02.sce @@ -0,0 +1,15 @@ +//Problem 4.02: A cell has an internal resistance of 0.02 ohms and an e.m.f. of 2.0 V. Calculate its terminal p.d. if it delivers (a) 5 A, (b) 50 A
+
+//initializing the variables:
+r = 0.02; // in ohms
+e = 2; // in volts
+I1 = 5; // in Amperes
+I2 = 50; // in Amperes
+
+//calculation:
+pd1 = e - (I1*r)
+pd2 = e - (I2*r)
+
+printf("\n\nResult\n\n")
+printf("\n(a)p.d %.1f Volts(V)",pd1)
+printf("\n(b)p.d %.1f Volts(V)\n",pd2)
\ No newline at end of file diff --git a/608/CH4/EX4.03/4_03.sce b/608/CH4/EX4.03/4_03.sce new file mode 100755 index 000000000..8ad6e200b --- /dev/null +++ b/608/CH4/EX4.03/4_03.sce @@ -0,0 +1,12 @@ +//Problem 4.03: The p.d. at the terminals of a battery is 25 V when no load is connected and 24 V when a load taking 10 A is connected. Determine the internal resistance of the battery.
+
+//initializing the variables:
+e1 = 25; // in volts
+e2 = 24; // in volts
+I2 = 10; // in Amperes
+
+//calculation:
+r = (e1 - e2)/I2
+
+printf("\n\nResult\n\n")
+printf("\n Resistance %.1f Ohms\n",r)
\ No newline at end of file diff --git a/608/CH4/EX4.04/4_04.sce b/608/CH4/EX4.04/4_04.sce new file mode 100755 index 000000000..16aa50bcd --- /dev/null +++ b/608/CH4/EX4.04/4_04.sce @@ -0,0 +1,17 @@ +//Problem 4.04: Ten 1.5 V cells, each having an internal resistance of 0.2 ohms, are connected in series to a load of 58 ohms . Determine(a) the current flowing in the circuit and (b) the p.d. at the battery terminals.
+
+//initializing the variables:
+r = 0.2; // in ohms
+n = 10; // no. of cells
+e = 1.5; // in volts
+R = 58; // in ohms
+
+//calculation:
+es = n*e
+rs = n*r
+I = es/(rs + R)
+pd = es - (I*rs)
+
+printf("\n\nResult\n\n")
+printf("\n (a)Current %.2f Amperes(A)",I)
+printf("\n (b)p.d %.1f Volts(V)\n",pd)
\ No newline at end of file diff --git a/608/CH40/EX40.01/40_01.sce b/608/CH40/EX40.01/40_01.sce new file mode 100755 index 000000000..16ba0886a --- /dev/null +++ b/608/CH40/EX40.01/40_01.sce @@ -0,0 +1,15 @@ +//Problem 40.01: A field plot between two metal plates is shown in Figure 40.9. The relative permeability of the dielectric is 2.8. Determine the capacitance per metre length of the system.
+
+//initializing the variables:
+e0 = 8.85E-12;
+er = 2.8;
+l = 1; // in m
+
+//calculation:
+//From Figure 40.9
+m = 16; // number of parallel squares measured along each equipotential
+n = 6; // the number of series squares measured along each line of force
+C = e0*er*l*m/n
+
+printf("\n\n Result \n\n")
+printf("\n capacitance is %.3E Farad.",C)
\ No newline at end of file diff --git a/608/CH40/EX40.02/40_02.sce b/608/CH40/EX40.02/40_02.sce new file mode 100755 index 000000000..05c6fb9aa --- /dev/null +++ b/608/CH40/EX40.02/40_02.sce @@ -0,0 +1,15 @@ +//Problem 40.02: A field plot for a cross-section of a concentric cable is shown in Figure 40.10. If the relative permeability of the dielectric is 3.4, determine the capacitance of a 100 m length of the cable.
+
+//initializing the variables:
+e0 = 8.85E-12;
+er = 3.4;
+l = 100; // in m
+
+//calculation:
+//From Figure 40.10
+m = 13; // number of parallel squares measured along each equipotential
+n = 4; // the number of series squares measured along each line of force
+C = e0*er*l*m/n
+
+printf("\n\n Result \n\n")
+printf("\n capacitance is %.3E Farad.",C)
\ No newline at end of file diff --git a/608/CH40/EX40.03/40_03.sce b/608/CH40/EX40.03/40_03.sce new file mode 100755 index 000000000..a894de01e --- /dev/null +++ b/608/CH40/EX40.03/40_03.sce @@ -0,0 +1,14 @@ +//Problem 40.03: A coaxial cable has an inner core radius of 0.5 mm and an outer conductor of internal radius 6.0 mm. Determine the capacitance per metre length of the cable if the dielectric has a relative permittivity of 2.7.
+
+//initializing the variables:
+e0 = 8.85E-12;
+er = 2.7;
+ri = 0.0005; // in m
+ro = 0.006; // in m
+
+//calculation:
+//capacitance C
+C = 2*%pi*e0*er/(log(ro/ri))
+
+printf("\n\n Result \n\n")
+printf("\n capacitance is %.3E Farad.",C)
\ No newline at end of file diff --git a/608/CH40/EX40.04/40_04.sce b/608/CH40/EX40.04/40_04.sce new file mode 100755 index 000000000..d35af8704 --- /dev/null +++ b/608/CH40/EX40.04/40_04.sce @@ -0,0 +1,14 @@ +//Problem 40.04: A single-core concentric cable has a capacitance of 80 pF per metre length. The relative permittivity of the dielectric is 3.5 and the core diameter is 8.0 mm. Determine the internal diameter of the sheath.
+
+//initializing the variables:
+C = 80E-12; // in Farads
+e0 = 8.85E-12;
+er = 3.5;
+d0 = 0.008; // in m
+
+//calculation:
+//internal diameter
+di = d0*(%e^(2*%pi*e0*er/C))
+
+printf("\n\n Result \n\n")
+printf("\n internal diameter is %.5f m.",di)
\ No newline at end of file diff --git a/608/CH40/EX40.05/40_05.sce b/608/CH40/EX40.05/40_05.sce new file mode 100755 index 000000000..13605ae35 --- /dev/null +++ b/608/CH40/EX40.05/40_05.sce @@ -0,0 +1,24 @@ +//Problem 40.05: A concentric cable has a core diameter of 32 mm and an inner sheath diameter of 80 mm. The core potential is 40 kV and the relative permittivity of the dielectric is 3.5. Determine (a) the capacitance per kilometre length of the cable, (b) the dielectric stress at a radius of 30 mm, and (c) the maximum and minimum values of dielectric stress.
+
+//initializing the variables:
+e0 = 8.85E-12;
+er = 3.5;
+di = 0.08; // in m
+d0 = 0.032; // in m
+r = 0.03; // in m
+V = 40000; // in Volts
+
+//calculation:
+//capacitance C
+C = 2*%pi*e0*er/(log(di/d0))
+//dielectric stress at radius r,
+E = V/(r*log(di/d0))
+//maximum dielectric stress,
+Emax = V/((d0/2)*(log((di/d0))))
+//minimum dielectric stress,
+Emin = V/((di/2)*(log((di/d0))))
+
+printf("\n\n Result \n\n")
+printf("\n capacitance is %.2E F/km",C*1E3)
+printf("\n dielectric stress at radius r is %.2E V/m",E)
+printf("\n maximum dielectric stress, is %.2E V/m minimum dielectric stress %.2E V/m",Emax, Emin)
\ No newline at end of file diff --git a/608/CH40/EX40.06/40_06.sce b/608/CH40/EX40.06/40_06.sce new file mode 100755 index 000000000..4b7e60ca2 --- /dev/null +++ b/608/CH40/EX40.06/40_06.sce @@ -0,0 +1,26 @@ +//Problem 40.06: A single-core concentric cable is to be manufactured for a 60 kV, 50 Hz transmission system. The dielectric used is paper which has a maximum permissible safe dielectric stress of 10 MV/m rms and a relative permittivity of 3.5. Calculate (a) the core and inner sheath radii for the most economical cable, (b) the capacitance per metre length, and (c) the charging current per kilometre run.
+
+//initializing the variables:
+e0 = 8.85E-12;
+er = 3.5;
+V = 60000; // in Volts
+f = 50; // in Hz
+Em = 10E6; // in V/m
+
+
+//calculation:
+//core radius, a
+a = V/Em
+//internal sheath radius,
+b = a*%e^1
+//capacitance
+C = 2*%pi*e0*er/(log(b/a))
+//Charging current
+I = V*2*%pi*f*C
+//charging current per kilometre
+Ipkm = I*1000
+
+printf("\n\n Result \n\n")
+printf("\n core radius is %.2E m and internal sheath radius %.2E m",a,b)
+printf("\n capacitance is %.2E F/m",C)
+printf("\n the charging current per kilometre %.2f A",Ipkm)
\ No newline at end of file diff --git a/608/CH40/EX40.07/40_07.sce b/608/CH40/EX40.07/40_07.sce new file mode 100755 index 000000000..eb6adfe5a --- /dev/null +++ b/608/CH40/EX40.07/40_07.sce @@ -0,0 +1,28 @@ +//Problem 40.07: A concentric cable has a core diameter of 25 mm and an inside sheath diameter of 80 mm. The relative permittivity of the dielectric is 2.5, the loss angle is 3.5 x 10-3 rad and the working voltage is 132 kV at 50 Hz frequency. Determine for a 1 km length of the cable (a) the capacitance, (b) the charging current and (c) the power loss.
+
+//initializing the variables:
+e0 = 8.85E-12;
+er = 2.5;
+di = 0.08; // in m
+d0 = 0.025; // in m
+r = 1000; // in m
+V = 132000; // in Volts
+f = 50; // in Hz
+del = 3.5E-3; // rad.
+
+//calculation:
+//core radius, a
+a = d0/2
+//internal sheath radius,
+b = di/2
+//capacitance
+C = 2*%pi*e0*er*1E3/(log(b/a))
+//Charging current
+I = V*2*%pi*f*C
+//power loss
+P = (2*%pi*f*C*tan(del))*V^2
+
+printf("\n\n Result \n\n")
+printf("\n (a)capacitance for a 1 km length is %.2E F",C)
+printf("\n (b)the charging current %.2E A/km",I)
+printf("\n (c)power loss %.0f W",P)
\ No newline at end of file diff --git a/608/CH40/EX40.08/40_08.sce b/608/CH40/EX40.08/40_08.sce new file mode 100755 index 000000000..6dc834e6d --- /dev/null +++ b/608/CH40/EX40.08/40_08.sce @@ -0,0 +1,18 @@ +//Problem 40.08: A concentric cable has a core diameter of 20 mm and a sheath inside diameter of 60 mm. The permittivity of the dielectric is 3.2. Using three equipotential surfaces within the dielectric, determine the capacitance of the cable per metre length by the method of curvilinear squares. Draw the field plot for the cable.
+
+//initializing the variables:
+e0 = 8.85E-12;
+er = 3.2;
+di = 0.06; // in m
+d0 = 0.020; // in m
+
+//calculation:
+//core radius, a
+a = d0/2
+//internal sheath radius,
+b = di/2
+//capacitance
+C = 2*%pi*e0*er/(log(b/a))
+
+printf("\n\n Result \n\n")
+printf("\n capacitance per m of length is %.2E F",C)
\ No newline at end of file diff --git a/608/CH40/EX40.09/40_09.sce b/608/CH40/EX40.09/40_09.sce new file mode 100755 index 000000000..5a247543e --- /dev/null +++ b/608/CH40/EX40.09/40_09.sce @@ -0,0 +1,17 @@ +//Problem 40.09: Two parallel wires, each of diameter 5 mm, are uniformly spaced in air at a distance of 50 mm between centres. Determine the capacitance of the line if the total length is 200 m.
+
+//initializing the variables:
+e0 = 8.85E-12;
+er = 1;
+D = 0.05; // in m
+d = 0.005; // in m
+l = 200; // in m
+
+//calculation:
+//capacitance
+C = %pi*e0*er/(log(D/(d/2)))
+//capacitance of a 200 m length
+C200 = C*l
+
+printf("\n\n Result \n\n")
+printf("\n capacitance of a 200 m length is %.2E F",C200)
\ No newline at end of file diff --git a/608/CH40/EX40.10/40_10.sce b/608/CH40/EX40.10/40_10.sce new file mode 100755 index 000000000..b9404c8c7 --- /dev/null +++ b/608/CH40/EX40.10/40_10.sce @@ -0,0 +1,25 @@ +//Problem 40.10: A single-phase circuit is composed of two parallel conductors, each of radius 4 mm, spaced 1.2 m apart in air. The p.d. between the conductors at a frequency of 50 Hz is 15 kV. Determine, for a 1 km length of line, (a) the capacitance of the conductors, (b) the value of charge carried by each conductor, and (c) the charging current.
+
+//initializing the variables:
+e0 = 8.85E-12;
+er = 1;
+D = 1.2; // in m
+r = 0.004; // in m
+f = 50; // in Hz
+V = 15000; // in Volts
+l = 1000; // in m
+
+//calculation:
+//capacitance
+C = %pi*e0*er/(log(D/r))
+//capacitance of a 1 km length
+Cpkm = C*l
+//Charge Q
+Q = Cpkm*V
+//Charging current
+I = V*2*%pi*f*Cpkm
+
+printf("\n\n Result \n\n")
+printf("\n capacitance per 1km length is %.2E F",Cpkm)
+printf("\n Charge Q is %.2E C",Q)
+printf("\n Charging current is %.3f A",I)
\ No newline at end of file diff --git a/608/CH40/EX40.11/40_11.sce b/608/CH40/EX40.11/40_11.sce new file mode 100755 index 000000000..18e41e1a2 --- /dev/null +++ b/608/CH40/EX40.11/40_11.sce @@ -0,0 +1,22 @@ +//Problem 40.11:The charging current for an 800 m run of isolated twin line is not to exceed 15 mA. The voltage between the lines is 10 kV at 50 Hz. If the line is air-insulated, determine (a) the maximum value required for the capacitance per metre length, and (b) the maximum diameter of each conductor if their distance between centres is 1.25 m.
+
+//initializing the variables:
+e0 = 8.85E-12;
+er = 1;
+I = 0.015; // in Amperes
+d = 1.25; // in m
+r = 800; // in m
+f = 50; // in Hz
+V = 10000; // in Volts
+
+//calculation:
+//capacitance
+C = I/(2*%pi*f*V)
+//required maximum value of capacitance
+Cmax = C/r
+//maximum diameter of each conductor
+D = 2*d/(%e^(%pi*e0*er/Cmax))
+
+printf("\n\n Result \n\n")
+printf("\n required maximum value of capacitance is %.2E F/m",Cmax)
+printf("\nthe maximum diameter of each conductor is %.4f m",D)
\ No newline at end of file diff --git a/608/CH40/EX40.12/40_12.sce b/608/CH40/EX40.12/40_12.sce new file mode 100755 index 000000000..3b58d43f0 --- /dev/null +++ b/608/CH40/EX40.12/40_12.sce @@ -0,0 +1,18 @@ +//Problem 40.12: Determine the energy stored in a 10 nF capacitor when charged to 1 kV, and the average power developed if this energy is dissipated in 10 μs.
+
+//initializing the variables:
+e0 = 8.85E-12;
+er = 1;
+C = 10E-9; // in Farad
+V = 1000; // in Volts
+t = 10E-6; // in sec
+
+//calculation:
+//energy stored,Wf
+Wf = C*V*V/2
+//average power developed
+Pav = Wf/t
+
+printf("\n\n Result \n\n")
+printf("\n the energy stored is %.2E J",Wf)
+printf("\nthe average power developed is %.0f W",Pav)
\ No newline at end of file diff --git a/608/CH40/EX40.13/40_13.sce b/608/CH40/EX40.13/40_13.sce new file mode 100755 index 000000000..4bf599745 --- /dev/null +++ b/608/CH40/EX40.13/40_13.sce @@ -0,0 +1,17 @@ +//Problem 40.13: A capacitor is charged with 5 mC. If the energy stored is 625 mJ, determine (a) the voltage across the plates and (b) the capacitance of the capacitor.
+
+//initializing the variables:
+e0 = 8.85E-12;
+er = 1;
+Q = 5E-3; // in Coulomb
+W = 0.625; // in Joules
+
+//calculation:
+//voltage across the plates
+V = 2*W/Q
+//Capacitance C
+C = Q/V
+
+printf("\n\n Result \n\n")
+printf("\n voltage across the plates is %.0f V",V)
+printf("\n Capacitance C is %.2E F",C)
\ No newline at end of file diff --git a/608/CH40/EX40.14/40_14.sce b/608/CH40/EX40.14/40_14.sce new file mode 100755 index 000000000..166ea3a9b --- /dev/null +++ b/608/CH40/EX40.14/40_14.sce @@ -0,0 +1,21 @@ +//Problem 40.14: A ceramic capacitor is to be constructed to have a capacitance of 0.01 μF and to have a steady working potential of 2.5 kV maximum. Allowing a safe value of field stress of 10 MV/m, determine (a) the required thickness of the ceramic dielectric, (b) the area of plate required if the relative permittivity of the ceramic is 10, and (c) the maximum energy stored by the capacitor.
+
+//initializing the variables:
+e0 = 8.85E-12;
+er = 10;
+C = 0.01E-6; // in Farad
+E = 10E6; // in V/m
+V = 2500; // in Volts
+
+//calculation:
+//thickness of ceramic dielectric,
+d = V/E
+//cross-sectional area of plate
+A = C*d/(e0*er)
+//Maximum energy stored,
+W = C*V*V/2
+
+printf("\n\n Result \n\n")
+printf("\n thickness of ceramic dielectric is %.2E m",d)
+printf("\n cross-sectional area of plate, is %.4f m2",A)
+printf("\n Maximum energy stored is %.4f J",W)
\ No newline at end of file diff --git a/608/CH40/EX40.15/40_15.sce b/608/CH40/EX40.15/40_15.sce new file mode 100755 index 000000000..b1f16ae8c --- /dev/null +++ b/608/CH40/EX40.15/40_15.sce @@ -0,0 +1,15 @@ +//Problem 40.15: A 400 pF capacitor is charged to a p.d. of 100 V. The dielectric has a cross-sectional area of 200 cm2 and a relative permittivity of 2.3. Calculate the energy stored per cubic metre of the dielectric.
+
+//initializing the variables:
+e0 = 8.85E-12;
+er = 2.3;
+A = 0.02; // in m2
+C = 400E-12; // in Farad
+V = 100; // in Volts
+
+//calculation:
+//energy stored per unit volume of dielectric,
+W = ((C*V)^2)/(2*e0*er*A^2)
+
+printf("\n\n Result \n\n")
+printf("\n energy stored per unit volume of dielectric is %.4f J/m3",W)
\ No newline at end of file diff --git a/608/CH40/EX40.16/40_16.sce b/608/CH40/EX40.16/40_16.sce new file mode 100755 index 000000000..518dcfa58 --- /dev/null +++ b/608/CH40/EX40.16/40_16.sce @@ -0,0 +1,14 @@ +//Problem 40.16: A coaxial cable has an inner core of radius 1.0 mm and an outer sheath of internal radius 4.0 mm. Determine the inductance of the cable per metre length. Assume that the relative permeability is unity.
+
+//initializing the variables:
+u0 = 4*%pi*1E-7;
+ur = 1;
+a = 0.001; // in m
+b = 0.004; // in m
+
+//calculation:
+//inductance L
+L = (u0*ur/(2*%pi))*(0.25 + log(b/a))
+
+printf("\n\n Result \n\n")
+printf("\n inductance L is %.2E H/m",L)
\ No newline at end of file diff --git a/608/CH40/EX40.17/40_17.sce b/608/CH40/EX40.17/40_17.sce new file mode 100755 index 000000000..972a834c4 --- /dev/null +++ b/608/CH40/EX40.17/40_17.sce @@ -0,0 +1,14 @@ +//Problem 40.17: A concentric cable has a core diameter of 10 mm. The inductance of the cable is 4 x 10^-7 H/m. Ignoring inductance due to internal linkages, determine the diameter of the sheath. Assume that the relative permeability is 1.
+
+//initializing the variables:
+u0 = 4*%pi*1E-7;
+ur = 1;
+da = 0.010; // in m
+L = 4E-7; // in H/m
+
+//calculation:
+//diameter of the sheath
+db = da*(%e^(L/(u0*ur/(2*%pi))))
+
+printf("\n\n Result \n\n")
+printf("\n diameter of the sheath is %.4f m",db)
\ No newline at end of file diff --git a/608/CH40/EX40.18/40_18.sce b/608/CH40/EX40.18/40_18.sce new file mode 100755 index 000000000..b69f1c60f --- /dev/null +++ b/608/CH40/EX40.18/40_18.sce @@ -0,0 +1,24 @@ +//Problem 40.18: A coaxial cable 7.5 km long has a core 10 mm diameter and a sheath 25 mm diameter, the sheath having negligible thickness. Determine for the cable (a) the inductance, assuming nonmagnetic materials, and (b) the capacitance, assuming a dielectric of relative permittivity 3.
+
+//initializing the variables:
+u0 = 4*%pi*1E-7;
+ur = 1;
+e0 = 8.85E-12;
+er = 3;
+da = 0.010; // in m
+db = 0.025; // in m
+l = 7500; // in m
+
+//calculation:
+//inductance per metre length
+L = (u0*ur/(2*%pi))*(0.25 + log(db/da))
+//Since the cable is 7500 m long,
+L7500 = L*7500
+//capacitance C
+C = 2*%pi*e0*er/(log(db/da))
+////Since the cable is 7500 m long,
+C7500 = C*7500
+
+printf("\n\n Result \n\n")
+printf("\ninductance is %.5f H",L7500)
+printf("\ncapCItance is %.2E F",C7500)
\ No newline at end of file diff --git a/608/CH40/EX40.19/40_19.sce b/608/CH40/EX40.19/40_19.sce new file mode 100755 index 000000000..6c1f52fde --- /dev/null +++ b/608/CH40/EX40.19/40_19.sce @@ -0,0 +1,16 @@ +//Problem 40.19: A single-phase power line comprises two conductors each with a radius 8.0 mm and spaced 1.2 m apart in air. Determine the inductance of the line per metre length ignoring internal linkages. Assume the relative permeability ur = 1.
+
+//initializing the variables:
+u0 = 4*%pi*1E-7;
+ur = 1;
+e0 = 8.85E-12;
+er = 3;
+D = 1.2; // in m
+a = 0.008; // in m
+
+//calculation:
+//inductance per metre length
+L = (u0*ur/(%pi))*(log(D/a))
+
+printf("\n\n Result \n\n")
+printf("\ninductance is %.2E H/m",L)
\ No newline at end of file diff --git a/608/CH40/EX40.20/40_20.sce b/608/CH40/EX40.20/40_20.sce new file mode 100755 index 000000000..5b14b779d --- /dev/null +++ b/608/CH40/EX40.20/40_20.sce @@ -0,0 +1,24 @@ +//Problem 40.20: Determine (a) the loop inductance, and (b) the capacitance of a 1 km length of single-phase twin line having conductors of diameter 10 mm and spaced 800 mm apart in air.
+
+//initializing the variables:
+u0 = 4*%pi*1E-7;
+ur = 1;
+e0 = 8.85E-12;
+er = 1;
+l = 1000; // in m
+D = 0.8; // in m
+a = 0.01/2; // in m
+
+//calculation:
+//inductance per metre length
+L = (u0*ur/(%pi))*(0.25 + log(D/a))
+//Since the cable is 1000 m long,
+L1k = L*l
+//capacitance C
+C = %pi*e0*er/(log(D/a))
+////Since the cable is 1000 m long,
+C1k = C*l
+
+printf("\n\n Result \n\n")
+printf("\ninductance is %.5f H",L1k)
+printf("\ncapcitance is %.2E F",C1k)
\ No newline at end of file diff --git a/608/CH40/EX40.21/40_21.sce b/608/CH40/EX40.21/40_21.sce new file mode 100755 index 000000000..e7769698f --- /dev/null +++ b/608/CH40/EX40.21/40_21.sce @@ -0,0 +1,14 @@ +//Problem 40.21: The total loop inductance of an isolated twin power line is 2.185 μH/m. The diameter of each conductor is 12 mm. Determine the distance between their centres.
+
+//initializing the variables:
+L = 2.185E-6; // in H/m
+u0 = 4*%pi*1E-7;
+ur = 1;
+a = 0.012/2; // in m
+
+//calculation:
+//distance D
+D = a*%e^((L*%pi)/(u0*ur) - 0.25)
+
+printf("\n\n Result \n\n")
+printf("\ndistance D is %.2f m",D)
\ No newline at end of file diff --git a/608/CH40/EX40.22/40_22.sce b/608/CH40/EX40.22/40_22.sce new file mode 100755 index 000000000..db2ba953c --- /dev/null +++ b/608/CH40/EX40.22/40_22.sce @@ -0,0 +1,17 @@ +//Problem 40.22: Calculate the value of the energy stored when a current of 50 mA is flowing in a coil of inductance 200 mH. What value of current would double the energy stored?
+
+//initializing the variables:
+L = 0.2; // in H
+I = 0.05; // in Amperes
+u0 = 4*%pi*1E-7;
+ur = 1;
+
+//calculation:
+//energy stored in inductor
+W = L*I*I/2
+//current I
+I = (2*2*W/L)^0.5
+
+printf("\n\n Result \n\n")
+printf("\nenergy stored in inductor is %.2E J",W)
+printf("\ncurrent I is %.2E A",I)
\ No newline at end of file diff --git a/608/CH40/EX40.23/40_23.sce b/608/CH40/EX40.23/40_23.sce new file mode 100755 index 000000000..9d6ade79c --- /dev/null +++ b/608/CH40/EX40.23/40_23.sce @@ -0,0 +1,18 @@ +//Problem 40.23: The airgap of a moving coil instrument is 2.0 mm long and has a cross-sectional area of 500 mm2. If the flux density is 50 mT, determine the total energy stored in the magnetic field of the airgap.
+
+//initializing the variables:
+B = 0.05; // in Tesla
+A = 500E-6; // in m2
+l = 0.002; // in m
+u0 = 4*%pi*1E-7;
+
+//calculation:
+//energy stored
+W = (B^2)/(2*u0)
+//Volume of airgap
+v = A*l
+//energy stored in airgap
+W = W*v
+
+printf("\n\n Result \n\n")
+printf("\nenergy stored in the airgap is %.2E J",W)
\ No newline at end of file diff --git a/608/CH40/EX40.24/40_24.sce b/608/CH40/EX40.24/40_24.sce new file mode 100755 index 000000000..6eb008765 --- /dev/null +++ b/608/CH40/EX40.24/40_24.sce @@ -0,0 +1,19 @@ +//Problem 40.24: Determine the strength of a uniform electric field if it is to have the same energy as that established by a magnetic field of flux density 0.8 T. Assume that the relative permeability of the magnetic field and the relative permittivity of the electric field are both unity.
+
+//initializing the variables:
+B = 0.8; // in Tesla
+A = 500E-6; // in m2
+l = 0.002; // in m
+u0 = 4*%pi*1E-7;
+ur = 1;
+e0 = 8.85E-12;
+er = 1;
+
+//calculation:
+//energy stored in mag. field
+W = (B^2)/(2*u0)
+//electric field
+E = (2*W/(e0*er))^0.5
+
+printf("\n\n Result \n\n")
+printf("\nelectric field strength is %.2E V/m",E)
\ No newline at end of file diff --git a/608/CH41/EX41.01/41_01.sce b/608/CH41/EX41.01/41_01.sce new file mode 100755 index 000000000..38aae30ed --- /dev/null +++ b/608/CH41/EX41.01/41_01.sce @@ -0,0 +1,26 @@ +//Problem 41.01: The ratio of output power to input power in a system is
+//(a)2 (b) 25 (c) 1000 and (d) 0.01
+//Determine the power ratio in each case (i) in decibels and (ii) in nepers.
+
+//initializing the variables:
+//ratio of output power to input power
+rp1 = 2;
+rp2 = 25;
+rp3 = 1000;
+rp4 = 0.01;
+
+//calculation:
+//power ratio in decibels
+rpd1 = 10*log10(rp1)
+rpd2 = 10*log10(rp2)
+rpd3 = 10*log10(rp3)
+rpd4 = 10*log10(rp4)
+//power ratio in nepers
+rpn1 = (log(rp1))/2
+rpn2 = (log(rp2))/2
+rpn3 = (log(rp3))/2
+rpn4 = (log(rp4))/2
+
+printf("\n\n Result \n\n")
+printf("\n power ratio in decibels are (a)%.0f dB (b)%.0f dB (c) %.0f dB and (d) %.0f dB",rpd1,rpd2,rpd3,rpd4)
+printf("\n power ratio in nepers are (a)%.3f Np (b)%.3f Np (c) %.3f Np and (d) %.3f Np",rpn1,rpn2,rpn3,rpn4)
\ No newline at end of file diff --git a/608/CH41/EX41.02/41_02.sce b/608/CH41/EX41.02/41_02.sce new file mode 100755 index 000000000..0d2eab963 --- /dev/null +++ b/608/CH41/EX41.02/41_02.sce @@ -0,0 +1,11 @@ +//Problem 41.02: 5% of the power supplied to a cable appears at the output terminals. Determine the attenuation in decibels.
+
+//initializing the variables:
+rp = 0.05; // power ratio P2/P1
+
+//calculation:
+//power ratio in decibels
+rpd = 10*log10(rp)
+
+printf("\n\n Result \n\n")
+printf("\nthe attenuation is %.0f dB",abs(rpd))
\ No newline at end of file diff --git a/608/CH41/EX41.03/41_03.sce b/608/CH41/EX41.03/41_03.sce new file mode 100755 index 000000000..ee1ba8d55 --- /dev/null +++ b/608/CH41/EX41.03/41_03.sce @@ -0,0 +1,12 @@ +//Problem 41.03: An amplifier has a gain of 15 dB. If the input power is 12 mW, determine the output power.
+
+//initializing the variables:
+gain = 1.5; // in dB
+Pi = 0.012; // in Watt
+
+//calculation:
+//output power
+Po = Pi*10^gain
+
+printf("\n\n Result \n\n")
+printf("\noutput power is %.4f W",Po)
\ No newline at end of file diff --git a/608/CH41/EX41.04/41_04.sce b/608/CH41/EX41.04/41_04.sce new file mode 100755 index 000000000..24d0b5cc0 --- /dev/null +++ b/608/CH41/EX41.04/41_04.sce @@ -0,0 +1,15 @@ +//Problem 41.04: The current output of an attenuator is 50 mA. If the current ratio of the attenuator is 1.32 Np, determine (a) the current input and (b) the current ratio expressed in decibels. Assume that the input and load resistances of the attenuator are equal.
+
+//initializing the variables:
+I2 = 0.05; // in Amperes
+rin = 1.32; // in Np
+
+//calculation:
+//current input, I1
+I1 = I2*%e^(rin)
+//current ratio in decibels
+rid = 20*log10(I2/I1)
+
+printf("\n\n Result \n\n")
+printf("\ncurrent input, I1 is %.4f A",I1)
+printf("\ncurrent ratio in decibels is %.2f dB",rid)
\ No newline at end of file diff --git a/608/CH41/EX41.05/41_05.sce b/608/CH41/EX41.05/41_05.sce new file mode 100755 index 000000000..08ee78cda --- /dev/null +++ b/608/CH41/EX41.05/41_05.sce @@ -0,0 +1,18 @@ +//Problem 41.05: Determine the characteristic impedance of each of the attenuator sections shown in Figure 41.9.
+
+//initializing the variables:
+R1a = 8; // in ohm
+R2a = 21; // in ohm
+R1b = 10; // in ohm
+R2b = 15; // in ohm
+R1c = 200; // in ohm
+R2c = 56.25; // in ohm
+
+//calculation:
+//for a T-section attenuator the characteristic impedance
+Roa = (R1a^2 + 2*R1a*R2a)^0.5
+Rob = (R1b^2 + 2*R1b*R2b)^0.5
+Roc = (R1c^2 + 2*R1c*R2c)^0.5
+
+printf("\n\n Result \n\n")
+printf("\nfor a T-sections attenuator the characteristic impedances are (a) %.0f ohm, (b) %.0f ohm and (c)%.0f ohm",Roa,Rob,Roc)
\ No newline at end of file diff --git a/608/CH41/EX41.06/41_06.sce b/608/CH41/EX41.06/41_06.sce new file mode 100755 index 000000000..5354bb465 --- /dev/null +++ b/608/CH41/EX41.06/41_06.sce @@ -0,0 +1,22 @@ +//Problem 41.06: A symmetrical pi-attenuator pad has a series arm of 500 ohm resistance and each shunt arm of 1 kohm resistance. Determine (a) the characteristic impedance, and (b) the attenuation (in dB) produced by the pad.
+
+//initializing the variables:
+R1 = 500; // in ohm
+R2 = 1000; // in ohm
+I1 = 1; // in ampere (lets say)
+
+//calculation:
+// for symmetrical pi-attenuator section
+//characteristic impedance, R0
+R0 = (R1*(R2^2)/(R1 + 2*R2))^0.5
+//current Ix
+Ix = (R2/(R2 + R1 + (R2*R0/(R2 + R0))))*I1
+//current I2
+I2 = (R2/(R2 + R0))*Ix
+ri = I1/I2; // retio of currents
+//attenuation
+attn = 20*log10(ri)
+
+printf("\n\n Result \n\n")
+printf("\n the characteristic impedance is %.0f ohm",R0)
+printf("\n attenuation is %.2f dB",attn)
\ No newline at end of file diff --git a/608/CH41/EX41.07/41_07.sce b/608/CH41/EX41.07/41_07.sce new file mode 100755 index 000000000..046d60bac --- /dev/null +++ b/608/CH41/EX41.07/41_07.sce @@ -0,0 +1,21 @@ +//Problem 41.07: For each of the attenuator networks shown in Figure 41.11, determine (a) the input resistance when the output port is open-circuited, (b) the input resistance when the output port is short-circuited, and (c) the characteristic impedance.
+
+//initializing the variables:
+R1 = 15; // in ohm
+R2 = 10; // in ohm
+R3 = 5; // in ohm
+
+//calculation:
+//For the T-network shown in Figure 41.11(i):
+Roct = R1 + R2
+Rsct = R1 + R1*R2/(R1 + R2)
+Rot = (Roct*Rsct)^0.5
+//For the Pi-network shown in Figure 41.11(ii):
+Rocpi = R3*(R1 + R3)/(R3 + R1 + R3)
+Rscpi = R3*R1/(R3 + R1)
+Ropi = (R1*(R3^2)/(R1 + 2*R3))^0.5
+
+printf("\n\n Result \n\n")
+printf("\n the input resistance when the output port is open-circuited is (a) %.0f ohm (b) %.0f ohm",Roct,Rocpi)
+printf("\n the input resistance when the output port is short-circuited is (a) %.0f ohm (b) %.2f ohm",Rsct, Rscpi)
+printf("\n the characteristic impedance is (a) %.1f ohm (b) %.2f ohm",Rot, Ropi)
\ No newline at end of file diff --git a/608/CH41/EX41.10/41_10.sce b/608/CH41/EX41.10/41_10.sce new file mode 100755 index 000000000..101d321ff --- /dev/null +++ b/608/CH41/EX41.10/41_10.sce @@ -0,0 +1,20 @@ +//Problem 41.10: The attenuator shown in Figure 41.15 feeds a matched load. Determine (a) the characteristic impedance R0, and (b) the insertion loss in decibels.
+
+//initializing the variables:
+R1 = 300; // in ohm
+R2 = 450; // in ohm
+I1 = 1; // in ampere (lets say)
+
+//calculation:
+//the characteristic impedance of a symmetric T-pad attenuator is given by
+R0 = (R1^2 + 2*R1*R2)^0.5
+//By current division
+//current I2
+I2 = (R2/(R2 + R1+ R0))*I1
+ri = I1/I2; // ratio of currents
+//insertion loss
+il = 20*log10(ri)
+
+printf("\n\n Result \n\n")
+printf("\n the characteristic impedance is %.0f ohm",R0)
+printf("\n insertion loss is %.2f dB",il)
\ No newline at end of file diff --git a/608/CH41/EX41.11/41_11.sce b/608/CH41/EX41.11/41_11.sce new file mode 100755 index 000000000..32d3d14a1 --- /dev/null +++ b/608/CH41/EX41.11/41_11.sce @@ -0,0 +1,27 @@ +//Problem 41.11: A 0–3 kohm rheostat is connected across the output of a signal generator of internal resistance 500 ohm. If a load of 2 kohm is connected across the rheostat, determine the insertion loss at a tapping of (a) 2 kohm, (b) 1 kohm.
+
+//initializing the variables:
+r = 500; // in ohm
+Rhm = 3000; // in ohm
+RL = 2000; // in ohm
+r1 = 2000; // in ohm
+r2 = 1000; // in ohm
+E = 1; // in volts (lets say)
+
+//calculation:
+//Without the rheostat in the circuit the voltage across the 2 kohm load, VL
+VL = (RL/(RL + r))*E
+//voltage V2 with 2kohm tapping
+V2 = ((RL*r1/(r1 + RL))/((RL*r1/(r1 + RL)) + Rhm - r1 + r))*E
+rv1 = VL/V2; // ratio of currents
+//insertion loss
+il1 = 20*log10(rv1)
+//voltage V1 with 1kohm tapping
+V1 = ((RL*r2/(r2 + RL))/((RL*r2/(r2 + RL)) + Rhm - r2 + r))*E
+rv2 = VL/V1; // ratio of currents
+//insertion loss
+il2 = 20*log10(rv2)
+
+printf("\n\n Result \n\n")
+printf("\n insertion loss for 2kohm tap is %.2f dB",il1)
+printf("\n insertion loss for 1kohm tap is %.2f dB",il2)
\ No newline at end of file diff --git a/608/CH41/EX41.12/41_12.sce b/608/CH41/EX41.12/41_12.sce new file mode 100755 index 000000000..b0864b6be --- /dev/null +++ b/608/CH41/EX41.12/41_12.sce @@ -0,0 +1,21 @@ +//Problem 41.12: A symmetrical pi-attenuator pad has a series arm of resistance 1000 ohm and shunt arms each of 500ohm . Determine (a) its characteristic impedance, and (b) the insertion loss (in decibels) when feeding a matched load..
+
+//initializing the variables:
+R1 = 1000; // in ohm
+R2 = 500; // in ohm
+I1 = 1; // in amperes (lets say)
+
+//calculation:
+//characteristic impedance of a symmetrical attenuator
+R0 = (R1*(R2^2)/(R1 + 2*R2))^0.5
+//current Ix
+Ix = (R2/(R2 + R1 + (R2*R0/(R2 + R0))))*I1
+//current I2
+I2 = (R2/(R2 + R0))*Ix
+ri = I1/I2; // retio of currents
+//insertion loss
+il = 20*log10(ri)
+
+printf("\n\n Result \n\n")
+printf("\n characteristic impedance is %.0f ohm",R0)
+printf("\n insertion loss is %.2f dB",il)
\ No newline at end of file diff --git a/608/CH41/EX41.13/41_13.sce b/608/CH41/EX41.13/41_13.sce new file mode 100755 index 000000000..dd15f0729 --- /dev/null +++ b/608/CH41/EX41.13/41_13.sce @@ -0,0 +1,21 @@ +//Problem 41.13: An asymmetrical T-section attenuator is shown in Figure 41.24. Determine for the section (a) the image impedances, and (b) the iterative impedances.
+
+//initializing the variables:
+R1 = 100; // in ohm
+R2 = 200; // in ohm
+R3 = 300; // in ohm
+I1 = 1; // in amperes (lets say)
+
+//calculation:
+//image impedance Roa
+Roa = ((R1 + R2)*(R2 + (R1*R3/(R1 + R3))))^0.5
+//image impedance Rob
+Rob = ((R1 + R3)*(R3 + (R1*R2/(R1 + R2))))^0.5
+//The iterative impedance at port 1
+Ri1 = (-1*R1 + (R1^2 - (-1*4*((R2*(R1 + R3)) + (R3*R1))))^0.5)/2
+//The iterative impedance at port 2
+Ri2 = (R1 + (R1^2 - (-1*4*((R3*(R1 + R2)) + (R2*R1))))^0.5)/2
+
+printf("\n\n Result \n\n")
+printf("\n image impedance are %.1f ohm and %.0f ohm ",Roa,Rob)
+printf("\n iterative impedances are %.1f ohm and %.1f ohm ",Ri1,Ri2)
\ No newline at end of file diff --git a/608/CH41/EX41.14/41_14.sce b/608/CH41/EX41.14/41_14.sce new file mode 100755 index 000000000..02c55ae78 --- /dev/null +++ b/608/CH41/EX41.14/41_14.sce @@ -0,0 +1,21 @@ +//Problem 41.14: An asymmetrical pi-section attenuator is shown in Figure 41.28. Determine for the section (a) the image impedances, and (b) the iterative impedances.
+
+//initializing the variables:
+R1 = 1000; // in ohm
+R2 = 2000; // in ohm
+R3 = 3000; // in ohm
+I1 = 1; // in amperes (lets say)
+
+//calculation:
+//image impedance Roa
+Roa = (((R3 + R2)*R1/(R1 + R2 + R3))*(R1*R3/(R1 + R3)))^0.5
+//image impedance Rob
+Rob = (((R3 + R1)*R2/(R1 + R2 + R3))*(R2*R3/(R2 + R3)))^0.5
+//The iterative impedance at port 1
+Ri1 = (-1*R1 + ((R1^2) - (-1*4*2*R2*R1))^0.5)/(2*2)
+//The iterative impedance at port 2
+Ri2 = (R1 + ((-1*R1)^2 - (-1*4*2*R2*R1))^0.5)/(2*2)
+
+printf("\n\n Result \n\n")
+printf("\n image impedance are %.0f ohm and %.0f ohm ",Roa,Rob)
+printf("\n iterative impedances are %.0f ohm and %.0f ohm ",Ri1,Ri2)
\ No newline at end of file diff --git a/608/CH41/EX41.15/41_15.sce b/608/CH41/EX41.15/41_15.sce new file mode 100755 index 000000000..5c1244148 --- /dev/null +++ b/608/CH41/EX41.15/41_15.sce @@ -0,0 +1,36 @@ +//Problem 41.15: A generator having an internal resistance of 500 ohm is connected to a 100 ohm load via an impedance-matching resistance pad as shown in Figure 41.33. Determine (a) the values of resistance R1 and R2, (b) the attenuation of the pad in decibels, and (c) its insertion loss.
+
+//initializing the variables:
+r = 500; // in ohm
+RL = 100; // in ohm
+E = 1; // in volts (lets say)
+
+//calculation:
+//res.
+R1 = (r*(r - RL))^0.5
+R2 = (r*RL^2/(r - RL))^0.5
+//current I1
+I1 = E/(r + R1 + R2*RL/(RL + R2))
+//current I2
+I2 = (R2/(R2 + RL))*I1
+//input power
+P1 = r*I1^2
+//output power
+P2 = RL*I2^2
+//attenuation
+attn = 10*log10(P1/P2)
+//Load current, IL
+IL = E/(r + RL)
+//voltage, VL
+VL = IL*RL
+//voltage, V1
+V1 = E - I1*r
+//voltage, V2
+V2 = V1 - I1*R1
+//insertion loss
+il = 20*log10(VL/V2)
+
+printf("\n\n Result \n\n")
+printf("\n R1 = %.1f ohm and R2 = %.1f ohm ",R1,R2)
+printf("\n attenuation is %.2f dB ",attn)
+printf("\n In decibels, the insertion loss is %.2f dB ",il)
\ No newline at end of file diff --git a/608/CH41/EX41.16/41_16.sce b/608/CH41/EX41.16/41_16.sce new file mode 100755 index 000000000..fd6c1b9b9 --- /dev/null +++ b/608/CH41/EX41.16/41_16.sce @@ -0,0 +1,19 @@ +//Problem 41.16: Five identical attenuator sections are connected in cascade. The overall attenuation is 70 dB and the voltage input to the first section is 20 mV. Determine (a) the attenuation of eac individual attenuation section, (b) the voltage output of the fina stage, and (c) the voltage output of the third stage.
+
+//initializing the variables:
+attnO = 70; // in dB
+n = 5; // numbers of identical atteneurs
+V1 = 0.02; // in Volts
+
+//calculation:
+//attenuation of each section
+attn = attnO/n
+//output of the final stage
+Vo = V1/(10^(attnO/20))
+//voltage output of the third stage
+V3 = V1/(10^(3*attn/20))
+
+printf("\n\n Result \n\n")
+printf("\n attenuation of each section = %.0f dB ",attn)
+printf("\n output of the final stage is %.2E V ",Vo)
+printf("\n voltage output of the third stage is %.2E V ",V3)
\ No newline at end of file diff --git a/608/CH42/EX42.01/42_01.sce b/608/CH42/EX42.01/42_01.sce new file mode 100755 index 000000000..d776ee867 --- /dev/null +++ b/608/CH42/EX42.01/42_01.sce @@ -0,0 +1,21 @@ +//Problem 42.01: Determine the cut-off frequency and the nominal impedance of each of the low-pass filter sections shown in Figure 42.19.
+
+//initializing the variables:
+L1 = 2*100E-3; // in Henry
+C1 = 0.2E-6; // in Fareads
+L2 = 0.4; // in Henry
+C2 = 2*200E-12; // in Fareads
+
+//calculation:
+//cut-off frequency
+fc1 = 1/(%pi*(L1*C1)^0.5)
+//nominal impedance
+R01 = (L1/C1)^0.5
+//cut-off frequency
+fc2 = 1/(%pi*(L2*C2)^0.5)
+//nominal impedance
+R02 = (L2/C2)^0.5
+
+printf("\n\n Result \n\n")
+printf("\n cut-off frequency %.0f Hz and the nominal impedance is %.0f ohm ",fc1, R01)
+printf("\n cut-off frequency %.0f Hz and the nominal impedance is %.0f ohm ",fc2, R02)
\ No newline at end of file diff --git a/608/CH42/EX42.02/42_02.sce b/608/CH42/EX42.02/42_02.sce new file mode 100755 index 000000000..17d7090d2 --- /dev/null +++ b/608/CH42/EX42.02/42_02.sce @@ -0,0 +1,15 @@ +//Problem 42.02: A filter section is to have a characteristic impedance at zero frequency of 600 ohm and a cut-off frequency at 5 MHz. Design (a) a low-pass T section filter, and (b) a low-pass pi section filter to meet these requirements.
+
+//initializing the variables:
+R0 = 600; // in ohm
+fc = 5E6; // in Hz
+
+//calculation:
+//capacitance
+C = 1/(%pi*R0*fc)
+//inductance
+L = R0/(%pi*fc)
+
+printf("\n\n Result \n\n")
+printf("\n A low-pass T section filter capcitance is %.2E farad and inductance is%.2E Henry",C, L/2)
+printf("\n A low-pass pi section filter capcitance is %.2E farad and inductance is%.2E Henry",C/2, L)
\ No newline at end of file diff --git a/608/CH42/EX42.03/42_03.sce b/608/CH42/EX42.03/42_03.sce new file mode 100755 index 000000000..a1cc9809d --- /dev/null +++ b/608/CH42/EX42.03/42_03.sce @@ -0,0 +1,16 @@ +//Problem 42.03: The nominal impedance of a low-pass pi section filter is 500 ohm and its cut-off frequency is at 100 kHz. Determine (a) the value of the characteristic impedance of the section at a frequency of 90 kHz, and (b) the value of the characteristic impedance of the equivalent low-pass T section filter.
+
+//initializing the variables:
+R0 = 500; // in ohm
+fc = 100000; // in Hz
+f = 90000; // in Hz
+
+//calculation:
+//characteristic impedance of the pi section
+Zpi = R0/(1 - (f/fc)^2)^0.5
+//characteristic impedance of the T section
+Zt = R0*(1 - (f/fc)^2)^0.5
+
+printf("\n\n Result \n\n")
+printf("\ncharacteristic impedance of the pi section is %.0f ohm",Zpi)
+printf("\ncharacteristic impedance of the T section is %.0f ohm",Zt)
\ No newline at end of file diff --git a/608/CH42/EX42.04/42_04.sce b/608/CH42/EX42.04/42_04.sce new file mode 100755 index 000000000..0e0b87b3a --- /dev/null +++ b/608/CH42/EX42.04/42_04.sce @@ -0,0 +1,17 @@ +//Problem 42.04: A low-pass section filter has a nominal impedanc of 600 ohm and a cut-off frequency of 2 MHz. Determine the frequency at which the characteristic impedance of the section is (a) 600 ohm (b) 1 kohm (c) 10kohm.
+
+//initializing the variables:
+R0 = 600; // in ohm
+fc = 2E6; // in Hz
+Z1 = 600; // in ohm
+Z2 = 1000; // in ohm
+Z3 = 10000; // in ohm
+
+//calculation:
+//frequency
+f1 = fc*(1 - (R0/Z1)^2)^0.5
+f2 = fc*(1 - (R0/Z2)^2)^0.5
+f3 = fc*(1 - (R0/Z3)^2)^0.5
+
+printf("\n\n Result \n\n")
+printf("\nfrequency at which the characteristic impedance of the section is 600 ohm is %.0f Hz and 1000 Ohm is %.2E Hz and 10000 ohm is %.3E Hz ",f1,f2,f3)
\ No newline at end of file diff --git a/608/CH42/EX42.05/42_05.sce b/608/CH42/EX42.05/42_05.sce new file mode 100755 index 000000000..93cb1c17f --- /dev/null +++ b/608/CH42/EX42.05/42_05.sce @@ -0,0 +1,21 @@ +//Problem 42.05: Determine for each of the high-pass filter sections shown in Figure 42.27 (i) the cut-off frequency, and (ii) the nominal impedance.
+
+//initializing the variables:
+L1 = 100E-3; // in Henry
+C1 = 0.2E-6/2; // in Fareads
+L2 = 200E-6/2; // in Henry
+C2 = 4000E-12; // in Fareads
+
+//calculation:
+//cut-off frequency
+fc1 = 1/(4*%pi*(L1*C1)^0.5)
+//nominal impedance
+R01 = (L1/C1)^0.5
+//cut-off frequency
+fc2 = 1/(4*%pi*(L2*C2)^0.5)
+//nominal impedance
+R02 = (L2/C2)^0.5
+
+printf("\n\n Result \n\n")
+printf("\n cut-off frequency %.0f Hz and the nominal impedance is %.0f ohm",fc1, R01)
+printf("\n cut-off frequency %.0f Hz and the nominal impedance is %.0f ohm ",fc2, R02)
\ No newline at end of file diff --git a/608/CH42/EX42.07/42_07.sce b/608/CH42/EX42.07/42_07.sce new file mode 100755 index 000000000..d302abc97 --- /dev/null +++ b/608/CH42/EX42.07/42_07.sce @@ -0,0 +1,20 @@ +//Problem 42.07: A low-pass T section filter having a cut-off frequency of 15 kHz is connected in series with a high-pass T section filter having a cut-off frequency of 10 kHz. The terminating impedance of the filter is 600 ohm .(a) Determine the values of the components comprising the composite filter.
+
+//initializing the variables:
+R0 = 600; // in ohm
+fc1 = 15000; // in Hz
+fc2 = 10000; // in Hz
+
+//calculation:
+//capacitance
+C1 = 1/(%pi*R0*fc1)
+//inductance
+L1 = R0/(%pi*fc1)
+//capacitance
+C2 = 1/(4*%pi*R0*fc2)
+//inductance
+L2 = R0/(4*%pi*fc2)
+
+printf("\n\n Result \n\n")
+printf("\n A low-pass T section filter capcitance is %.2E farad and inductance is%.2E Henry",C1, L1/2)
+printf("\n A high-pass pi section filter capcitance is %.2E farad and inductance is%.2E Henry",2*C2, L2)
\ No newline at end of file diff --git a/608/CH42/EX42.08/42_08.sce b/608/CH42/EX42.08/42_08.sce new file mode 100755 index 000000000..1ab25fd1a --- /dev/null +++ b/608/CH42/EX42.08/42_08.sce @@ -0,0 +1,17 @@ +//Problem 42.08: A high-pass T section filter has a cut-off frequency of 500 Hz and a nominal impedance of 600 ohm . Determine the frequency at which the characteristic impedance of the section is (a) zero, (b) 300 ohm , (c) 590 ohm .
+
+//initializing the variables:
+R0 = 600; // in ohm
+fc = 500; // in Hz
+Z1 = 0; // in ohm
+Z2 = 300; // in ohm
+Z3 = 590; // in ohm
+
+//calculation:
+//frequency
+f1 = fc
+f2 = fc/(1 - (Z2/R0)^2)^0.5
+f3 = fc/(1 - (Z3/R0)^2)^0.5
+
+printf("\n\n Result \n\n")
+printf("\nfrequency at which the characteristic impedance of the section is 0 ohm is %.0f Hz and 300 Ohm is %.1f Hz and 590 ohm is %.0f Hz ",f1,f2,f3)
\ No newline at end of file diff --git a/608/CH42/EX42.09/42_09.sce b/608/CH42/EX42.09/42_09.sce new file mode 100755 index 000000000..d1dfdf9f4 --- /dev/null +++ b/608/CH42/EX42.09/42_09.sce @@ -0,0 +1,20 @@ +//Problem 42.09: The propagation coefficients of two filter networks are given by (a) r = (1.25 + i0.52) (b) r = 1.794/_-39.4° Determine for each (i) the attenuation coefficient, and (ii) the phase shift coefficient.
+
+//initializing the variables:
+r1 = 1.25 + %i*0.52; // propagation coefficients
+rr = 1.794; // propagation coefficients
+thetar = -39.4; // in ddegrees
+
+//calculation:
+//r
+r2 = rr*cos(thetar*%pi/180) + %i*rr*sin(thetar*%pi/180)
+//attenuation coefficient
+a1 = real(r1)
+a2 = real(r2)
+//phase shift coefficient
+b1 = imag(r1)
+b2 = imag(r2)
+
+printf("\n\n Result \n\n")
+printf("\nattenuation coefficient are for (a) is %.2f N and for (b) is %.3f N ",a1,a2)
+printf("\nphase shift coefficient are for (a) is %.2f rad and for (b) is %.3f rad ",b1,b2)
\ No newline at end of file diff --git a/608/CH42/EX42.10/42_10.sce b/608/CH42/EX42.10/42_10.sce new file mode 100755 index 000000000..ad5e4a60a --- /dev/null +++ b/608/CH42/EX42.10/42_10.sce @@ -0,0 +1,36 @@ +//Problem 42.10: The current input to a filter section is 24/_10° mA and the current output is 8/_-45° mA. Determine for the section (a) the attenuation coefficient, (b) the phase shift coefficient, and (c) the propagation coefficient. (d) If five such sections are cascaded determine the output current of the fifth stage and the overall propagation constant of the network.
+
+//initializing the variables:
+ri1 = 0.024; // in amperes
+ri2 = 0.008; // in amperes
+thetai1 = 10; // in ddegrees
+thetai2 = -45; // in ddegrees
+
+//calculation:
+//currents
+I1 = ri1*cos(thetai1*%pi/180) + %i*ri1*sin(thetai1*%pi/180)
+I2 = ri2*cos(thetai2*%pi/180) + %i*ri2*sin(thetai2*%pi/180)
+//ir
+ir = I1/I2
+irmag = ri1/ri2
+thetai = thetai1-thetai2
+//attenuation coefficient
+a = log(irmag)
+//phase shift coefficient
+b = thetai*%pi/180
+//propagation coefficient
+r = a + %i*b
+//output current of the fifth stage
+I6 = I1/(ir^5)
+x = ir^5
+xmg = (real(x)^2 + imag(x)^2)^0.5
+//overall attenuation coefficient
+ad = log(xmg)
+//overall phase shift coefficient
+bd = atan(imag(x)/real(x)) + 2*%pi
+
+printf("\n\n Result \n\n")
+printf("\nattenuation coefficient is %.3f N ",a)
+printf("\nphase shift coefficient is %.3f rad ",b)
+printf("\npropagation coefficient is %.3f + (%.3f)i ",a,b)
+printf("\nthe output current of the fifth stage is %.2E + (%.2E)i A and the overall propagation coefficient is %.2f + (%.2f)i",real(I6),imag(I6),ad,bd)
\ No newline at end of file diff --git a/608/CH42/EX42.11/42_11.sce b/608/CH42/EX42.11/42_11.sce new file mode 100755 index 000000000..b932cd799 --- /dev/null +++ b/608/CH42/EX42.11/42_11.sce @@ -0,0 +1,25 @@ +//Problem 42.11: For the low-pass T section filter shown in Figure 42.34 determine (a) the attenuation coefficient, (b) the phase shift coefficient and (c) the propagation coefficient r .
+
+//initializing the variables:
+XL = %i*5; // in ohms
+Xc = -1*%i*10; // in ohms
+RL = 12; // in ohms
+I1 = 1; // in amperes (lets say)
+
+//calculation:
+//current I2
+I2 = (Xc/(Xc + XL + RL))*I1
+//current ratio
+Ir = I1/I2
+Irmg = (real(Ir)^2 + imag(Ir)^2)^0.5
+//attenuation coefficient
+a = log(Irmg)
+//phase shift coefficient
+b = atan(imag(Ir)/real(Ir))
+//propagation coefficient
+r = a + %i*b
+
+printf("\n\n Result \n\n")
+printf("\nattenuation coefficient is %.3f N ",a)
+printf("\nphase shift coefficient is %.3f rad ",b)
+printf("\npropagation coefficient is %.3f + (%.3f)i ",a,b)
\ No newline at end of file diff --git a/608/CH42/EX42.12/42_12.sce b/608/CH42/EX42.12/42_12.sce new file mode 100755 index 000000000..e56e00b86 --- /dev/null +++ b/608/CH42/EX42.12/42_12.sce @@ -0,0 +1,15 @@ +//Problem 42.12: Determine for the filter section shown in Figure 42.40, (a) the time delay for the signal to pass through the filter, assuming the phase shift is small, and (b) the time delay for a signal to pass through the section at the cut-off frequency.
+
+//initializing the variables:
+L = 2*0.5; // in Henry
+C = 2E-9; // in Farad
+
+//calculation:
+//time delay
+t = (L*C)^0.5
+//time delay at the cut-off frequency
+tfc = t*%pi/2
+
+printf("\n\n Result \n\n")
+printf("\n time delay is %.2E sec ",t)
+printf("\ntime delay at the cut-off frequency is %.2E sec",tfc)
\ No newline at end of file diff --git a/608/CH42/EX42.13/42_13.sce b/608/CH42/EX42.13/42_13.sce new file mode 100755 index 000000000..f4bbfa708 --- /dev/null +++ b/608/CH42/EX42.13/42_13.sce @@ -0,0 +1,21 @@ +//Problem 42.13: A filter network comprising n identical sections passes signals of all frequencies up to 500 kHz and provides a total delay of 9.55 μs. If the nominal impedance of the circuit into which the filter is inserted is 1 kohm , determine (a) the values of the elements in each section, and (b) the value of n.
+
+//initializing the variables:
+fc = 500000; // in Hz
+t1 = 9.55E-6; // in secs
+R0 = 1000; // in ohm
+
+//calculation:
+//for a low-pass filter section, capacitance
+C = 1/(%pi*R0*fc)
+//inductance
+L = R0/(%pi*fc)
+//time delay
+t2 = (L*C)^0.5
+//number of cascaded sections required
+n = t1/t2
+
+printf("\n\n Result \n\n")
+printf("\n for low-pass T section inductance is %.2E H and capacitance is %.2E F",L/2,C)
+printf("\n for low-pass pi section inductance is %.2E H and capacitance is %.2E F",L,C/2)
+printf("\nnumber of cascaded sections required is %.0f",n)
\ No newline at end of file diff --git a/608/CH42/EX42.14/42_14.sce b/608/CH42/EX42.14/42_14.sce new file mode 100755 index 000000000..adeaaca84 --- /dev/null +++ b/608/CH42/EX42.14/42_14.sce @@ -0,0 +1,19 @@ +//Problem 42.14: A filter network consists of 8 sections in cascade having a nominal impedance of 1 kohm . If the total delay time is 4 μs, determine the component values for each section if the filter is (a) a low-pass T network, and (b) a high-pass pi network.
+
+//initializing the variables:
+n = 8; // sections in cascade
+R0 = 1000; // in ohm
+t1 = 4E-6; // in secs
+
+
+//calculation:
+//time delay
+t2 = t1/n
+//capacitance
+C = t2/R0
+//inductance
+L = t2*R0
+
+printf("\n\n Result \n\n")
+printf("\n for low-pass T section inductance is %.2E H and capacitance is %.2E F",L/2,C)
+printf("\n for high-pass pi section inductance is %.2E H and capacitance is %.2E F",2*L,C)
\ No newline at end of file diff --git a/608/CH43/EX43.01/43_01.sce b/608/CH43/EX43.01/43_01.sce new file mode 100755 index 000000000..f7d000731 --- /dev/null +++ b/608/CH43/EX43.01/43_01.sce @@ -0,0 +1,19 @@ +//Problem 43.01: A and B are two coils in close proximity. A has 1200 turns and B has 1000 turns. When a current of 0.8 A flows in coil A a flux of 100 μWb links with coil A and 75% of this flux links coil B. Determine (a) the self inductance of coil A, and (b) the mutual inductance.
+
+//initializing the variables:
+Na = 1200;
+Nb = 1000;
+Ia = 0.8; // in amperes
+Phia = 100E-6; // in Wb
+xb = 0.75;
+
+//calculation:
+//self inductance of coil A
+La = Na*Phia/Ia
+//mutual inductance, M
+Phib = xb*Phia
+M = Nb*Phib/Ia
+
+printf("\n\n Result \n\n")
+printf("\n self inductance of coil A is %.2f H",La)
+printf("\n mutual inductance, M is %.2E H",M)
\ No newline at end of file diff --git a/608/CH43/EX43.02/43_02.sce b/608/CH43/EX43.02/43_02.sce new file mode 100755 index 000000000..9c547f6f7 --- /dev/null +++ b/608/CH43/EX43.02/43_02.sce @@ -0,0 +1,16 @@ +//Problem 43.02: Two circuits have a mutual inductance of 600 mH. A current of 5 A in the primary is reversed in 200 ms. Determine the e.m.f. induced in the secondary, assuming the current changes at a uniform rate.
+
+//initializing the variables:
+M = 600E-3; // in Henry
+Ia = 5; // in amperes
+dt = 0.2; // in secs
+
+//calculation:
+//change of current
+dIa = 2*Ia
+dIadt = dIa/dt
+//secondary induced e.m.f., E2
+E2 = -1*M*dIadt
+
+printf("\n\n Result \n\n")
+printf("\n secondary induced e.m.f., E2 is %.0f V",E2)
\ No newline at end of file diff --git a/608/CH43/EX43.03/43_03.sce b/608/CH43/EX43.03/43_03.sce new file mode 100755 index 000000000..027857493 --- /dev/null +++ b/608/CH43/EX43.03/43_03.sce @@ -0,0 +1,13 @@ +//Problem 43.03: Two coils have self inductances of 250 mH and 400 mH respectively. Determine the magnetic coupling coefficient of the pair of coils if their mutual inductance is 80 mH.
+
+//initializing the variables:
+La = 250E-3; // in Henry
+Lb = 400E-3; // in Henry
+M = 80E-3; // in Henry
+
+//calculation:
+//coupling coefficient,
+k = M/(La*Lb)^0.5
+
+printf("\n\n Result \n\n")
+printf("\n coupling coefficient, is %.3f",k)
\ No newline at end of file diff --git a/608/CH43/EX43.04/43_04.sce b/608/CH43/EX43.04/43_04.sce new file mode 100755 index 000000000..790b8225f --- /dev/null +++ b/608/CH43/EX43.04/43_04.sce @@ -0,0 +1,19 @@ +//Problem 43.04: Two coils, X and Y, having self inductances of 80 mH and 60 mH respectively, are magnetically coupled. Coil X has 200 turns and coil Y has 100 turns. When a current of 4A flows in coil X the change of flux in coil Y is 5 mWb. Determine (a) the mutual inductance between the coils, and (b) the coefficient of coupling.
+
+//initializing the variables:
+Lx = 80E-3; // in Henry
+Ly = 60E-3; // in Henry
+Nx = 200; // turns
+Ny = 100; // turns
+Ix = 4; // in Amperes
+Phiy = 0.005; // in Wb
+
+//calculation:
+//mutual inductance, M
+M = Ny*Phiy/(2*Ix)
+//coupling coefficient,
+k = M/(Lx*Ly)^0.5
+
+printf("\n\n Result \n\n")
+printf("\n mutual inductance, M is %.2E H",M)
+printf("\n coupling coefficient, is %.3f",k)
\ No newline at end of file diff --git a/608/CH43/EX43.05/43_05.sce b/608/CH43/EX43.05/43_05.sce new file mode 100755 index 000000000..f02e3b25e --- /dev/null +++ b/608/CH43/EX43.05/43_05.sce @@ -0,0 +1,16 @@ +//Problem 43.05: Two coils connected in series have self inductance of 40 mH and 10 mH respectively. The total inductance of the circuit is found to be 60 mH. Determine (a) the mutual inductance between the two coils, and (b) the coefficient of coupling.
+
+//initializing the variables:
+La = 40E-3; // in Henry
+Lb = 10E-3; // in Henry
+L = 60E-3; // in Henry
+
+//calculation:
+//mutual inductance, M
+M = (L - La - Lb)/2
+//coupling coefficient,
+k = M/(La*Lb)^0.5
+
+printf("\n\n Result \n\n")
+printf("\n mutual inductance, M is %.2E H",M)
+printf("\n coupling coefficient, is %.3f",k)
\ No newline at end of file diff --git a/608/CH43/EX43.06/43_06.sce b/608/CH43/EX43.06/43_06.sce new file mode 100755 index 000000000..4ea859133 --- /dev/null +++ b/608/CH43/EX43.06/43_06.sce @@ -0,0 +1,22 @@ +//Problem 43.06: Two mutually coupled coils X and Y are connected in series to a 240 V d.c. supply. Coil X has a resistance of 5 ohm and an inductance of 1 H. Coil Y has a resistance of 10 ohm and an inductance of 5 H. At a certain instant after the circuit is connected, the current is 8 A and increasing at a rate of 15 A/s. Determine (a) the mutual inductance between the coils and (b) the coefficient of coupling.
+
+//initializing the variables:
+V = 240; // in Volts
+Ra = 5; // in Ohm
+La = 1; // in Henry
+Rb = 10; // in Ohm
+Lb = 5; // in Henry
+I = 8; // in amperes
+dIdt = 15; // in A/sec
+
+//calculation:
+//Kirchhoff’s voltage law
+L = (V - I*(Ra + Rb))/dIdt
+//mutual inductance, M
+M = (L - La - Lb)/2
+//coupling coefficient,
+k = M/(La*Lb)^0.5
+
+printf("\n\n Result \n\n")
+printf("\n mutual inductance, M is %.0f H",M)
+printf("\n coupling coefficient, is %.3f",k)
\ No newline at end of file diff --git a/608/CH43/EX43.07/43_07.sce b/608/CH43/EX43.07/43_07.sce new file mode 100755 index 000000000..9919eaa3c --- /dev/null +++ b/608/CH43/EX43.07/43_07.sce @@ -0,0 +1,22 @@ +//Problem 43.07: Two coils are connected in series and their effective inductance is found to be 15 mH. When the connection to one coil is reversed, the effective inductance is found to be 10 mH. If the coefficient of coupling is 0.7, determine (a) the self inductance of each coil, and (b) the mutual inductance.
+
+//initializing the variables:
+k = 0.7; // coefficient of coupling
+L1 = 15E-3; // in Henry
+L2 = 10E-3; // in Henry
+
+//calculation:
+//L1 = La + Lb + 2*k*(La*Lb)^0.5
+//L2 = La + Lb - 2*k*(La*Lb)^0.5
+//self inductance of coils
+a = ((L1 - (L1 + L2)/2)/(2*k))^2
+La1 =((L1 + L2)/2 + (((L1 + L2)/2)^2 - 4*a)^0.5)/2
+La2 =((L1 + L2)/2 - (((L1 + L2)/2)^2 - 4*a)^0.5)/2
+Lb1 = (L1 + L2)/2 - La1
+Lb2 = (L1 + L2)/2 - La2
+//mutual inductance, M
+M = (L1 - L2)/4
+
+printf("\n\n Result \n\n")
+printf("\nself inductance of coils are %.2E H and %.2E H",La1, Lb1)
+printf("\n mutual inductance, M is %.2E H",M)
diff --git a/608/CH43/EX43.08/43_08.sce b/608/CH43/EX43.08/43_08.sce new file mode 100755 index 000000000..7372dc26e --- /dev/null +++ b/608/CH43/EX43.08/43_08.sce @@ -0,0 +1,22 @@ +//Problem 43.08: For the circuit shown in Figure 43.7, determine the p.d. E2 which appears across the open-circuited secondary winding, given that E1 D 8 sin 2500t volts.
+
+//initializing the variables:
+E1 = 8; // in Volts
+thetae1 = 0; // in degrees
+w = 2500; // in rad/sec
+R = 15; // in ohm
+L = 5E-3; // in Henry
+M = 0.1E-3; // in Henry
+
+//calculation:
+//voltage
+E1 = E1*cos(thetae1*%pi/180) + %i*E1*sin(thetae1*%pi/180)
+//Impedance of primary
+Z1 = R + %i*w*L
+//Primary current I1
+I1 = E1/Z1
+//E2
+E2 = %i*w*M*I1
+
+printf("\n\n Result \n\n")
+printf("\nE2 is %.2f + (%.2f)i V",real(E2), imag(E2))
\ No newline at end of file diff --git a/608/CH43/EX43.09/43_09.sce b/608/CH43/EX43.09/43_09.sce new file mode 100755 index 000000000..7be58297a --- /dev/null +++ b/608/CH43/EX43.09/43_09.sce @@ -0,0 +1,17 @@ +//Problem 43.09: Two coils x and y, with negligible resistance, have self inductances of 20 mH and 80 mH respectively, and the coefficient of coupling between them is 0.75. If a sinusoidal alternating p.d. of 5 V is applied to x, determine the magnitude of the open circuit e.m.f. induced in y.
+
+//initializing the variables:
+Lx = 20E-3; // in Henry
+Ly = 80E-3; // in Henry
+k = 0.75; // coupling coeff.
+Ex = 5; // in Volts
+
+//calculation:
+//mutual inductance
+M = k*(Lx*Ly)^0.5
+//magnitude of the open circuit e.m.f. induced
+Ey = M*Ex/Lx
+
+printf("\n\n Result \n\n")
+printf("\n mutual inductance is %.2f H",M)
+printf("\n magnitude of the open circuit e.m.f. induced is %.2f V",Ey)
\ No newline at end of file diff --git a/608/CH43/EX43.10/43_10.sce b/608/CH43/EX43.10/43_10.sce new file mode 100755 index 000000000..dcc0bfa6b --- /dev/null +++ b/608/CH43/EX43.10/43_10.sce @@ -0,0 +1,32 @@ +//Problem 43.10: For the circuit shown in Figure 43.9, determine the value of the secondary current I2 if E1 = 2/_0° volts and the frequency is 1000/pi Hz.
+
+//initializing the variables:
+E1 = 2; // in Volts
+thetae1 = 0; // in degrees
+f = 1000/%pi; // in Hz
+R1 = 4; // in ohm
+R2 = 16; // in ohm
+R3 = 16; // in ohm
+R4 = 50; // in ohm
+L = 10E-3; // in Henry
+M = 2E-3; // in Henry
+
+//calculation:
+w = 2*%pi*f
+//voltage
+E1 = E1*cos(thetae1*%pi/180) + %i*E1*sin(thetae1*%pi/180)
+//R1e is the real part of Z1e
+R1e = R1 + R2 + ((R3 + R4)*(M^2)*(w^2))/((R3 + R4)^2 + (w*L)^2)
+//X1e is the imaginary part of Z1e
+X1e = w*L - (L*(M^2)*(w^3))/((R3 + R4)^2 + (w*L)^2)
+Z1e = R1e + %i*X1e
+Z2e = R3 + R4 + %i*w*L
+//primary current, I1
+I1 = E1/Z1e
+//E2
+E2 = %i*w*M*I1
+//secondary current I2
+I2 = E2/Z2e
+
+printf("\n\n Result \n\n")
+printf("\n secondary current I2 is %.2E +(%.2E)i A",real(I2), imag(I2))
\ No newline at end of file diff --git a/608/CH43/EX43.11/43_11.sce b/608/CH43/EX43.11/43_11.sce new file mode 100755 index 000000000..a9ce29fdc --- /dev/null +++ b/608/CH43/EX43.11/43_11.sce @@ -0,0 +1,38 @@ +//Problem 43.11: For the coupled circuit shown in Figure 43.10, calculate (a) the self impedance of the primary circuit, (b) the self impedance of the secondary circuit, (c) the impedance reflected into the primary circuit, (d) the effective primary impedance, (e) the primary current, and (f) the secondary current + +//initializing the variables: +E1 = 50; // in Volts +thetae1 = 0; // in degrees +w = 500; // in rad/sec +R1 = 300; // in ohm +L1 = 0.2; // in Henry +L2 = 0.5; // in Henry +L3 = 0.3; // in Henry +R2 = 500; // in ohm +C = 5E-6; // in farad +M = 0.2; // in Henry + +//calculation: +//voltage +E1 = E1*cos(thetae1*%pi/180) + %i*E1*sin(thetae1*%pi/180) +// Self impedance of primary circuit +Z1 = R1 + %i*w*(L1 + L2) +//Self impedance of secondary circuit, +Z2 = R2 + %i*(w*L3 - 1/(w*C)) +//reflected impedance, Zr +Zr = (w*M)^2/Z2 +//Effective primary impedance, +Z1e = Z1 + Zr +//Primary current I1 +I1 = E1/Z1e +//Secondary current I2 +E2 = %i*w*M*I1 +I2 = E2/Z2 + +printf("\n\n Result \n\n") +printf("\n Self impedance of primary circuit, Z1 is %.0f + (%.0f)i ohm",real(Z1), imag(Z1)) +printf("\n Self impedance of secondary circuit, Z2 is %.0f + (%.0f)i ohm",real(Z2), imag(Z2)) +printf("\n reflected impedance, Zr is %.0f +(%.0f)i ohm",real(Zr), imag(Zr)) +printf("\n Effective primary impedance Z1(eff) is %.0f +(%.0f)i ohm",real(Z1e), imag(Z1e)) +printf("\n primary current I1 is %.2f +(%.2f)i A",real(I1), imag(I1)) +printf("\n secondary current I2 is %.2f +(%.2f)i A",real(I2), imag(I2)) diff --git a/608/CH43/EX43.12/43_12.sce b/608/CH43/EX43.12/43_12.sce new file mode 100755 index 000000000..b51a00705 --- /dev/null +++ b/608/CH43/EX43.12/43_12.sce @@ -0,0 +1,41 @@ +//Problem 43.12:For the circuit shown in Figure 43.12 each winding is tuned to resonate at the same frequency. Determine (a) the reso-nant frequency, (b) the value of capacitor C2 , (c) the effective primary impedance, (d) the primary current, (e) the voltage across capacitor C2 and (f) the coefficient of coupling. + +//initializing the variables: +E1 = 20; // in Volts +thetae1 = 0; // in degrees +R1 = 15; // in ohm +C1 = 400E-12; // in farad +R2 = 30; // in ohm +L1 = 0.001; // in Henry +L2 = 0.0002; // in Henry +R3 = 50; // in ohm +M = 10E-6; // in Henry + +//calculation: +//voltage +E1 = E1*cos(thetae1*%pi/180) + %i*E1*sin(thetae1*%pi/180) +//the resonant frequency, fr +fr = 1/(2*%pi*(L1*C1)^0.5) +w = 2*%pi*fr +//The secondary is also tuned to a resonant frequency +//capacitance,C2 +C2 = 1/(L2*(2*%pi*fr)^2) +//the effective primary impedance Z1eff +Z1e = R1 + R2 + ((w*M)^2)/R3 +//Primary current I1 +I1 = E1/Z1e +//Secondary current I2 +E2 = %i*w*M*I1 +I2 = E2/R3 +//voltage across capacitor C2 +Vc2 = I2*-1*%i/(w*C2) +//coefficient of coupling, k +k = M/(L1*L2)^0.5 + +printf("\n\n Result \n\n") +printf("\n the resonant frequency,fr is %.0f Hz",fr) +printf("\n capacitance,C2 is %.2E F",C2) +printf("\n Effective primary impedance Z1(eff) is %.0f +(%.0f)i ohm",real(Z1e), imag(Z1e)) +printf("\n primary current I1 is %.2f +(%.2f)i A",real(I1), imag(I1)) +printf("\n voltage across capacitor C2 is %.0f +(%.0f)i V",real(Vc2), imag(Vc2)) +printf("\n coefficient of coupling, k is %.4f",k) diff --git a/608/CH43/EX43.13/43_13.sce b/608/CH43/EX43.13/43_13.sce new file mode 100755 index 000000000..3da123109 --- /dev/null +++ b/608/CH43/EX43.13/43_13.sce @@ -0,0 +1,26 @@ +//Problem 43.13:For the coupled circuit shown in Figure 43.16, determiine the values of currents I1 and I2. + +//initializing the variables: +E1 = 250; // in Volts +thetae1 = 0; // in degrees +R1 = %i*50; // in ohm +R2 = 10; // in ohm +R3 = 10; // in ohm +R4 = %i*50; // in ohm +R5 = 50; // in ohm +M = %i*10; // in ohm + +//calculation: +//voltage +E1 = E1*cos(thetae1*%pi/180) + %i*E1*sin(thetae1*%pi/180) +//Applying Kirchhoff’s voltage law to the primary circuit gives +//(R1 + R2)*I1 - M*I2 = E1 +//Applying Kirchhoff’s voltage law to the secondary circuit gives +//-1*M*I1 + ( R3 + R4 + R5)*I2 = 0 +//solving these two +I2 = E1/((R1 + R2)*(R3 + R4 + R5)/(M) + (-1*M)) +I1 = I2*(R3 + R4 + R5)/(M) + +printf("\n\n Result \n\n") +printf("\n primary current I1 is %.2f +(%.2f)i A",real(I1), imag(I1)) +printf("\n secondary current I2 is %.2f +(%.2f)i A",real(I2), imag(I2)) diff --git a/608/CH43/EX43.14/43_14.sce b/608/CH43/EX43.14/43_14.sce new file mode 100755 index 000000000..7c4e0ff70 --- /dev/null +++ b/608/CH43/EX43.14/43_14.sce @@ -0,0 +1,35 @@ +//Problem 43.14:The circuit diagram of an air-cored transformer winding is shown in Figure 43.17. The coefficient of coupling between primary and secondary windings is 0.70. Determine for the circuit (a) the mutual inductance M, (b) the primary current I1 and (c) the secondary terminal p.d.
+
+//initializing the variables:
+re = 40; // in Volts
+thetae1 = 0; // in degrees
+R1 = 5; // in ohm
+L1 = 0.001; // in Henry
+L2 = 0.006; // in Henry
+R2 = 40; // in ohm
+rzl = 200; // in ohm
+thetazl = -60; // in degrees
+k = 0.70
+f = 20000; // in Hz
+
+//calculation:
+w = 2*%pi*f
+//voltage
+E1 = re*cos(thetae1*%pi/180) + %i*re*sin(thetae1*%pi/180)
+//impedance
+ZL = rzl*cos(thetazl*%pi/180) + %i*rzl*sin(thetazl*%pi/180)
+//mutual inductance, M
+M = k*(L1*L2)^0.5
+//Applying Kirchhoff’s voltage law to the primary circuit gives
+//(R1 + %i*w*L1)*I1 - %i*w*M*I2 = E1
+//Applying Kirchhoff’s voltage law to the secondary circuit gives
+//-1*%i*w*M*I1 + ( R2 + ZL + %i*w*L2)*I2 = 0
+//solving these two
+I1 = E1/((R1 +%i*w*L1) - (%i*w*M)^2/(R2 + ZL + %i*w*L2))
+//secondary terminal p.d.
+pd = I2*ZL
+
+printf("\n\n Result \n\n")
+printf("\n mutual induction M is %.2E H",M)
+printf("\n primary current I1 is %.2f +(%.2f)i A",real(I1), imag(I1))
+printf("\n secondary terminal p.d. is %.2f +(%.2f)i V",real(pd), imag(pd))
diff --git a/608/CH43/EX43.15/43_15.sce b/608/CH43/EX43.15/43_15.sce new file mode 100755 index 000000000..0e17c8943 --- /dev/null +++ b/608/CH43/EX43.15/43_15.sce @@ -0,0 +1,29 @@ +//Problem 43.15:A mutual inductor is used to couple a 20 ohm resistive load to a 50/_0° V generator as shown in Figure 43.18. The generator has an internal resistance of 5 ohm and the mutual inductor parameters are R1 = 20 ohm , L1 = 0.2 H, R2 = 25 ohm , L2 = 0.4 H and M = 0.1 H. The supply frequency is 75/pi Hz. Determine (a) the generator current I1 and (b) the load current I2 . + +//initializing the variables: +E1 = 50; // in Volts +thetae1 = 0; // in degrees +r = 5; // in ohm +R1 = 20; // in ohm +L1 = 0.2; // in Henry +L2 = 0.4; // in Henry +R2 = 25; // in ohm +RL = 20; // in ohm +M = 0.1; // in Henry +f = 75/%pi; // in Hz + +//calculation: +w = 2*%pi*f +//voltage +E1 = E1*cos(thetae1*%pi/180) + %i*E1*sin(thetae1*%pi/180) +//Applying Kirchhoff’s voltage law to the primary circuit gives +//(r + R1 + %i*w*L1)*I1 - %i*w*M*I2 = E1 +//Applying Kirchhoff’s voltage law to the secondary circuit gives +//-1*%i*w*M*I1 + ( R2 + RL + %i*w*L2)*I2 = 0 +//solving these two +I2 = E1/((r + R1 + %i*w*L1)*(R2 + RL + %i*w*L2)/(%i*w*M) + (-1*%i*w*M)) +I1 = I2*(R2 + RL + %i*w*L2)/(%i*w*M) + +printf("\n\n Result \n\n") +printf("\n primary current I1 is %.2f +(%.2f)i A",real(I1), imag(I1)) +printf("\n load current I2 is %.2f +(%.2f)i A",real(I2), imag(I2)) diff --git a/608/CH43/EX43.16/43_16.sce b/608/CH43/EX43.16/43_16.sce new file mode 100755 index 000000000..b15ee2112 --- /dev/null +++ b/608/CH43/EX43.16/43_16.sce @@ -0,0 +1,41 @@ +//Problem 43.16:The mutual inductor of problem 43.15 is connected to the circuit of Figure 43.19. Determine the source and load currents for (a) the windings as shown (i.e. with the dots adjacent), and (b) with one winding reversed (i.e. with the dots at opposite ends). + +//initializing the variables: +E1 = 50; // in Volts +thetae1 = 0; // in degrees +r = 5; // in ohm +R1 = 20; // in ohm +L1 = 0.2; // in Henry +R = 8; // in ohm +L = 0.1; // in Henry +L2 = 0.4; // in Henry +R2 = 25; // in ohm +RL = 20; // in ohm +M = 0.1; // in Henry +f = 75/%pi; // in Hz + +//calculation: +w = 2*%pi*f +//voltage +E1 = E1*cos(thetae1*%pi/180) + %i*E1*sin(thetae1*%pi/180) +//Applying Kirchhoff’s voltage law to the primary circuit gives +//(r + R1 + %i*w*L1 + R + %i*w*L)*I1 - (%i*w*M + R + %i*w*L)*I2 = E1 +//Applying Kirchhoff’s voltage law to the secondary circuit gives +//-1*(%i*w*M + R + %i*w*L)*I1 + (R2 + RL + %i*w*L2 + R + %i*w*L)*I2 = 0 +//solving these two +I2 = E1/((r + R1 + %i*w*L1 + R + %i*w*L)*(R2 + RL + %i*w*L2 + R + %i*w*L)/((%i*w*M + R + %i*w*L)) + (-1*(%i*w*M + R + %i*w*L))) +I1 = I2*(R2 + RL + %i*w*L2 + R + %i*w*L)/(%i*w*M + R + %i*w*L) +//reversing +//Applying Kirchhoff’s voltage law to the primary circuit gives +//(r + R1 + %i*w*L1 + R + %i*w*L)*I1r - (-1*%i*w*M + R + %i*w*L)*I2r = E1 +//Applying Kirchhoff’s voltage law to the secondary circuit gives +//-1*(-1*%i*w*M + R + %i*w*L)*I1r + (R2 + RL + %i*w*L2 + R + %i*w*L)*I2r = 0 +//solving these two +I2r = E1/((r + R1 + %i*w*L1 + R + %i*w*L)*(R2 + RL + %i*w*L2 + R + %i*w*L)/((-1*%i*w*M + R + %i*w*L)) + (-1*(-1*%i*w*M + R + %i*w*L))) +I1r = I2r*(R2 + RL + %i*w*L2 + R + %i*w*L)/((-1*%i*w*M + R + %i*w*L)) + +printf("\n\n Result \n\n") +printf("\n primary current I1 is %.2f +(%.2f)i A",real(I1), imag(I1)) +printf("\n load current I2 is %.2f +(%.2f)i A",real(I2), imag(I2)) +printf("\n reversed primary current I1r is %.2f +(%.2f)i A",real(I1r), imag(I1r)) +printf("\n reversed load current I2r is %.2f +(%.2f)i A",real(I2r), imag(I2r)) diff --git a/608/CH44/EX44.01/44_01.sce b/608/CH44/EX44.01/44_01.sce new file mode 100755 index 000000000..6f9669c26 --- /dev/null +++ b/608/CH44/EX44.01/44_01.sce @@ -0,0 +1,16 @@ +//Problem 44.01:A parallel-wire air-spaced transmission line operating at 1910 Hz has a phase shift of 0.05 rad/km. Determine (a) the wavelength on the line, and (b) the speed of transmission of a signal. + +//initializing the variables: +f = 1910; // in Hz +b = 0.05; // in rad/km + +//calculation: +w = 2*%pi*f +//wavelength +Y = 2*%pi/b +//speed of transmission +u = f*Y + +printf("\n\n Result \n\n") +printf("\n wavelength Y is %.1f km",Y) +printf("\n speed of transmission %.2E km/sec",u) diff --git a/608/CH44/EX44.02/44_02.sce b/608/CH44/EX44.02/44_02.sce new file mode 100755 index 000000000..25f12181b --- /dev/null +++ b/608/CH44/EX44.02/44_02.sce @@ -0,0 +1,20 @@ +//Problem 44.02:A transmission line has an inductance of 4 mH/loop km and a capacitance of 0.004 μF/km. Determine, for a frequency of operation of 1 kHz, (a) the phase delay, (b) the wavelength on the line, and (c) the velocity of propagation (in metres per second) of the signal. + +//initializing the variables: +L = 0.004; // in Henry/loop +C = 0.004E-6; // in F/loop +f = 1000; // in Hz + +//calculation: +w = 2*%pi*f +//phase delay +b = w*(L*C)^0.5 +//wavelength +Y = 2*%pi/b +//speed of transmission +u = f*Y + +printf("\n\n Result \n\n") +printf("\n phase delay is %.3f rad/km",b) +printf("\n wavelength Y is %.1f km",Y) +printf("\n speed of transmission %.2E km/sec",u) diff --git a/608/CH44/EX44.03/44_03.sce b/608/CH44/EX44.03/44_03.sce new file mode 100755 index 000000000..50ed97cfb --- /dev/null +++ b/608/CH44/EX44.03/44_03.sce @@ -0,0 +1,17 @@ +//Problem 44.03: When operating at a frequency of 2 kHz, a cable has an attenuation of 0.25 Np/km and a phase shift of 0.20 rad/km. If a 5 V rms signal is applied at the sending end, determine the voltage at a point 10 km down the line, assuming that the termination is equal to the characteristic impedance of the line. + +//initializing the variables: +a = 0.25; // in Np/km +b = 0.20; // in rad/km +Vs = 5; // in Volts +n = 10; // in km +f = 2000; // in Hz + +//calculation: +w = 2*%pi*f +//the voltage 10 km down the line +r = a + %i*b +VR = Vs*%e^(-1*n*r) + +printf("\n\n Result \n\n") +printf("\n the voltage 10 km down the line is %.2f +(%.2f)i A",real(VR), imag(VR)) diff --git a/608/CH44/EX44.04/44_04.sce b/608/CH44/EX44.04/44_04.sce new file mode 100755 index 000000000..fccee4414 --- /dev/null +++ b/608/CH44/EX44.04/44_04.sce @@ -0,0 +1,24 @@ +//Problem 44.04: A transmission line 5 km long has a characteristic impedance of 800/_-25° ohm. At a particular frequency, the attenuation coefficient of the line is 0.5 Np/km and the phase shift coefficient is 0.25 rad/km. Determine the magnitude and phase of the current at the receiving end, if the sending end voltage is 2.0/_0° V r.m.s. + +//initializing the variables: +a = 0.5; // in Np/km +b = 0.25; // in rad/km +rvs = 2; // in Volts +thetavs = 0; // in degrees +rzo = 800; // in ohm +thetazo = -25; // in degrees +n = 5; // in km + +//calculation: +//voltage +Vs = rvs*cos(thetavs*%pi/180) + %i*rvs*sin(thetavs*%pi/180) +//characteristic impedance +Zo = rzo*cos(thetazo*%pi/180) + %i*rzo*sin(thetazo*%pi/180) +// receiving end voltage +r = a + %i*b +VR = Vs*%e^(-1*n*r) +//Receiving end current, +IR = VR/Zo + +printf("\n\n Result \n\n") +printf("\n Receiving end current, IR is %.2E +(%.2E)i A",real(IR), imag(IR)) diff --git a/608/CH44/EX44.05/44_05.sce b/608/CH44/EX44.05/44_05.sce new file mode 100755 index 000000000..56499c4d4 --- /dev/null +++ b/608/CH44/EX44.05/44_05.sce @@ -0,0 +1,16 @@ +//Problem 44.05: The voltages at the input and at the output of a transmission line properly terminated in its characteristic impedance are 8.0 V and 2.0 V rms respectively. Determine the output voltage if the length of the line is doubled. + +//initializing the variables: +Vs = 8; // in Volts +VR = 2; // in Volts +x = 2; + +//calculation: +// receiving end voltage VR = Vs*e^(-nr) +//e^-nr = p +p = VR/Vs +//If the line is doubled in length, then +VR = Vs*(p)^2 + +printf("\n\n Result \n\n") +printf("\n Receiving end voltage If the line is doubled in length, VR is %.2f +(%.2f)i V",real(VR), imag(VR)) diff --git a/608/CH44/EX44.06/44_06.sce b/608/CH44/EX44.06/44_06.sce new file mode 100755 index 000000000..ab22c6dd6 --- /dev/null +++ b/608/CH44/EX44.06/44_06.sce @@ -0,0 +1,19 @@ +//Problem 44.06: At a frequency of 1.5 kHz the open-circuit impedance of a length of transmission line is 800/_-50° ohm and the short-circuit impedance is 413/_-20° ohm. Determine the characteristic impedance of the line at this frequency. + +//initializing the variables: +rzoc = 800; // in ohm +thetazoc = -50; // in degrees +rzsc = 413; // in ohm +thetazsc = -20; // in degrees +f = 1500; // in Hz + +//calculation: +//open circuit impedance +Zoc = rzoc*cos(thetazoc*%pi/180) + %i*rzoc*sin(thetazoc*%pi/180) +//short circuit impedance +Zsc = rzsc*cos(thetazsc*%pi/180) + %i*rzsc*sin(thetazsc*%pi/180) +//characteristic impedance Zo +Zo = (Zoc*Zsc)^0.5 + +printf("\n\n Result \n\n") +printf("\n characteristic impedance Zo is %.0f +(%.0f)i ohm",real(Zo), imag(Zo)) diff --git a/608/CH44/EX44.07/44_07.sce b/608/CH44/EX44.07/44_07.sce new file mode 100755 index 000000000..f0b823cf1 --- /dev/null +++ b/608/CH44/EX44.07/44_07.sce @@ -0,0 +1,16 @@ +//Problem 44.07: A transmission line has the following primary constants: resistance R = 15 ohm/loop km, inductance L = 3.4 mH/loop km, conductance G = 3 μS/km and capacitance C = 10 nF/km. Determine the characteristic impedance of the line when the frequency is 2 kHz. + +//initializing the variables: +R = 15; // in ohm/loop km +L = 0.0034; // in H/loop km +C = 10E-9; // in F/km +G = 3E-6; // in S/km +f = 2000; // in Hz + +//calculation: +w = 2*%pi*f +//characteristic impedance Zo +Zo = ((R + %i*w*L)/(G + %i*w*C))^0.5 + +printf("\n\n Result \n\n") +printf("\n characteristic impedance Zo is %.1f +(%.1f)i ohm",real(Zo), imag(Zo)) diff --git a/608/CH44/EX44.08/44_08.sce b/608/CH44/EX44.08/44_08.sce new file mode 100755 index 000000000..432f1c5cc --- /dev/null +++ b/608/CH44/EX44.08/44_08.sce @@ -0,0 +1,27 @@ +//Problem 44.08: A transmission line having negligible losses has primary line constants of inductance L = 0.5 mH/loop km and capacitance C = 0.12 μF/km. Determine, at an operating frequency of 400 kHz, (a) the characteristic impedance, (b) the propagation coefficient, (c) the wavelength on the line, and (d) the velocity of propagation, in metres per second, of a signal. + +//initializing the variables: +L = 0.0005; // in H/loop km +C = 0.12E-6; // in F/km +f = 400000; // in Hz + +//calculation: +w = 2*%pi*f +//characteristic impedance Zo +Zo = (L/C)^0.5 +//the propagation coefficient +r = %i*w*(L*C)^0.5 +//the attenuation coefficient +a = real(r) +//the phaseshift coefficient +b = imag(r) +//wavelength +Y = 2*%pi/b +//velocity of propagation +u = f*Y + +printf("\n\n Result \n\n") +printf("\n characteristic impedance Zo is %.1f +(%.1f)i ohm",real(Zo), imag(Zo)) +printf("\n propagation coefficient is %.2f +(%.2f)i",a,b) +printf("\n wavelength Y is %.3f km",Y) +printf("\n speed of transmission %.2E km/sec",u) diff --git a/608/CH44/EX44.09/44_09.sce b/608/CH44/EX44.09/44_09.sce new file mode 100755 index 000000000..a0c05429f --- /dev/null +++ b/608/CH44/EX44.09/44_09.sce @@ -0,0 +1,25 @@ +//Problem 44.09: At a frequency of 1 kHz the primary constants of a transmission line are resistance R = 25 ohm/loop km, inductance L = 5 mH/loop km, capacitance C = 0.04 μF/km and conductance G = 80 μS/km. Determine for the line (a) the characteristic impedance, (b) the propagation coefficient, (c) the attenuation coefficient and (d) the phase-shift coefficient. + +//initializing the variables: +R = 25; // in ohm/loop km +L = 0.005; // in H/loop km +C = 0.04E-6; // in F/km +G = 80E-6; // in S/km +f = 1000; // in Hz + +//calculation: +w = 2*%pi*f +//characteristic impedance Zo +Zo = ((R + %i*w*L)/(G + %i*w*C))^0.5 +//the propagation coefficient +r =((R + %i*w*L)*(G + %i*w*C))^0.5 +//the attenuation coefficient +a = real(r) +//the phaseshift coefficient +b = imag(r) + +printf("\n\n Result \n\n") +printf("\n characteristic impedance Zo is %.1f +(%.1f)i ohm",real(Zo), imag(Zo)) +printf("\n propagation coefficient is %.4f +(%.4f)i",a,b) +printf("\n attenuation coefficient is %.4f Np/km",a) +printf("\n the phaseshift coefficient %.4f rad/km",b) diff --git a/608/CH44/EX44.10/44_10.sce b/608/CH44/EX44.10/44_10.sce new file mode 100755 index 000000000..c20ce7f7a --- /dev/null +++ b/608/CH44/EX44.10/44_10.sce @@ -0,0 +1,39 @@ +//Problem 44.10: An open wire line is 300 km long and is terminated in its characteristic impedance. At the sending end is a generator having an open-circuit e.m.f. of 10.0 V, an internal impedance of (400 + j0) ohmand a frequency of 1 kHz. If the line primary constants are R = 8 ohm/loop km, L = 3 mH/loop km, C = 7500 pF/km and G = 0.25 μS/km, determine (a) the characteristic impedance, (b) the propagation coefficient, (c) the attenuation and phase-shift coefficients, (d) the sending-end current, (e) the receiving-end current, (f) the wavelength on the line, and (g) the speed of transmission of signal. + +//initializing the variables: +R = 8; // in ohm/loop km +L = 0.003; // in H/loop km +C = 7500E-12; // in F/km +G = 0.25E-6; // in S/km +f = 1000; // in Hz +n = 300; // in km +Zg = 400 + %i*0; // in ohm +Vg = 10; // in Volts + +//calculation: +w = 2*%pi*f +//characteristic impedance Zo +Zo = ((R + %i*w*L)/(G + %i*w*C))^0.5 +//the propagation coefficient +r = ((R + %i*w*L)*(G + %i*w*C))^0.5 +//the attenuation coefficient +a = real(r) +//the phaseshift coefficient +b = imag(r) +//the sending-end current, +Is = Vg/(Zg + Zo) +//the receiving-end current, +IR = Is*%e^(-1*n*r) +//wavelength +Y = 2*%pi/b +//velocity of propagation +u = f*Y + +printf("\n\n Result \n\n") +printf("\n characteristic impedance Zo is %.1f +(%.1f)i ohm",real(Zo), imag(Zo)) +printf("\n propagation coefficient is %.2f +(%.2f)i",a,b) +printf("\n attenuation coefficient is %.4f Np/km and the phaseshift coefficient %.4f rad/km",a,b) +printf("\n sending-end current Is is %.3E +(%.3E)i A",real(Is), imag(Is)) +printf("\n receiving-end current IR is %.3E +(%.3E)i A",real(IR), imag(IR)) +printf("\n wavelength Y is %.3f km",Y) +printf("\n speed of transmission %.2E km/sec",u) diff --git a/608/CH44/EX44.11/44_11.sce b/608/CH44/EX44.11/44_11.sce new file mode 100755 index 000000000..5eb36e54b --- /dev/null +++ b/608/CH44/EX44.11/44_11.sce @@ -0,0 +1,15 @@ +//Problem 44.11: An underground cable has the following primary constants: resistance R = 10 ohm/loop km, inductance L = 1.5 mH/loop km, conductance G = 1.2 μS/km and capacitance C = 0.06 μF/km. Determine by how much the inductance should be increased to satisfy the condition for minimum distortion. + +//initializing the variables: +R = 10; // in ohm/loop km +L = 0.0015; // in H/loop km +C = 0.06E-6; // in F/km +G = 1.2E-6; // in S/km + +//calculation: +//the condition for minimum distortion is given by LG = CR, from which, +Lm = C*R/G +dL = Lm - L + +printf("\n\n Result \n\n") +printf("\n inductance should be increased by %.2E H/loop km for minimum distortion",dL) diff --git a/608/CH44/EX44.12/44_12.sce b/608/CH44/EX44.12/44_12.sce new file mode 100755 index 000000000..c6920783e --- /dev/null +++ b/608/CH44/EX44.12/44_12.sce @@ -0,0 +1,28 @@ +//Problem 44.12: A cable has the following primary constants: resistance R = 80 ohm/loop km, conductance, G = 2 μS/km, and capacitance C = 5 nF/km. Determine, for minimum distortion at a frequency of 1.5 kHz (a) the value of inductance per loop kilometre required, (b) the propagation coefficient, (c) the velocity of propagation of signal, and (d) the wavelength on the line + +//initializing the variables: +R = 80; // in ohm/loop km +C = 5E-9; // in F/km +G = 2E-6; // in S/km +f = 1500; // in Hz + +//calculation: +w = 2*%pi*f +//the condition for minimum distortion is given by LG = CR, from which, inductance +L = C*R/G +//attenuation coefficient, +a = (R*G)^0.5 +//phase shift coefficient, +b = w*(L*C)^0.5 +//propagation coefficient, +r = a + %i*b +//velocity of propagation, +u = 1/(L*C)^0.5 +//wavelength +Y = u/f + +printf("\n\n Result \n\n") +printf("\n inductance is %.2f H",L) +printf("\n propagation coefficient is %.4f +(%.4f)i",a,b) +printf("\n wavelength Y is %.2f km",Y) +printf("\n speed of transmission %.2E km/sec",u) diff --git a/608/CH44/EX44.13/44_13.sce b/608/CH44/EX44.13/44_13.sce new file mode 100755 index 000000000..afce2db58 --- /dev/null +++ b/608/CH44/EX44.13/44_13.sce @@ -0,0 +1,27 @@ +//Problem 44.13: A cable which has a characteristic impedance of 75 ohm is terminated in a 250 ohm resistive load. Assuming that the cable has negligible losses and the voltage measured across the terminating load is 10 V, calculate the value of (a) the reflection coefficient for the line, (b) the incident current, (c) the incident voltage, (d) the reflected current, and (e) the reflected voltage. + +//initializing the variables: +Zo = 75; // in ohm +ZR = 250; // in ohm +VR = 10; // in Volts + +//calculation: +//reflection coefficient +p = (Zo - ZR)/(Zo + ZR) +//Current flowing in the terminating load +IR = VR/ZR +//incident current, Ii +Ii = IR/(1 + p) +//incident voltage, Vi +Vi = Ii*Zo +//reflected current, Ir +Ir = IR - Ii +//reflected voltage, Vr +Vr = -1*Ir*Zo + +printf("\n\n Result \n\n") +printf("\n reflection coefficient is %.3f",p) +printf("\n incident current, Ii is %.4f A",Ii) +printf("\n incident voltage, Vi is %.2f V",Vi) +printf("\n reflected current, Ir is %.4f A",Ir) +printf("\n reflected voltage, Vr is %.2f V",Vr) diff --git a/608/CH44/EX44.14/44_14.sce b/608/CH44/EX44.14/44_14.sce new file mode 100755 index 000000000..4a0ca40e9 --- /dev/null +++ b/608/CH44/EX44.14/44_14.sce @@ -0,0 +1,16 @@ +//Problem 44.14: A long transmission line has a characteristic impedance of 500 - j40 ohm and is terminated in an impedance of (a) 500 + j40 ohm and (b) 600 + j20 ohm. Determine the magnitude of the reflection coefficient in each case. + +//initializing the variables: +Zo = 500 - %i*40; // in ohm +ZR1 = 500 + %i*40; // in ohm +ZR2 = 600 + %i*20; // in ohm + +//calculation: +//reflection coefficient +p1 = (Zo - ZR1)/(Zo + ZR1) +p2 = (Zo - ZR2)/(Zo + ZR2) +p1mag = (real(p1)^2 + imag(p1)^2)^0.5 +p2mag = (real(p2)^2 + imag(p2)^2)^0.5 + +printf("\n\n Result \n\n") +printf("\n reflection coefficient (a)%.3f and (b)%.3f",p1mag, p2mag) diff --git a/608/CH44/EX44.15/44_15.sce b/608/CH44/EX44.15/44_15.sce new file mode 100755 index 000000000..e53d9fb1d --- /dev/null +++ b/608/CH44/EX44.15/44_15.sce @@ -0,0 +1,24 @@ +//Problem 44.15: A loss-free transmission line has a characteristic impedance of 500/_0° and is connected to an aerial of impedance 320 + j240 ohm. Determine (a) the magnitude of the ratio of the reflected to the incident voltage wave, and (b) the incident voltage if the reflected voltage is 20/_35° V + +//initializing the variables: +rzo = 500; // in ohm +thetazo = 0; // in degrees +ZR = 320 + %i*240; // in ohm +rvr = 20; // in volts +thetavr = 35; // in degrees + +//calculation: +//voltage +VR = rvr*cos(thetavr*%pi/180) + %i*rvr*sin(thetavr*%pi/180) +//characteristic impedance +Zo = rzo*cos(thetazo*%pi/180) + %i*rzo*sin(thetazo*%pi/180) +//the ratio of the reflected to the incident voltage +//vr = VR/Vi +vr = (ZR - Zo)/(Zo + ZR) +vrmag = (real(vr)^2 + imag(vr)^2)^0.5 +//incident voltage, Vi +Vi = VR/vr + +printf("\n\n Result \n\n") +printf("\n the magnitude of the ratio Vr : Vi is %.3f",vrmag) +printf("\n incident voltage, Vi is %.2f +(%.2f)i V",real(Vi), imag(Vi)) diff --git a/608/CH44/EX44.16/44_16.sce b/608/CH44/EX44.16/44_16.sce new file mode 100755 index 000000000..a9f1f6bbd --- /dev/null +++ b/608/CH44/EX44.16/44_16.sce @@ -0,0 +1,19 @@ +//Problem 44.16: A transmission line has a characteristic impedance of 600/_0° and negligible loss. If the terminating impedance of the line is 400 + j250 ohm, determine (a) the reflection coefficient and (b) the standing-wave ratio. + +//initializing the variables: +rzo = 600; // in ohm +thetazo = 0; // in degrees +ZR = 400 + %i*250; // in ohm + +//calculation: +//characteristic impedance +Zo = rzo*cos(thetazo*%pi/180) + %i*rzo*sin(thetazo*%pi/180) +//reflection coefficient +p = (Zo - ZR)/(Zo + ZR) +pmag = (real(p)^2 + imag(p)^2)^0.5 +//standing-wave ratio, +s = (1 + pmag)/(1 - pmag) + +printf("\n\n Result \n\n") +printf("\n reflection coefficient, is %.4f +(%.4f)i V",real(p), imag(p)) +printf("\n standing-wave ratio, s is %.3f",s) diff --git a/608/CH44/EX44.17/44_17.sce b/608/CH44/EX44.17/44_17.sce new file mode 100755 index 000000000..12e9fcee1 --- /dev/null +++ b/608/CH44/EX44.17/44_17.sce @@ -0,0 +1,22 @@ +//Problem 44.17: A low-loss transmission line has a mismatched load such that the reflection coefficient at the termination is 0.2/_-120°. The characteristic impedance of the line is 80 ohm. Calculate (a) the standing-wave ratio, (b) the load impedance, and (c) the incident current flowing if the reflected current is 10 mA. + +//initializing the variables: +rp = 0.2; +thetap = -120; // in degrees +Zo = 80; // in ohm +Ir = 0.01; // in Amperes + +//calculation: +//reflection coefficient +p = rp*cos(thetap*%pi/180) + %i*rp*sin(thetap*%pi/180) +//standing-wave ratio, +s = (1 + rp)/(1 - rp) +//load impedance ZR +ZR = Zo*(1 - p)/(1 + p) +//incident current +Ii = Ir*(s + 1)/(s - 1) + +printf("\n\n Result \n\n") +printf("\n standing-wave ratio, s is %.1f",s) +printf("\n load impedance ZR is %.2f +(%.2f)i ohm",real(ZR), imag(ZR)) +printf("\n incident current is %.3f A",Ii) diff --git a/608/CH44/EX44.18/44_18.sce b/608/CH44/EX44.18/44_18.sce new file mode 100755 index 000000000..1f7bcdca1 --- /dev/null +++ b/608/CH44/EX44.18/44_18.sce @@ -0,0 +1,12 @@ +//Problem 44.18: The standing-wave ratio on a mismatched line is calculated as 1.60. If the incident power arriving at the termination is 200 mW, determine the value of the reflected power. + +//initializing the variables: +s = 1.6; +Pi = 0.2; // in Watts + +//calculation: +//reflected power, Pr +Pr = Pi*((s - 1)/(s + 1))^2 + +printf("\n\n Result \n\n") +printf("\n reflected power, Pr is %.5f W",Pr) diff --git a/608/CH45/EX45.01/45_01.sce b/608/CH45/EX45.01/45_01.sce new file mode 100755 index 000000000..37fbae79e --- /dev/null +++ b/608/CH45/EX45.01/45_01.sce @@ -0,0 +1,25 @@ +//Problem 45.01: A 500 nF capacitor is connected in series with a 100 kohm resistor and the circuit is connected to a 50 V, d.c. supply. Calculate (a) the initial value of current flowing, (b) the value of current 150 ms after connection, (c) the value of capacitor voltage 80 ms after connection, and (d) the time after connection when the resistor voltage is 35 V. + +//initializing the variables: +C = 500E-9; // in Farad +R = 100000; // in Ohm +V = 50; // in VOlts +ti = 0.15; // in sec +tc = 0.08; // in sec +Vrt = 35; // in Volts + +//calculation: +//Initial current, +i0 = (V/R) +//when time t = 150ms current is +i150 = (V/R)*%e^(-1*ti/(R*C)) +//capacitor voltage, Vc +Vc = V*(1 - %e^(-1*tc/(R*C))) +//time, t +tvr = -1*R*C*log(Vrt/V) + +printf("\n\n Result \n\n") +printf("\n initial value of current flowing is %.2E A",i0) +printf("\n current flowing at t = 150ms is %.2E A",i150) +printf("\n value of capacitor voltage at t = 80ms is %.2f V",Vc) +printf("\n the time after connection when the resistor voltage is 35 V is %.4f sec",tvr) diff --git a/608/CH45/EX45.02/45_02.sce b/608/CH45/EX45.02/45_02.sce new file mode 100755 index 000000000..abe0fa15b --- /dev/null +++ b/608/CH45/EX45.02/45_02.sce @@ -0,0 +1,14 @@ +//Problem 45.02: A d.c. voltage supply of 200 V is connected across a 5 μF capacitor as shown in Figure 45.5. When the supply is suddenly cut by opening switch S, the capacitor is left isolated except for a parallel resistor of 2 Mohm. Calculate the p.d. across the capacitor after 20 s. + +//initializing the variables: +C = 5E-6; // in Farad +R = 2000000; // in Ohm +V = 200; // in VOlts +tc = 20; // in sec + +//calculation: +//capacitor voltage, Vc +Vc = V*(%e^(-1*tc/(R*C))) + +printf("\n\n Result \n\n") +printf("\n value of capacitor voltage at t = 20s is %.2f V",Vc) diff --git a/608/CH45/EX45.03/45_03.sce b/608/CH45/EX45.03/45_03.sce new file mode 100755 index 000000000..703ee814b --- /dev/null +++ b/608/CH45/EX45.03/45_03.sce @@ -0,0 +1,29 @@ +//Problem 45.03: A coil of inductance 50 mH and resistance 5 ohm is connected to a 110 V, d.c. supply. Determine (a) the final value of current, (b) the value of current after 4 ms, (c) the value of the voltage across the resistor after 6 ms, (d) the value of the voltage across the inductance after 6 ms, and (e) the time when the current reaches 15 A. + +//initializing the variables: +L = 0.05; // in Henry +R = 5; // in Ohm +V = 110; // in VOlts +ti = 0.004; // in sec +tvr = 0.006; // in sec +tvl = 0.006; // in sec +it = 15; // in amperes + +//calculation: +//steady state current i +i = V/R +//when time t = 4ms current is +i4 = (V/R)*(1 - %e^(-1*ti*R/L)) +//resistor voltage, VR +VR6 = V*(1 - %e^(-1*tvr*R/L)) +//inductor voltage, VL +VL6 = V*(%e^(-1*tvl*R/L)) +//time, t +ti = (-1*L/R)*log(1 - it*R/V) + +printf("\n\n Result \n\n") +printf("\n steady state current i is %.0f A",i) +printf("\n when time t = 4ms current is is %.2f A",i4) +printf("\n value of resistor voltage at t = 6ms is %.2f V",VR6) +printf("\n value of inductor voltage at t = 6ms is %.2f V",VL6) +printf("\n the time after connection when the current is 15 A is %.5f sec",ti) diff --git a/608/CH45/EX45.04/45_04.sce b/608/CH45/EX45.04/45_04.sce new file mode 100755 index 000000000..d71ea6b36 --- /dev/null +++ b/608/CH45/EX45.04/45_04.sce @@ -0,0 +1,17 @@ +//Problem 45.04: In the circuit shown in Figure 45.8, a current of 5 A flows from the supply source. Switch S is then opened. Determine (a) the time for the current in the 2 H inductor to fall to 200 mA, and (b) the maximum voltage appearing across the resistor. + +//initializing the variables: +i = 5; // in Amperes +L = 2 // in Henry +i1 = 0.2; // in Amperes +R = 10; // in Ohm + +//calculation: +//time t +ti = (-1*L/R)*log(i1/i) +//voltage across the resistor is a maximum +VRm = i*R + +printf("\n\n Result \n\n") +printf("\n time t for the current in the 2 H inductor to fall to 200 mA is %.3f sec",ti) +printf("\n max voltage across the resistor is %.0f V",VRm) diff --git a/608/CH45/EX45.05/45_05.sce b/608/CH45/EX45.05/45_05.sce new file mode 100755 index 000000000..f9554f693 --- /dev/null +++ b/608/CH45/EX45.05/45_05.sce @@ -0,0 +1,30 @@ +//Problem 45.05: A series L –R–C circuit has inductance, L = 2 mH, resistance, R = 1 kohm and capacitance, C = 5 μF. (a) Determine whether the circuit is over, critical or underdamped. (b) If C = 5 nF, determine the state of damping. + +//initializing the variables: +L = 0.002 // in Henry +R = 1000; // in Ohm +C1 = 5E-6; // in farad +C2 = 5E-9; // in farad + +//calculation: +a = (R/(2*L))^2 +b = 1/(L*C1) +if (a>b) then + s1 = "overdamped"; +elseif (a<b) then + s1 = "underdamped"; +else + s1 = "critically damped"; +end +c = 1/(L*C2) +if (a>c) then + s2 = "overdamped"; +elseif (a<c) then + s2 = "underdamped"; +else + s2 = "critically damped"; +end + +printf("\n\n Result \n\n") +printf("\n circuit is %s",s1) +printf("\n if C = 5 nF, circuit is %s",s2)
\ No newline at end of file diff --git a/608/CH45/EX45.06/45_06.sce b/608/CH45/EX45.06/45_06.sce new file mode 100755 index 000000000..2630b6c0c --- /dev/null +++ b/608/CH45/EX45.06/45_06.sce @@ -0,0 +1,14 @@ +//Problem 45.06: In the circuit of problem 45.05, what value of capacitance will give critical damping ? + + +//initializing the variables: +L = 0.002 // in Henry +R = 1000; // in Ohm + +//calculation: +a = (R/(2*L))^2 +//for critically damped +C = 4*L/R^2 + +printf("\n\n Result \n\n") +printf("\n capacitance C is %.2E F",C) diff --git a/608/CH5/EX5.01/5_01.sce b/608/CH5/EX5.01/5_01.sce new file mode 100755 index 000000000..6caaf286c --- /dev/null +++ b/608/CH5/EX5.01/5_01.sce @@ -0,0 +1,21 @@ +//Problem 5.01: For the circuit shown in Figure 5.2, determine (a) the battery voltage V, (b) the total resistance of the circuit, and (c) the values of resistance of resistors R1, R2 and R3, given that the p.d.’s across R1, R2 and R3 are 5 V, 2 V and 6 V respectively.
+
+//initializing the variables:
+V1 = 5; // in volts
+V2 = 2; // in volts
+V3 = 6; // in volts
+I = 4; // in Amperes
+
+//calculation:
+Vt = V1 + V2 + V3
+Rt = Vt/I
+R1 = V1/I
+R2 = V2/I
+R3 = V3/I
+
+printf("\n\nResult\n\n")
+printf("\n (a) Total Voltage %.0f Volts(V)",Vt)
+printf("\n (b)Total Resistance %.2f Ohms",Rt)
+printf("\n (c)Resistance(R1) %.2f Ohms",R1)
+printf("\n (c)Resistance(R2) %.1f Ohms",R2)
+printf("\n (c)Resistance(R3) %.1f Ohms",R3)
\ No newline at end of file diff --git a/608/CH5/EX5.02/5_02.sce b/608/CH5/EX5.02/5_02.sce new file mode 100755 index 000000000..34616e2f1 --- /dev/null +++ b/608/CH5/EX5.02/5_02.sce @@ -0,0 +1,17 @@ +//Problem 5.02: For the circuit shown in Figure 5.3, determine the p.d. across resistor R3. If the total resistance of the circuit is 100 ohms, determine the current flowing through resistor R1. Find also the value of resistor R2
+
+//initializing the variables:
+V1 = 10; // in volts
+V2 = 4; // in volts
+Vt = 25; // in volts
+Rt = 100; // in ohms
+
+//calculation:
+V3 = Vt - V1 - V2
+I = Vt/Rt
+R2 = V2/I
+
+printf("\n\nResult\n\n")
+printf("\n (a)Voltage(V3) %.0f Volts(V)",V3)
+printf("\n (b)current %.2f Amperes(A)",I)
+printf("\n (c)Resistance(R2) %.0f Ohms",R2)
\ No newline at end of file diff --git a/608/CH5/EX5.03/5_03.sce b/608/CH5/EX5.03/5_03.sce new file mode 100755 index 000000000..ec3444378 --- /dev/null +++ b/608/CH5/EX5.03/5_03.sce @@ -0,0 +1,18 @@ +//Problem 5.03: A 12 V battery is connected in a circuit having three series-connected resistors having resistances of 4 ohms, 9 ohms and 11 ohms. Determine the current flowing through, and the p.d. across the 9 ohms resistor. Find also the power dissipated in the 11 ohms resistor.
+
+//initializing the variables:
+Vt = 12; // in volts
+R1 = 4; // in ohms
+R2 = 9; // in ohms
+R3 = 11; // in ohms
+
+//calculation:
+Rt = R1 + R2 + R3
+I = Vt/Rt
+V9 = I*R2
+P11 = I*I*R3
+
+printf("\n\nResult\n\n")
+printf("\n (a)current %.1f Amperes(A)",I)
+printf("\n (b)Voltage(V2) %.1f Volts(V)",V9)
+printf("\n (c)Power %.2f Watt(W)",P11)
\ No newline at end of file diff --git a/608/CH5/EX5.04/5_04.sce b/608/CH5/EX5.04/5_04.sce new file mode 100755 index 000000000..3205d6d61 --- /dev/null +++ b/608/CH5/EX5.04/5_04.sce @@ -0,0 +1,14 @@ +//Problem 5.04:Determine the value of voltage V shown in Figure 5.6.
+
+//initializing the variables:
+Vt = 50; // in volts
+R1 = 4; // in ohms
+R2 = 6; // in ohms
+
+//calculation:
+Rt = R1 + R2
+I = Vt/Rt
+V2 = I*R2
+
+printf("\n\nResult\n\n")
+printf("\n Voltage(V) %.0f Volts(V)\n",V2)
\ No newline at end of file diff --git a/608/CH5/EX5.05/5_05.sce b/608/CH5/EX5.05/5_05.sce new file mode 100755 index 000000000..cbf5e8b82 --- /dev/null +++ b/608/CH5/EX5.05/5_05.sce @@ -0,0 +1,17 @@ +//Problem 5.05:Two resistors are connected in series across a 24 V supply and a current of 3 A flows in the circuit. If one of the resistors has a resistance of 2 ohms determine (a) the value of the other resistor, and (b) the p.d. across the 2 resistor. If the circuit is connected for 50 hours, how much energy is used?
+
+//initializing the variables:
+Vt = 24; // in volts
+R1 = 2; // in ohms
+I = 3; // in Amperes
+t = 50; // in hrs
+
+//calculation:
+V1 = I*R1
+R2 = [Vt-(I*R1)]/I
+E = Vt*I*t
+
+printf("\n\nResult\n\n")
+printf("\n (a)Voltage(V1) %.0f Volts(V)",V1)
+printf("\n (b)Resistance(R2) %.0f Ohms",R2)
+printf("\n (a)Energy(E) %.2E Wh",E)
\ No newline at end of file diff --git a/608/CH5/EX5.06/5_06.sce b/608/CH5/EX5.06/5_06.sce new file mode 100755 index 000000000..c4d37cc11 --- /dev/null +++ b/608/CH5/EX5.06/5_06.sce @@ -0,0 +1,16 @@ +//Problem 5.06:For the circuit shown in Figure 5.10, determine (a) the reading on the ammeter, and (b) the value of resistor R2
+
+//initializing the variables:
+R1 = 5; // in ohms
+R3 = 20; // in ohms
+I1 = 8; // in Amperes
+It = 11; // in Amperes
+
+//calculation:
+Vt = I1*R1
+I3 = Vt/R3
+R2 = Vt/[It - I1 - I3]
+
+printf("\n\nResult\n\n")
+printf("\n (a)Ammeter Reading %.0f Amperes(A)",I3)
+printf("\n (b)Resistance(R2) %.0f Ohms\n",R2)
\ No newline at end of file diff --git a/608/CH5/EX5.07/5_07.sce b/608/CH5/EX5.07/5_07.sce new file mode 100755 index 000000000..4a9da6f33 --- /dev/null +++ b/608/CH5/EX5.07/5_07.sce @@ -0,0 +1,14 @@ +//Problem 5.07: Two resistors, of resistance 3 ohms and 6 ohms, are connected in parallel across a battery having a voltage of 12 V. Determine (a) the total circuit resistance and (b) the current flowing in the 3 ohms resistor.
+
+//initializing the variables:
+R1 = 3; // in ohms
+R2 = 6; // in ohms
+Vt = 12; // in volts
+
+//calculation:
+Rt = R1*R2/[R1 + R2]
+I1 = [Vt/R1]
+
+printf("\n\nResult\n\n")
+printf("\n (a)Total Resistance %.0f Ohms",Rt)
+printf("\n (b)Current(I1) %.0f Amperes(A)\n",I1)
\ No newline at end of file diff --git a/608/CH5/EX5.08/5_08.sce b/608/CH5/EX5.08/5_08.sce new file mode 100755 index 000000000..2b934a8a6 --- /dev/null +++ b/608/CH5/EX5.08/5_08.sce @@ -0,0 +1,17 @@ +//Problem 5.08: For the circuit shown in Figure 5.12, find (a) the value of the supply voltage V and (b) the value of current I.
+
+//initializing the variables:
+R1 = 10; // in ohms
+R2 = 20; // in ohms
+R3 = 60; // in ohms
+I2 = 3; // in Amperes
+
+//calculation:
+Vt = I2*R2
+I1 = Vt/R1
+I3 = Vt/R3
+I = I1 +I2 + I3
+
+printf("\n\nResult\n\n")
+printf("\n (a)Voltage(V) %.0f Volts(V)",Vt)
+printf("\n (b)Total Current(I) %.0f Amperes(A)",I)
\ No newline at end of file diff --git a/608/CH5/EX5.10/5_10.sce b/608/CH5/EX5.10/5_10.sce new file mode 100755 index 000000000..5b10e0267 --- /dev/null +++ b/608/CH5/EX5.10/5_10.sce @@ -0,0 +1,17 @@ +//Problem 5.10: Find the equivalent resistance for the circuit shown in Figure 5.17.
+
+//initializing the variables:
+R1 = 1; // in ohms
+R2 = 2.2; // in ohms
+R3 = 3; // in ohms
+R4 = 6; // in ohms
+R5 = 18; // in ohms
+R6 = 4; // in ohms
+
+
+//calculation:
+R0 = 1/[(1/3) + (1/6) + (1/18)]
+Rt = R1 + R2 + R0 + R6
+
+printf("\n\nResult\n\n")
+printf("\n Equivalent Resistance %.0f Ohms\n",Rt)
\ No newline at end of file diff --git a/608/CH5/EX5.11/5_11.sce b/608/CH5/EX5.11/5_11.sce new file mode 100755 index 000000000..5df293083 --- /dev/null +++ b/608/CH5/EX5.11/5_11.sce @@ -0,0 +1,26 @@ +//Problem 5.11: For the series-parallel arrangement shown in Figure 5.19, find (a) the supply current, (b) the current flowing through each resistor and (c) the p.d. across each resistor.
+
+//initializing the variables:
+R1 = 2.5; // in ohms
+R2 = 6; // in ohms
+R3 = 2; // in ohms
+R4 = 4; // in ohms
+Vt = 200; // in volts
+
+//calculation:
+R0 = 1/[(1/R2) + (1/R3)]
+Rt = R1 + R0 + R4
+It = Vt/Rt
+I1 = It
+I4 = It
+I2 = R3*It/(R3+R2)
+I3 = It - I2
+V1 = I1*R1
+V2 = I2*R2
+V3 = I3*R3
+V4 = I4*R4
+
+printf("\n\nResult\n\n")
+printf("\n (a)Total Current Supply %.0f Amperes(A)",It)
+printf("\n (b)Current through resistors (R1, R2, R3, R4)%.0f, %.2f, %.2f, %.0f Amperes(A) respectively",I1, I2, I3, I4)
+printf("\n (c)voltage across resistors (R1, R2, R3, R4)%.1f, %.1f, %.1f, %.0f Volts(V) respectively",V1, V2, V3, V4)
\ No newline at end of file diff --git a/608/CH5/EX5.12/5_12.sce b/608/CH5/EX5.12/5_12.sce new file mode 100755 index 000000000..a21866b65 --- /dev/null +++ b/608/CH5/EX5.12/5_12.sce @@ -0,0 +1,23 @@ +//Problem 5.12: For the circuit shown in Figure 5.21 calculate (a) the value of resistor Rx such that the total power dissipated in the circuit is 2.5 kW, and (b) the current flowing in each of the four resistors.
+
+//initializing the variables:
+R1 = 15; // in ohms
+R2 = 10; // in ohms
+R3 = 38; // in ohms
+Vt = 250; // in volts
+P = 2500; // in Watt
+
+//calculation:
+It = P/Vt
+I2 = R1*It/(R1+R2)
+I1 = It - I2
+Re1 = 1/[(1/R1) + (1/R2)]
+Rt = Vt/It
+Re2 = Rt - Re1
+Rx = 1/[(1/Re2) - (1/R3)]
+I4 = R3*It/(R3+Rx)
+I3 = It - I4
+
+printf("\n\nResult\n\n")
+printf("\n (a)Resistance (Rx) %.0f Ohms",Rx)
+printf("\n (b)Current through resistors (R1, R2, R3, R4): %.0f, %.0f, %.0f, %.0f Amperes(A) respectively",I1, I2, I3, I4)
\ No newline at end of file diff --git a/608/CH5/EX5.13/5_13.sce b/608/CH5/EX5.13/5_13.sce new file mode 100755 index 000000000..c6cee399f --- /dev/null +++ b/608/CH5/EX5.13/5_13.sce @@ -0,0 +1,21 @@ +//Problem 5.13: For the arrangement shown in Figure 5.22, find the current Ix
+
+//initializing the variables:
+R1 = 8; // in ohms
+R2 = 2; // in ohms
+R3 = 1.4; // in ohms
+R4 = 9; // in ohms
+R5 = 2; // in ohms
+Vt = 17; // in volts
+
+//calculation:
+R01 = R1*R2/(R1 + R2)
+R02 = R01 + R3
+R03 = R4*R02/(R4 +R02)
+Rt = R5 + R03
+It = Vt/Rt
+I1 = R4*It/(R4 + R02)
+Ix = R2*I1/(R1 + R2)
+
+printf("\n\nResult\n\n")
+printf("\n Current(Ix) %.1f Amperes(A)\n",Ix)
\ No newline at end of file diff --git a/608/CH5/EX5.14/5_14.sce b/608/CH5/EX5.14/5_14.sce new file mode 100755 index 000000000..f6329d22e --- /dev/null +++ b/608/CH5/EX5.14/5_14.sce @@ -0,0 +1,11 @@ +//Problem 5.14: If three identical lamps are connected in parallel and the combined resistance is 150 ohms, find the resistance of one lamp.
+
+//initializing the variables:
+Rt = 150; // in ohms
+n = 3; // no. of identical lamp
+
+//calculation:
+R = Rt*3 // (1/Rt)=(1/R)+(1/R)+(1/R)
+
+printf("\n\nResult\n\n")
+printf("\n Resistance %.0f Ohms\n",R)
\ No newline at end of file diff --git a/608/CH5/EX5.15/5_15.sce b/608/CH5/EX5.15/5_15.sce new file mode 100755 index 000000000..efc6db413 --- /dev/null +++ b/608/CH5/EX5.15/5_15.sce @@ -0,0 +1,12 @@ +//Problem 5.15: Three identical lamps A, B and C are connected in series across a 150 V supply. State (a) the voltage across each lamp, and (b) the effect of lamp C failing.
+
+//initializing the variables:
+//series connection
+n = 3; // no. of identical lamp
+Vt = 150; // in volts
+
+//calculation:
+V = Vt/3 // Since each lamp is identical, then V volts across each.
+
+printf("\n\nResult\n\n")
+printf("\n Voltage across each resistor = %.0f Volts(V)",V)
\ No newline at end of file diff --git a/608/CH6/EX6.01/6_01.sce b/608/CH6/EX6.01/6_01.sce new file mode 100755 index 000000000..9871a8995 --- /dev/null +++ b/608/CH6/EX6.01/6_01.sce @@ -0,0 +1,15 @@ +//Problem 6.01:(a) Determine the p.d. across a 4 μF capacitor when charged with 5 mC. (b) Find the charge on a 50 pF capacitor when the voltage applied to it is 2 kV.
+
+//initializing the variables:
+C1 = 4E-6; // in Farad
+C2 = 50E-12; // in Farad
+Q1 = 5E-3; // in Coulomb
+V2 = 2000; // in volts
+
+//calculation:
+V1 = Q1/C1
+Q2 = C2*V2
+
+printf("\n\nResult\n\n")
+printf("\n (a)P.d %.0f Volts(V)",V1)
+printf("\n (b)Charge(Q) %.2E Coulomb(C)",Q2)
\ No newline at end of file diff --git a/608/CH6/EX6.02/6_02.sce b/608/CH6/EX6.02/6_02.sce new file mode 100755 index 000000000..3a8095f3f --- /dev/null +++ b/608/CH6/EX6.02/6_02.sce @@ -0,0 +1,13 @@ +//Problem 6.02: A direct current of 4 A flows into a previously uncharged 20 μF capacitor for 3 ms. Determine the pd between the plates.
+
+//initializing the variables:
+I = 4; // in amperes
+C = 20E-6; // in Farad
+t = 3E-3; // in sec
+
+//calculation:
+Q = I*t
+V = Q/C
+
+printf("\n\nResult\n\n")
+printf("\n (a)P.d %.0f Volts(V)\n",V)
\ No newline at end of file diff --git a/608/CH6/EX6.03/6_03.sce b/608/CH6/EX6.03/6_03.sce new file mode 100755 index 000000000..532dcc96f --- /dev/null +++ b/608/CH6/EX6.03/6_03.sce @@ -0,0 +1,13 @@ +//Problem 6.03:A 5 μF capacitor is charged so that the pd between its plates is 800 V. Calculate how long the capacitor can provide an average discharge current of 2 mA.
+
+//initializing the variables:
+I = 2E-3; // in amperes
+C = 5E-6; // in Farad
+V = 800; // in volts
+
+//calculation:
+Q = C*V
+t = Q/I
+
+printf("\n\nResult\n\n")
+printf("\n capacitor can provide an average discharge current of 2mA for %.0f Sec\n",t)
\ No newline at end of file diff --git a/608/CH6/EX6.04/6_04.sce b/608/CH6/EX6.04/6_04.sce new file mode 100755 index 000000000..540cf85da --- /dev/null +++ b/608/CH6/EX6.04/6_04.sce @@ -0,0 +1,15 @@ +//Problem 6.04: Two parallel rectangular plates measuring 20 cm by 40 cm carry an electric charge of 0.2 μC. Calculate the electric flux density. If the plates are spaced 5 mm apart and the voltage between them is 0.25 kV determine the electric field strength.
+
+//initializing the variables:
+Q = 0.2E-6; // in Coulomb
+A = 800E-4; // in m2
+d = 0.005; // in m
+V = 250; // in Volts
+
+//calculation:
+D = Q/A
+E = V/d
+
+printf("\n\nResult\n\n")
+printf("\n (a)Electric flux density D %.2E C/m2",D)
+printf("\n (b)Electric field strength E %.2E V/m\n",E)
\ No newline at end of file diff --git a/608/CH6/EX6.05/6_05.sce b/608/CH6/EX6.05/6_05.sce new file mode 100755 index 000000000..c0744941f --- /dev/null +++ b/608/CH6/EX6.05/6_05.sce @@ -0,0 +1,12 @@ +//Problem 6.05: The flux density between two plates separated by mica of relative permittivity 5 is 2 μC/m2. Find the voltage gradient between the plates.
+
+//initializing the variables:
+D = 2E-6; // in μC/m2
+e0 = 8.85E-12; // in F/m
+er = 5;
+
+//calculation:
+E = D/(e0*er)
+
+printf("\n\nResult\n\n")
+printf("\n Electric field strength E %.2E V/m\n",E)
\ No newline at end of file diff --git a/608/CH6/EX6.06/6_06.sce b/608/CH6/EX6.06/6_06.sce new file mode 100755 index 000000000..333c8db51 --- /dev/null +++ b/608/CH6/EX6.06/6_06.sce @@ -0,0 +1,18 @@ +//Problem 6.06: Two parallel plates having a pd of 200 V between them are spaced 0.8 mm apart. What is the electric field strength? Find also the flux density when the dielectric between the plates is (a) air, and (b) polythene of relative permittivity 2.3
+
+//initializing the variables:
+d = 0.8E-3; // in m
+e0 = 8.85E-12; // in F/m
+era = 1; // for air
+erp = 2.3; // for polythene
+V =200; // in Volts
+
+//calculation:
+E = V/d
+//for air
+Da = E*e0*era
+//for polythene
+Dp = E*e0*erp
+printf("\n\nResult\n\n")
+printf("\n (a)Electric flux density for air %.2E C/m2",Da)
+printf("\n (b)Electric flux density for polythene %.2E C/m2\n",Dp)
\ No newline at end of file diff --git a/608/CH6/EX6.07/6_07.sce b/608/CH6/EX6.07/6_07.sce new file mode 100755 index 000000000..a3c6b2f09 --- /dev/null +++ b/608/CH6/EX6.07/6_07.sce @@ -0,0 +1,16 @@ +//Problem 6.07: (a) A ceramic capacitor has an effective plate area of 4 cm2 separated by 0.1 mm of ceramic of relative permittivity 100. Calculate the capacitance of the capacitor in picofarads. (b) If the capacitor in part (a) is given a charge of 1.2 μC what will be the pd between the plates?
+
+//initializing the variables:
+A = 4E-4; // in m2
+d = 0.1E-3; // in m
+e0 = 8.85E-12; // in F/m
+er = 100;
+Q = 1.2E-6; // in coulomb
+
+//calculation:
+C = e0*er*A/d
+V = Q/C
+
+printf("\n\nResult\n\n")
+printf("\n (a)Capacitance %.2E F",C)
+printf("\n (b)P.d.= %.0f Volts(V)\n",V)
\ No newline at end of file diff --git a/608/CH6/EX6.08/6_08.sce b/608/CH6/EX6.08/6_08.sce new file mode 100755 index 000000000..a4805ab0e --- /dev/null +++ b/608/CH6/EX6.08/6_08.sce @@ -0,0 +1,13 @@ +//Problem 6.08: A waxed paper capacitor has two parallel plates, each of effective area 800 cm2. If the capacitance of the capacitor is 4425 pF determine the effective thickness of the paper if its relative permittivity is 2.5
+
+//initializing the variables:
+A = 800E-4; // in m2
+C = 4425E-12; // in Farads
+e0 = 8.85E-12; // in F/m
+er = 2.5;
+
+//calculation:
+d = e0*er*A/C
+
+printf("\n\nResult\n\n")
+printf("\n Thickness %.2E m\n",d)
\ No newline at end of file diff --git a/608/CH6/EX6.09/6_09.sce b/608/CH6/EX6.09/6_09.sce new file mode 100755 index 000000000..1bd7c3bfa --- /dev/null +++ b/608/CH6/EX6.09/6_09.sce @@ -0,0 +1,16 @@ +//Problem 6.09: A parallel plate capacitor has nineteen interleaved plates each 75 mm by 75 mm separated by mica sheets 0.2 mm thick. Assuming the relative permittivity of the mica is 5, calculate the capacitance of the capacitor.
+
+//initializing the variables:
+n = 19; // no. of plates
+L = 75E-3; // in m
+B = 75E-3; // in m
+d = 0.2E-3; // in m
+e0 = 8.85E-12; // in F/m
+er = 5;
+
+//calculation:
+A = L*B
+C = e0*er*A*(n-1)/d
+
+printf("\n\nResult\n\n")
+printf("\n Capacitance %.2E F\n",C)
\ No newline at end of file diff --git a/608/CH6/EX6.10/6_10.sce b/608/CH6/EX6.10/6_10.sce new file mode 100755 index 000000000..90c963513 --- /dev/null +++ b/608/CH6/EX6.10/6_10.sce @@ -0,0 +1,15 @@ +//Problem 6.10: Calculate the equivalent capacitance of two capacitors of 6 μF and 4 μF connected (a) in parallel and (b) in series
+
+//initializing the variables:
+C1 = 6E-6; // in Farads
+C2 = 4E-6; // in Farads
+
+//calculation:
+// in Parallel
+Cp = C1 +C2
+// in Series
+Cs = 1/[(1/C1) + (1/C2)]
+
+printf("\n\nResult\n\n")
+printf("\n (a)Capacitance in parallel %.2E F",Cp)
+printf("\n (b)Capacitance in Series %.2E F\n",Cs)
\ No newline at end of file diff --git a/608/CH6/EX6.11/6_11.sce b/608/CH6/EX6.11/6_11.sce new file mode 100755 index 000000000..d4ebe45f2 --- /dev/null +++ b/608/CH6/EX6.11/6_11.sce @@ -0,0 +1,12 @@ +//Problem 6.11: What capacitance must be connected in series with a 30 μF capacitor for the equivalent capacitance to be 12 μF?
+
+//initializing the variables:
+C1 = 30E-6; // in Farads
+Cs = 12E-6; // in Farads
+
+//calculation:
+// in Series
+C2 = 1/[(1/Cs) - (1/C1)]
+
+printf("\n\nResult\n\n")
+printf("\n (a)Capacitance in series %.2E F\n",C2)
\ No newline at end of file diff --git a/608/CH6/EX6.12/6_12.sce b/608/CH6/EX6.12/6_12.sce new file mode 100755 index 000000000..221413751 --- /dev/null +++ b/608/CH6/EX6.12/6_12.sce @@ -0,0 +1,22 @@ +//Problem 6.12: Capacitances of 1 μF, 3 μF, 5 μF and 6 μF are connected in parallel to a direct voltage supply of 100 V. Determine (a) the equivalent circuit capacitance, (b) the total charge and (c) the charge on each capacitor.
+
+//initializing the variables:
+C1 = 1E-6; // in Farads
+C2 = 3E-6; // in Farads
+C3 = 5E-6; // in Farads
+C4 = 6E-6; // in Farads
+Vt = 100; // in Volts
+
+//calculation:
+// in Parallel
+Cp = C1 + C2 + C3 + C4
+Qt = Vt*Cp
+Q1 = C1*Vt
+Q2 = C2*Vt
+Q3 = C3*Vt
+Q4 = C4*Vt
+
+printf("\n\nResult\n\n")
+printf("\n (a)Equivalent Capacitance in Parallel %.2E F",Cp)
+printf("\n (b)Total charge %.2E C",Qt)
+printf("\n (c)Charge on each capacitors (C1, C2, C3, C4) %.2E C, %.2E C, %.2E C, %.2E C respectively",Q1,Q2,Q3,Q4)
\ No newline at end of file diff --git a/608/CH6/EX6.13/6_13.sce b/608/CH6/EX6.13/6_13.sce new file mode 100755 index 000000000..40c3caced --- /dev/null +++ b/608/CH6/EX6.13/6_13.sce @@ -0,0 +1,20 @@ +//Problem 6.13: Capacitances of 3 μF, 6 μF and 12 μF are connected in series across a 350 V supply. Calculate (a) the equivalent circuit capacitance, (b) the charge on each capacitor and (c) the pd across each capacitor.
+
+//initializing the variables:
+C1 = 3E-6; // in Farads
+C2 = 6E-6; // in Farads
+C3 = 12E-6; // in Farads
+Vt = 350; // in Volts
+
+//calculation:
+// in series
+Cs = 1/[(1/C1) + (1/C2) + (1/C3)]
+Qt = Vt*Cs
+V1 = Qt/C1
+V2 = Qt/C2
+V3 = Qt/C3
+
+printf("\n\nResult\n\n")
+printf("\n (a)Equivalent Capacitance in Series %.2E F",Cs)
+printf("\n (b)Charge on each capacitors (C1, C2, C3) %.2E C",Qt)
+printf("\n (b)P.d Across each capacitors (C1, C2, C3) %.0f V, %.0f V, %.0f V respectively",V1, V2, V3)
\ No newline at end of file diff --git a/608/CH6/EX6.14/6_14.sce b/608/CH6/EX6.14/6_14.sce new file mode 100755 index 000000000..a25565e6b --- /dev/null +++ b/608/CH6/EX6.14/6_14.sce @@ -0,0 +1,16 @@ +//Problem 6.14: A capacitor is to be constructed so that its capacitance is 0.2 μF and to take a p.d. of 1.25 kV across its terminals. The dielectric is to be mica which, after allowing a safety factor of 2, has a dielectric strength of 50 MV/m. Find (a) the thickness of the mica needed, and (b) the area of a plate assuming a two-plate construction. (Assume r for mica to be 6)
+
+//initializing the variables:
+C = 0.2E-6; // in Farads
+V = 1250; // in Volts
+E = 50E6; // in V/m
+e0 = 8.85E-12; // in F/m
+er = 6;
+
+//calculation:
+d = V/E
+A = C*d/e0/er
+
+printf("\n\nResult\n\n")
+printf("\n (a)Thickness %.2E m",d)
+printf("\n (b)Area of plate is %.2E m2 \n",A)
\ No newline at end of file diff --git a/608/CH6/EX6.15/6_15.sce b/608/CH6/EX6.15/6_15.sce new file mode 100755 index 000000000..c49052ef5 --- /dev/null +++ b/608/CH6/EX6.15/6_15.sce @@ -0,0 +1,14 @@ +//Problem 6.15: (a) Determine the energy stored in a 3 μF capacitor when charged to 400 V. (b) Find also the average power developed if this energy is dissipated in a time of 10 μs
+
+//initializing the variables:
+C = 3E-6; // in Farads
+V = 400; // in Volts
+t = 10E-6; // in secs
+
+//calculation:
+W = C*V*V/2
+P = W/t
+
+printf("\n\nResult\n\n")
+printf("\n (a)Energy stored %.2f J",W)
+printf("\n (b)Power developed %.2E W ",P)
\ No newline at end of file diff --git a/608/CH6/EX6.16/6_16.sce b/608/CH6/EX6.16/6_16.sce new file mode 100755 index 000000000..6a83421db --- /dev/null +++ b/608/CH6/EX6.16/6_16.sce @@ -0,0 +1,11 @@ +//Problem 6.16: A 12 μF capacitor is required to store 4 J of energy. Find the pd to which the capacitor must be charged.
+
+//initializing the variables:
+C = 12E-6; // in Farads
+W = 4; // in Joules
+
+//calculation:
+V = (2*W/C)^0.5
+
+printf("\n\nResult\n\n")
+printf("\n P.d %.1f V\n",V)
\ No newline at end of file diff --git a/608/CH6/EX6.17/6_17.sce b/608/CH6/EX6.17/6_17.sce new file mode 100755 index 000000000..561c5a25a --- /dev/null +++ b/608/CH6/EX6.17/6_17.sce @@ -0,0 +1,13 @@ +//Problem 6.17: A capacitor is charged with 10 mC. If the energystored is 1.2 J find (a) the voltage and (b) the capacitance.
+
+//initializing the variables:
+W = 1.2; // in Joules
+Q = 10E-3; // in Coulomb
+
+//calculation:
+V = 2*W/Q
+C = Q/V
+
+printf("\n\nResult\n\n")
+printf("\n (a)P.d %.0f V",V)
+printf("\n (b)Capacitance %.2E F\n",C)
\ No newline at end of file diff --git a/608/CH7/EX7.01/7_01.sce b/608/CH7/EX7.01/7_01.sce new file mode 100755 index 000000000..51b920009 --- /dev/null +++ b/608/CH7/EX7.01/7_01.sce @@ -0,0 +1,13 @@ +//Problem 7.01: A magnetic pole face has a rectangular section having dimensions 200 mm by 100 mm. If the total flux emerging from the pole is 150 μWb, calculate the flux density.
+
+//initializing the variables:
+Phi = 150E-6; // in Wb
+l = 200E-3; // in m
+b = 100E-3; // in m
+
+//calculation:
+A = l*b
+B = Phi/A
+
+printf("\n\nResult\n\n")
+printf("\n Flux density %.2E T\n",B)
\ No newline at end of file diff --git a/608/CH7/EX7.02/7_02.sce b/608/CH7/EX7.02/7_02.sce new file mode 100755 index 000000000..ffa2abcc1 --- /dev/null +++ b/608/CH7/EX7.02/7_02.sce @@ -0,0 +1,13 @@ +//Problem 7.02: The maximum working flux density of a lifting electromagnet is 1.8 T and the effective area of a pole face is circular in cross-section. If the total magnetic flux produced is 353 mWb, determine the radius of the pole face.
+
+//initializing the variables:
+Phi = 353E-3; // in Wb
+B = 1.8; // in tesla
+Pi = 3.14;
+
+//calculation:
+A = Phi/B
+r = (A/Pi)^0.5
+
+printf("\n\nResult\n\n")
+printf("\n radius of the pole face is %.2E m\n",r)
\ No newline at end of file diff --git a/608/CH7/EX7.03/7_03.sce b/608/CH7/EX7.03/7_03.sce new file mode 100755 index 000000000..1b9eae165 --- /dev/null +++ b/608/CH7/EX7.03/7_03.sce @@ -0,0 +1,14 @@ +//Problem 7.03: A magnetizing force of 8000 A/m is applied to a circular magnetic circuit of mean diameter 30 cm by passing a current through a coil wound on the circuit. If the coil is uniformly wound around the circuit and has 750 turns, find the current in the coil.
+
+//initializing the variables:
+H = 8000; // in A/m
+d = 0.30; // in m
+N = 750; // no. of turns
+Pi = 3.14;
+
+//calculation:
+l = Pi*d
+I = H*l/N
+
+printf("\n\nResult\n\n")
+printf("\n current I = %.2f Ampere(A)\n",I)
\ No newline at end of file diff --git a/608/CH7/EX7.04/7_04.sce b/608/CH7/EX7.04/7_04.sce new file mode 100755 index 000000000..e96bc62bf --- /dev/null +++ b/608/CH7/EX7.04/7_04.sce @@ -0,0 +1,13 @@ +//Problem 7.04: A flux density of 1.2 T is produced in a piece of cast steel by a magnetizing force of 1250 A/m. Find the relative permeability of the steel under these conditions.
+
+//initializing the variables:
+B = 1.2; // in Tesla
+H = 1250; // in A/m
+Pi = 3.14;
+u0 = 4*Pi*1E-7;
+
+//calculation:
+ur = B/(u0*H)
+
+printf("\n\nResult\n\n")
+printf("\n relative permeability of the steel = %.0f \n",ur)
\ No newline at end of file diff --git a/608/CH7/EX7.05/7_05.sce b/608/CH7/EX7.05/7_05.sce new file mode 100755 index 000000000..9ae618cf5 --- /dev/null +++ b/608/CH7/EX7.05/7_05.sce @@ -0,0 +1,14 @@ +//Problem 7.05: Determine the magnetic field strength and the mmf required to produce a flux density of 0.25 T in an air gap of length 12 mm.
+
+//initializing the variables:
+B = 0.25; // in Tesla
+u0 = 4*%pi*1E-7;
+l = 12E-3; // in m
+
+//calculation:
+H = B/u0
+mmf = H*l
+
+printf("\n\nResult\n\n")
+printf("\n (a)Magnetic field strength H = %.0f A/m",H)
+printf("\n (b)mmf = %.0f A\n",mmf)
\ No newline at end of file diff --git a/608/CH7/EX7.06/7_06.sce b/608/CH7/EX7.06/7_06.sce new file mode 100755 index 000000000..32fcc1009 --- /dev/null +++ b/608/CH7/EX7.06/7_06.sce @@ -0,0 +1,18 @@ +//Problem 7.06: A coil of 300 turns is wound uniformly on a ring of non-magnetic material. The ring has a mean circumference of 40 cm and a uniform cross sectional area of 4 cm2. If the current in the coil is 5 A, calculate (a) the magnetic field strength, (b) the flux density and (c) the total magnetic flux in the ring.
+
+//initializing the variables:
+N = 300; // no. of turns
+l = 0.40; // in m
+A = 4E-4; // in m2
+I = 5; // in Amperes
+u0 = 4*%pi*1E-7;
+ur = 1
+//calculation:
+H = N*I/l
+B = u0*ur*H
+Phi = B*A
+
+printf("\n\nResult\n\n")
+printf("\n (a)Magnetic field strength H = %.0f A/m\",H)
+printf("\n (b)Flux Density = %.2E T",B)
+printf("\n (c)total magnetic flux = %.2E Wb",Phi)
\ No newline at end of file diff --git a/608/CH7/EX7.07/7_07.sce b/608/CH7/EX7.07/7_07.sce new file mode 100755 index 000000000..f9374aac0 --- /dev/null +++ b/608/CH7/EX7.07/7_07.sce @@ -0,0 +1,17 @@ +//Problem 7.07: An iron ring of mean diameter 10 cm is uniformly wound with 2000 turns of wire. When a current of 0.25 A is passed through the coil a flux density of 0.4 T is set up in the iron. Find (a) the magnetizing force and (b) the relative permeability of the iron under these conditions.
+
+//initializing the variables:
+N = 2000; // no. of turns
+d = 0.10; // in m
+B = 0.4; // in Tesla
+I = 0.25; // in Amperes
+u0 = 4*%pi*1E-7;
+
+//calculation:
+l = %pi*d
+H = N*I/l
+ur = B/(u0*H)
+
+printf("\n\nResult\n\n")
+printf("\n (a)Magnetic field strength H = %.0f A/m",H)
+printf("\n (b)relative permeability of the iron = %.0f ",ur)
\ No newline at end of file diff --git a/608/CH7/EX7.08/7_08.sce b/608/CH7/EX7.08/7_08.sce new file mode 100755 index 000000000..87550b0e9 --- /dev/null +++ b/608/CH7/EX7.08/7_08.sce @@ -0,0 +1,15 @@ +//Problem 7.08: A uniform ring of cast iron has a cross-sectional area of 10 cm2 and a mean circumference of 20 cm. Determine the mmf necessary to produce a flux of 0.3 mWb in the ring.
+
+//initializing the variables:
+A = 10E-4; // in m2
+l = 0.20; // in m
+Phi = 0.3E-3; // in Wb
+u0 = 4*%pi*1E-7;
+
+//calculation:
+B = Phi/A
+// from the magnetisation curve, corresponding the value of B
+H = 1000
+mmf = H*l
+printf("\n\nResult\n\n")
+printf("\n (a)mmf = %.0f A\n",mmf)
\ No newline at end of file diff --git a/608/CH7/EX7.09/7_09.sce b/608/CH7/EX7.09/7_09.sce new file mode 100755 index 000000000..c42a9b292 --- /dev/null +++ b/608/CH7/EX7.09/7_09.sce @@ -0,0 +1,15 @@ +//Problem 7.09: Determine the reluctance of a piece of mumetal of length 150 mm and cross-sectional area 1800 mm2 when the relative permeability is 4000. Find also the absolute permeability of the mumetal.
+
+//initializing the variables:
+A = 18E-4; // in m2
+l = 0.15; // in m
+u0 = 4*%pi*1E-7;
+ur = 4000;
+
+//calculation:
+S = l/(u0*ur*A)
+u = u0*ur
+
+printf("\n\nResult\n\n")
+printf("\n (a)Reluctance S = %.0f /H",S)
+printf("\n (b)Absolute permeability, μ = %.2E H/m\n",u)
\ No newline at end of file diff --git a/608/CH7/EX7.10/7_10.sce b/608/CH7/EX7.10/7_10.sce new file mode 100755 index 000000000..6932f5ec3 --- /dev/null +++ b/608/CH7/EX7.10/7_10.sce @@ -0,0 +1,18 @@ +//Problem 7.10: A mild steel ring has a radius of 50 mm and a crosssectional area of 400 mm2. A current of 0.5 A flows in a coil wound uniformly around the ring and the flux produced is 0.1 mWb. If the relative permeability at this value of current is 200 find (a) the reluctance of the mild steel and (b) the number of turns on the coil.
+
+//initializing the variables:
+A = 4E-4; // in m2
+r = 0.05; // in m
+I = 0.5; // in Amperes
+Phi = 0.1E-3; // in Wb
+u0 = 4*%pi*1E-7;
+ur = 200;
+
+//calculation:
+l = 2*%pi*r
+S = l/(u0*ur*A)
+N = S*Phi/I
+
+printf("\n\nResult\n\n")
+printf("\n (a)Reluctance S = %.2E /H",S)
+printf("\n (b)number of turns on the coil = %.0f ",N)
\ No newline at end of file diff --git a/608/CH7/EX7.11/7_11.sce b/608/CH7/EX7.11/7_11.sce new file mode 100755 index 000000000..b5f8a6664 --- /dev/null +++ b/608/CH7/EX7.11/7_11.sce @@ -0,0 +1,22 @@ +//Problem 7.11: A closed magnetic circuit of cast steel contains a 6 cm long path of cross-sectional area 1 cm2 and a 2 cm path of cross-sectional area 0.5 cm2. A coil of 200 turns is wound around the 6 cm length of the circuit and a current of 0.4 A flows. Determine the flux density in the 2 cm path, if the relative permeability of the cast steel is 750.
+
+//initializing the variables:
+A1 = 1E-4; // in m2
+A2 = 0.5E-4; // in m2
+l1 = 0.06; // in m
+l2 = 0.02; // in m
+N1 = 200; // no. of turns about 6 cm coil
+I = 0.4; // in Amperes
+u0 = 4*%pi*1E-7;
+ur = 750;
+
+//calculation:
+//Reluctance
+S1 = l1/(u0*ur*A1) // for 6 cm
+S2 = l2/(u0*ur*A2) // for 2 cm
+St = S1 + S2
+Phi = N1*I/St
+B2 = Phi/A2
+
+printf("\n\nResult\n\n")
+printf("\n flux density in the 2 cm path = %.2f T\n",B2)
\ No newline at end of file diff --git a/608/CH7/EX7.12/7_12.sce b/608/CH7/EX7.12/7_12.sce new file mode 100755 index 000000000..af23fee2a --- /dev/null +++ b/608/CH7/EX7.12/7_12.sce @@ -0,0 +1,22 @@ +//Problem 7.12: A silicon iron ring of cross-sectional area 5 cm2 has a radial air gap of 2 mm cut into it. If the mean length of the silicon iron path is 40 cm, calculate the magnetomotive force to produce a flux of 0.7 mWb.
+
+//initializing the variables:
+A = 5E-4; // in m2
+l = 0.4; // in m
+r = 2E-3; // in m
+u0 = 4*%pi*1E-7;
+Phi = 0.7E-3; // in Wb
+
+//calculation:
+//For the silicon iron:
+B = Phi/A
+//From the B–H curve for silicon iron, corresponding to value of B
+Hs = 1650
+mmfs = Hs*l
+//For the air gap:
+Ha = B/u0
+mmfa = Ha*r
+mmft = mmfs + mmfa
+
+printf("\n\nResult\n\n")
+printf("\n Total mmf to produce a flux of 0.7 mWb = %.0f A\n",mmft)
\ No newline at end of file diff --git a/608/CH7/EX7.13/7_13.sce b/608/CH7/EX7.13/7_13.sce new file mode 100755 index 000000000..08cbef41d --- /dev/null +++ b/608/CH7/EX7.13/7_13.sce @@ -0,0 +1,32 @@ +//Problem 7.13: Figure 7.4 shows a ring formed with two differentmaterials—cast steel and mild steel. The dimensions are:
+// mean length cross-sectional area
+// Mild steel 400 mm 500 mm2
+// Cast steel 300 mm 312.5 mm2
+//Find the total mmf required to cause a flux of 500 μWb in the magnetic circuit. Determine also the total circuit reluctance.
+
+//initializing the variables:
+A1 = 5E-4; // in m2
+A2 = 3.125E-4; // in m2
+l1 = 0.4; // in m
+l2 = 0.3; // in m
+u0 = 4*%pi*1E-7;
+Phi = 0.5E-3; // in Wb
+
+//calculation:
+//For the mild steel:
+B1 = Phi/A1
+//From the B–H curve for Mild steel, corresponding to value of B
+Hm = 1400
+mmfm = Hm*l1
+//For the cast steel:
+B2 = Phi/A2
+//From the B–H curve for cast steel steel, corresponding to value of B
+Hc = 4800
+mmfc = Hc*l2
+mmft = mmfm + mmfc
+//Reluctance
+S = mmft/Phi
+
+printf("\n\nResult\n\n")
+printf("\n Total mmf to produce a flux of 0.5 mWb = %.0f A",mmft)
+printf("\n Total circuit reluctance = %.2E /H",S)
\ No newline at end of file diff --git a/608/CH7/EX7.14/7_14.sce b/608/CH7/EX7.14/7_14.sce new file mode 100755 index 000000000..b3953c0ce --- /dev/null +++ b/608/CH7/EX7.14/7_14.sce @@ -0,0 +1,27 @@ +//Problem 7.14: A section through a magnetic circuit of uniform cross-sectional area 2 cm2 is shown in Figure 7.5. The cast steel core has a mean length of 25 cm. The air gap is 1 mm wide and the coil has 5000 turns. The B–H curve for cast steel is shown on page 78. Determine the current in the coil to produce a flux density of 0.80 T in the air gap, assuming that all the flux passes through both parts of the magnetic circuit.
+
+//initializing the variables:
+A = 2E-4; // in m2
+l1 = 0.25; // in m
+l2 = 0.001; // in m
+u0 = 4*%pi*1E-7;
+N = 5000; // no. of turns
+B = 0.8; // in tesla
+ua = 1; // for air
+
+//calculation:
+//for the core
+//From the B–H curve for Mild steel, corresponding to value of B = 0.8
+H = 750
+ur = B/(u0*H)
+S1 = l1/(u0*ur*A)
+//For the air gap:
+S2 = l2/(u0*ua*A)
+St = S1 + S2
+//flux
+Phi = B*A
+//current
+I = St*Phi/N
+
+printf("\n\nResult\n\n")
+printf("\n current = %.3f A",I)
\ No newline at end of file diff --git a/608/CH8/EX8.02/8_02.sce b/608/CH8/EX8.02/8_02.sce new file mode 100755 index 000000000..5bfaee267 --- /dev/null +++ b/608/CH8/EX8.02/8_02.sce @@ -0,0 +1,16 @@ +//Problem 8.02: A conductor carries a current of 20 A and is at rightangles to a magnetic field having a flux density of 0.9 T. If the length of the conductor in the field is 30 cm, calculate the force acting on the conductor. Determine also the value of the force if the conductor is inclined at an angle of 30° to the direction of the field.
+
+//initializing the variables:
+B = 0.9; // in tesla
+I = 20; // in Amperes
+l = 0.30; // in m
+alpha = 30; // in degree
+u0 = 4*%pi*1E-7;
+
+//calculation:
+F1 = B*I*l
+F2 = B*I*l*sin(alpha*%pi/180)
+
+printf("\n\nResult\n\n")
+printf("\n (a)Force when the conductor is at right angles to the field = %.1f N",F1)
+printf("\n (b)Force when the conductor is at 30° angle to the field = %.1f N",F2)
\ No newline at end of file diff --git a/608/CH8/EX8.03/8_03.sce b/608/CH8/EX8.03/8_03.sce new file mode 100755 index 000000000..79400d816 --- /dev/null +++ b/608/CH8/EX8.03/8_03.sce @@ -0,0 +1,13 @@ +//Problem 8.03: Determine the current required in a 400 mm length of conductor of an electric motor, when the conductor is situated at right-angles to a magnetic field of flux density 1.2 T, if a force of 1.92 N is to be exerted on the conductor.
+
+//initializing the variables:
+F = 1.92; // in newton
+B = 1.2; // in tesla
+l = 0.40; // in m
+u0 = 4*%pi*1E-7;
+
+//calculation:
+I = F/(B*l)
+
+printf("\n\nResult\n\n")
+printf("\n (a)Current I = %.0f Amperes(A)",I)
\ No newline at end of file diff --git a/608/CH8/EX8.04/8_04.sce b/608/CH8/EX8.04/8_04.sce new file mode 100755 index 000000000..2f910de6d --- /dev/null +++ b/608/CH8/EX8.04/8_04.sce @@ -0,0 +1,16 @@ +//Problem 8.04: A conductor 350 mm long carries a current of 10 A and is at right-angles to a magnetic field lying between two circular pole faces each of radius 60 mm. If the total flux between the pole faces is 0.5 mWb, calculate the magnitude of the force exerted on the conductor.
+
+//initializing the variables:
+r = 0.06; // in m
+I = 10; // in Amperes
+l = 0.35; // in m
+Phi = 0.5E-3; // in Wb
+u0 = 4*%pi*1E-7;
+
+//calculation:
+A = %pi*r*r
+B = Phi/A
+F = B*I*l
+
+printf("\n\nResult\n\n")
+printf("\n (a)Force F = %.3f N\n",F)
\ No newline at end of file diff --git a/608/CH8/EX8.06/8_06.sce b/608/CH8/EX8.06/8_06.sce new file mode 100755 index 000000000..697f7cb9c --- /dev/null +++ b/608/CH8/EX8.06/8_06.sce @@ -0,0 +1,20 @@ +//Problem 8.06: A coil is wound on a rectangular former of width 24 mm and length 30 mm. The former is pivoted about an axis passing through the middle of the two shorter sides and is placed in a uniform magnetic field of flux density 0.8 T, the axis being perpendicular to the field. If the coil carries a current of 50 mA, determine the force on each coil side (a) for a single-turn coil, (b) for a coil wound with 300 turns.
+
+//initializing the variables:
+N1 = 1; // for a single-turn coil
+N2 = 300; // no. of turns
+b = 0.024; // in m
+B = 0.8; // in Tesla
+I = 0.05; // in Amperes
+l = 0.030; // in m
+u0 = 4*%pi*1E-7;
+
+//calculation:
+//For a single-turn coil,
+F1 = N1*B*I*l
+//for a coil wound with 300 turns.
+F2 = N2*B*I*l
+
+printf("\n\nResult\n\n")
+printf("\n (a)For a single-turn coil, force on each coil side = %.4f N",F1)
+printf("\n (b)For a 300-turn coil, force on each coil side = %.2f N",F2)
\ No newline at end of file diff --git a/608/CH8/EX8.07/8_07.sce b/608/CH8/EX8.07/8_07.sce new file mode 100755 index 000000000..8e97411f5 --- /dev/null +++ b/608/CH8/EX8.07/8_07.sce @@ -0,0 +1,13 @@ +//Problem 8.07: An electron in a television tube has a charge of 1.6 ð 1019 coulombs and travels at 3 ð 107 m/s perpendicular to a field of flux density 18.5 μT. Determine the force exerted on the electron in the field.
+
+//initializing the variables:
+Q = 1.6E-19; // in Coulomb
+v = 3E7; // in m/s
+B = 18.5E-6; // in Tesla
+u0 = 4*%pi*1E-7;
+
+//calculation:
+F = Q*v*B
+
+printf("\n\nResult\n\n")
+printf("\n Force exerted on the electron in the field. = %.2E N",F)
\ No newline at end of file diff --git a/608/CH9/EX9.01/9_01.sce b/608/CH9/EX9.01/9_01.sce new file mode 100755 index 000000000..2f37a2687 --- /dev/null +++ b/608/CH9/EX9.01/9_01.sce @@ -0,0 +1,16 @@ +//Problem 9.01: A conductor 300 mm long moves at a uniform speed of 4 m/s at right-angles to a uniform magnetic field of flux density 1.25 T. Determine the current flowing in the conductor when (a) its ends are open-circuited, (b) its ends are connected to a load of 20 ohm resistance.
+
+//initializing the variables:
+l = 0.3; // in m
+v = 4; // in m/s
+B = 1.25; // in Tesla
+R = 20; // in ohms
+u0 = 4*%pi*1E-7;
+
+//calculation:
+E = B*l*v
+I2 = E/R
+
+printf("\n\nResult\n\n")
+printf("\n (a)If the ends of the conductor are open circuited no current will flow even though %.1f V has been induced",E)
+printf("\n (b)From Ohm’s law, I = %.3f Ampere",I2)
\ No newline at end of file diff --git a/608/CH9/EX9.02/9_02.sce b/608/CH9/EX9.02/9_02.sce new file mode 100755 index 000000000..358f53d6b --- /dev/null +++ b/608/CH9/EX9.02/9_02.sce @@ -0,0 +1,14 @@ +//Problem 9.02: At what velocity must a conductor 75 mm long cut a magnetic field of flux density 0.6 T if an e.m.f. of 9 V is to be induced in it? Assume the conductor, the field and the direction of motion are mutually perpendicular.
+
+//initializing the variables:
+l = 0.075; // in m
+E = 9; // in Volts
+B = 0.6; // in Tesla
+R = 20; // in ohms
+u0 = 4*%pi*1E-7;
+
+//calculation:
+v = E/(B*l)
+
+printf("\n\nResult\n\n")
+printf("\n velocity v = %.0f m/s\n",v)
\ No newline at end of file diff --git a/608/CH9/EX9.03/9_03.sce b/608/CH9/EX9.03/9_03.sce new file mode 100755 index 000000000..f936284e9 --- /dev/null +++ b/608/CH9/EX9.03/9_03.sce @@ -0,0 +1,22 @@ +//Problem 9.03: A conductor moves with a velocity of 15 m/s at an angle of (a) 90°, (b) 60° and (c) 30° to a magnetic field produced between two square-faced poles of side length 2 cm. If the flux leaving a pole face is 5 μWb, find the magnitude of the induced e.m.f. in each case.
+
+//initializing the variables:
+l = 0.02; // in m
+b = 0.02; // in m
+v = 15; // in m/s
+R = 20; // in ohms
+Phi = 5E-6; // in Wb
+u0 = 4*%pi*1E-7;
+a1 = 90; // in degrees
+a2 = 60; // in degrees
+a3 = 30; // in degrees
+
+//calculation:
+A = l*b
+B = Phi/A
+E90 = B*l*v*sin(a1*%pi/180)
+E60 = B*l*v*sin(a2*%pi/180)
+E30 = B*l*v*sin(a3*%pi/180)
+
+printf("\n\nResult\n\n")
+printf("\n Induced e.m.f. at angles 90°, 60°, 30° are %.2E V, %.2E V, %.3E V respectively\n",E90,E60,E30)
\ No newline at end of file diff --git a/608/CH9/EX9.04/9_04.sce b/608/CH9/EX9.04/9_04.sce new file mode 100755 index 000000000..6425a1165 --- /dev/null +++ b/608/CH9/EX9.04/9_04.sce @@ -0,0 +1,14 @@ +//Problem 9.04: The wing span of a metal aeroplane is 36 m. If the aeroplane is flying at 400 km/h, determine the e.m.f. induced between its wing tips. Assume the vertical component of the earth’s magnetic field is 40 μT
+
+//initializing the variables:
+s = 36; // in m
+v = 400; // in km/h
+u0 = 4*%pi*1E-7;
+B = 40E-6; // in Tesla
+
+//calculation:
+v0 = v*5/18
+E = B*s*v0
+
+printf("\n\nResult\n\n")
+printf("\n Induced e.m.f. = %.2f V\n",E)
\ No newline at end of file diff --git a/608/CH9/EX9.06/9_06.sce b/608/CH9/EX9.06/9_06.sce new file mode 100755 index 000000000..2631ee7b2 --- /dev/null +++ b/608/CH9/EX9.06/9_06.sce @@ -0,0 +1,13 @@ +//Problem 9.06: Determine the e.m.f. induced in a coil of 200 turns when there is a change of flux of 25 mWb linking with it in 50 ms
+
+//initializing the variables:
+N = 200; // no. of turns
+dt = 0.050; // change of time in sec
+u0 = 4*%pi*1E-7;
+dPhi = 0.025; // change of flux in Wb
+
+//calculation:
+E = -1*N*dPhi/dt
+
+printf("\n\nResult\n\n")
+printf("\n Induced e.m.f. = %.0f V\n",E)
\ No newline at end of file diff --git a/608/CH9/EX9.07/9_07.sce b/608/CH9/EX9.07/9_07.sce new file mode 100755 index 000000000..c7d94810a --- /dev/null +++ b/608/CH9/EX9.07/9_07.sce @@ -0,0 +1,14 @@ +//Problem 9.07: A flux of 400 μWb passing through a 150-turn coil is reversed in 40 ms. Find the average e.m.f. induced.
+
+//initializing the variables:
+N = 150; // no. of turns
+dt = 0.040; // change of time in sec
+u0 = 4*%pi*1E-7;
+dPhi = 800E-6; // change of flux in Wb
+
+//calculation:
+//Since the flux reverses, the flux changes from C400 μWb to 400 μWb, a total change of flux of 800 μWb
+E = -1*N*dPhi/dt
+
+printf("\n\nResult\n\n")
+printf("\n Induced e.m.f. = %.0f V\n",E)
\ No newline at end of file diff --git a/608/CH9/EX9.08/9_08.sce b/608/CH9/EX9.08/9_08.sce new file mode 100755 index 000000000..0da1613ea --- /dev/null +++ b/608/CH9/EX9.08/9_08.sce @@ -0,0 +1,12 @@ +//Problem 9.08: Calculate the e.m.f. induced in a coil of inductance 12 H by a current changing at the rate of 4 A/s
+
+//initializing the variables:
+L = 12; // in Henry
+u0 = 4*%pi*1E-7;
+dIdt = 4; // change of current with change in time in A/s
+
+//calculation:
+E = -1*L*dIdt
+
+printf("\n\nResult\n\n")
+printf("\n Induced e.m.f. = %.0f V\n",E)
\ No newline at end of file diff --git a/608/CH9/EX9.09/9_09.sce b/608/CH9/EX9.09/9_09.sce new file mode 100755 index 000000000..90a6252fd --- /dev/null +++ b/608/CH9/EX9.09/9_09.sce @@ -0,0 +1,12 @@ +//Problem 9.09: An e.m.f. of 1.5 kV is induced in a coil when a current of 4 A reduced to 0 A in a 8 ms. find the inductance of the coil.
+
+//initializing the variables:
+E = 1500; // in Volts
+dt = 0.008; // Change of time in sec
+dI = 4; // change of current in A/s
+
+//calculation:
+L = abs(E)*dt/dI
+
+printf("\n\n Result \n\n")
+printf("\n Inductance L = %.0f H\n",L)
\ No newline at end of file diff --git a/608/CH9/EX9.10/9_10.sce b/608/CH9/EX9.10/9_10.sce new file mode 100755 index 000000000..1139ee923 --- /dev/null +++ b/608/CH9/EX9.10/9_10.sce @@ -0,0 +1,11 @@ +//Problem 9.10: An 8 H inductor has a current of 3 A flowing through it. How much energy is stored in the magnetic field of the inductor?
+
+//initializing the variables:
+L = 8; // in Henry
+I = 3; // in Amperes
+
+//calculation:
+W = L*I*I/2
+
+printf("\n\n Result \n\n")
+printf("\n Energy stored, W = %.0f J\n",W)
\ No newline at end of file diff --git a/608/CH9/EX9.11/9_11.sce b/608/CH9/EX9.11/9_11.sce new file mode 100755 index 000000000..691199ecd --- /dev/null +++ b/608/CH9/EX9.11/9_11.sce @@ -0,0 +1,12 @@ +//Problem 9.11: Calculate the coil inductance when a current of 4 A in a coil of 800 turns produces a flux of 5 mWb linking with the coil.
+
+//initializing the variables:
+I = 4; // in Amperes
+N = 800; //turns
+Phi = 0.005; // in Wb
+
+//calculation:
+L = N*Phi/I
+
+printf("\n\n Result \n\n")
+printf("\n Inductance L = %.0f H\n",L)
\ No newline at end of file diff --git a/608/CH9/EX9.12/9_12.sce b/608/CH9/EX9.12/9_12.sce new file mode 100755 index 000000000..00aec320e --- /dev/null +++ b/608/CH9/EX9.12/9_12.sce @@ -0,0 +1,19 @@ +//Problem 9.12: A flux of 25 mWb links with a 1500 turn coil when a current of 3 A passes through the coil. Calculate (a) the inductance of the coil, (b) the energy stored in the magnetic field, and (c) the average e.m.f. induced if the current falls to zero in 150 ms.
+
+//initializing the variables:
+I1 = 3; // in Amperes
+I2 = 0; // in Amperes
+dt = 0.150; // in secs
+N = 1500; //turns
+Phi = 0.025; // in Wb
+
+//calculation:
+L = N*Phi/I1
+W = L*I1*I1/2
+dI = I1 - I2
+E = -1*L*dI/dt
+
+printf("\n\n Result \n\n")
+printf("\n (a)Inductance L = %.1f H",L)
+printf("\n (b)energy stored W = %.2f J",W)
+printf("\n (c)e.m.f. induced = %.0f V",E)
\ No newline at end of file diff --git a/608/CH9/EX9.13/9_13.sce b/608/CH9/EX9.13/9_13.sce new file mode 100755 index 000000000..cfcbbcbfa --- /dev/null +++ b/608/CH9/EX9.13/9_13.sce @@ -0,0 +1,17 @@ +//Problem 9.13: A 750 turn coil of inductance 3 H carries a current of 2 A. Calculate the flux linking the coil and the e.m.f. induced in the coil when the current collapses to zero in 20 ms
+
+//initializing the variables:
+I1 = 2; // in Amperes
+I2 = 0; // in Amperes
+dt = 0.020; // in secs
+N = 750; //turns
+L = 3; // in Henry
+
+//calculation:
+Phi = L*I1/N
+dI = I1 - I2
+E = -1*L*dI/dt
+
+printf("\n\n Result \n\n")
+printf("\n (a)Flux = %.3f Wb",Phi)
+printf("\n (b)e.m.f. induced = %.0f V",E)
\ No newline at end of file diff --git a/608/CH9/EX9.14/9_14.sce b/608/CH9/EX9.14/9_14.sce new file mode 100755 index 000000000..5cd2fdfe1 --- /dev/null +++ b/608/CH9/EX9.14/9_14.sce @@ -0,0 +1,12 @@ +//Problem 9.14: Calculate the mutual inductance between two coils when a current changing at 200 A/s in one coil induces an e.m.f. of 1.5 V in the other.
+
+//initializing the variables:
+dI1dt = 200; // change of current with change in time in A/s
+N = 2; // no. of coils
+E2 = 1.5; // in Volts
+
+//calculation:
+M = abs(E2)/dI1dt
+
+printf("\n\n Result \n\n")
+printf("\n mutual inductance, M = %.4f H\n", M)
\ No newline at end of file diff --git a/608/CH9/EX9.15/9_15.sce b/608/CH9/EX9.15/9_15.sce new file mode 100755 index 000000000..f4599dcb7 --- /dev/null +++ b/608/CH9/EX9.15/9_15.sce @@ -0,0 +1,12 @@ +//Problem 9.15: The mutual inductance between two coils is 18 mH. Calculate the steady rate of change of current in one coil to induce an e.m.f. of 0.72 V in the other.
+
+//initializing the variables:
+M = 0.018; // in Henry
+N = 2; // no. of coils
+E2 = 0.72; // in Volts
+
+//calculation:
+dI1dt = abs(E2)/M
+
+printf("\n\n Result \n\n")
+printf("\n rate of change of current dI1/dt = %.0f A/s\n", dI1dt)
\ No newline at end of file diff --git a/608/CH9/EX9.16/9_16.sce b/608/CH9/EX9.16/9_16.sce new file mode 100755 index 000000000..d6937679f --- /dev/null +++ b/608/CH9/EX9.16/9_16.sce @@ -0,0 +1,17 @@ +//Problem 9.16: Two coils have a mutual inductance of 0.2 H. If the current in one coil is changed from 10 A to 4 A in 10 ms, calculate (a) the average induced e.m.f. in the second coil, (b) the change of flux linked with the second coil if it is wound with 500 turns.
+
+//initializing the variables:
+M = 0.2; // in Henry
+I1 = 10; // in Amperes
+I2 = 4; // in Amperes
+dt = 0.010; // in secs
+N = 500; // turns
+
+//calculation:
+dI1dt = (I1 -I2)/dt
+E2 = -1*dI1dt*M
+dPhi = abs(E2)*dt/N
+
+printf("\n\n Result \n\n")
+printf("\n (a)Induced e.m.f. E2 = %.0f V", E2)
+printf("\n (b)change of flux = %.4f Wb", dPhi)
\ No newline at end of file |
