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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/CH43/EX43.13 | |
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Diffstat (limited to '608/CH43/EX43.13')
-rwxr-xr-x | 608/CH43/EX43.13/43_13.sce | 26 |
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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)) |