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.