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| author | priyanka | 2015-06-24 15:03:17 +0530 |
|---|---|---|
| committer | priyanka | 2015-06-24 15:03:17 +0530 |
| commit | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (patch) | |
| tree | ab291cffc65280e58ac82470ba63fbcca7805165 /608/CH20/EX20.23/20_23.sce | |
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initial commit / add all books
Diffstat (limited to '608/CH20/EX20.23/20_23.sce')
| -rwxr-xr-x | 608/CH20/EX20.23/20_23.sce | 27 |
1 files changed, 27 insertions, 0 deletions
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)
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