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+//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)
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