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Diffstat (limited to '608/CH39/EX39.02/39_02.sce')
-rwxr-xr-x | 608/CH39/EX39.02/39_02.sce | 17 |
1 files changed, 17 insertions, 0 deletions
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)
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