initial commit / add all books

17 files changed, 388 insertions, 0 deletions

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
new 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