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Determine using superposition theorem, the voltage across the 4 ohm resistor shown in Fig.

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Superposition theorem needs one source acting at a time.

Step I : De-acting current source.

The circuit is redrawn after this change in Fig.

I1 = 10/(2 + (4x(8 + 2))/(4 + (8 + 2))) = 10/(2 + 40/14) = 2.059amp

I2 = (2.059 x 10)/14 = 1.471 amp,  in downward direction

Step II : De-activate the voltage source. The circuit is redrawn after the change, in Fig.  With the currents marked as shown. Id = 2Ic relating the voltage drops in Loop ADC.

Thus Ib = 3 Ic.

Resistance of parallel combination of 2 and 4 ohms = (2 x 4)/(2 + 4) = 1.333 Ω

Resistance for flow of Ib = 8 + 1.333 = 9.333 Ω

The 5-amp current from the sources gets divided into Ib (= 3 Ic) and Ia, at the node F.

Ib = 3Ic = 2.0/(2.0 + 9.333) x 5 = 0.8824

∴ Ic = 0.294 amp, in downward direction. 

Step III. Apply superposition theorem, for finding the total current into the 4-ohm resistor 

= Current due to Current source + Current due to Voltage source 

= 0.294 + 1.471 = 1.765 amp in downward direction. 

Check. In the branch AD, 

The voltage source drives a current from A to D of 2.059 amp, and the current source drives a current of Id (= 2Ic) which is 0.588 amp, from D to A. 

The net current in branch AD 

= 2.059 − 0.588 = 1.471 amp ...eqn. (a) 

With respect to O, A is at a potential of + 10 volts. 

Potential of D with respect to O 

= (net current in resistor) × 4 

= 1.765 × 4 = + 7.06 volts 

Between A and D, the potential difference is (10 − 7.06) volts 

Hence, the current through this branch 

= (10 - 7.06)/2 = 1.47 amp from A to D − = ...eqn (b) 

This is the same as eqn. (a) and hence checks the result, obtained previously.

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