Step 1:

The total circuit current is .

Circuit voltage is .

Frequency of the circuit is 50 Hz.

Find the total circuit impedance .

Impedance of the circuit is .

Step 2:

Find the equivalent branch impedance of branch B.

Resistor R₂ and capacitor C are in series.

Impedance offered by capacitor is .

The equivalent branch impedance of branch B :

.

.

Step 3:

Find the equivalent branch impedance of branch A.

Branch A and branch B are in parallel.

Total Impedance of the circuit is .

So the branch impedance of branch A is .

Step 4:

Find the branch current I_A.

Use the current division rule : .

Therefore the branch current I_A = 3.0468 A.

Step 1:

Find the current through 12 resistor.

Super Position Theorem :

Take one voltage source at a time and replace all other with short or internal resistance.

Equivalent resistance :

5 resistor is in parallel with 12 resistor.

Resistance R₁ is in series with 7 resistor.

3 resistor is in parallel with 10.5294 resistor.

Resistance R₃ is in series with 4 resistor.

Equivalent resistance is .

Step 2:

Apply Kirchoff law for the loop 1.

Apply Kirchoff law for the loop 2.

Apply Kirchoff law for the loop 3.

Solve equations (1), (2) and (3).

, and .

Current across resistor 12 is .

Current across resistor 12 is A.

Step 2:

Super Position Theorem :

Take one voltage source at a time and replace all other with short or internal resistance.

Apply Kirchoff law for the loop 1.

Apply Kirchoff law for the loop 2.

Apply Kirchoff law for the loop 3.

Solve equations (1), (2) and (3).

, and .

Current across resistor 12 is .

Current across resistor 12 is A. (Negative sign indicates opposite direction)

Step 3:

Current through Load resistor 12 =

             = Current produced by 12 resistor due to 5 V + Current produced by 12 resistor due to 10V

             = 0.4658-0.233

             = 0.2328

Current through 12 resistor is 0.2328 A.

Solution :

Current through 12 resistor is 0.2328 A.