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author | priyanka | 2015-06-24 15:03:17 +0530 |
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committer | priyanka | 2015-06-24 15:03:17 +0530 |
commit | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (patch) | |
tree | ab291cffc65280e58ac82470ba63fbcca7805165 /608/CH26/EX26.07 | |
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initial commit / add all books
Diffstat (limited to '608/CH26/EX26.07')
-rwxr-xr-x | 608/CH26/EX26.07/26_07.sce | 37 |
1 files changed, 37 insertions, 0 deletions
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)
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