Question

Figure shows a two branched parallel circuit with

$R_A = 20\Omega$, $X_L = 15 \Omega$, $R_B = 6\Omega$ and $X_C = 8\Omega$.

The total current supplied by alternating source V is $I = \sqrt{29}$ A.

15. Current passing through the inductor ($X_L$) is :-

• A. Zero
• B. 4A
• C. 2 A
• D. 5 A

Answer 1

Answer: (C)

2 A

Answer 2

The circuit consists of two parallel branches.

Branch 1: $R_A = 20\Omega$ and $X_L = 15\Omega$. Impedance $Z_1 = 20 + j15$, and $|Z_1| = \sqrt{20^2 + 15^2} = 25\Omega$.

Branch 2: $R_B = 6\Omega$ and $X_C = 8\Omega$. Impedance $Z_2 = 6 - j8$, and $|Z_2| = \sqrt{6^2 + (-8)^2} = 10\Omega$.

Total impedance $|Z_{total}| = \frac{50\sqrt{29}}{29}\Omega$.

Voltage across the parallel circuit $V = I |Z_{total}| = \sqrt{29} \times \frac{50\sqrt{29}}{29} = 50$ V.

Current through the inductor is the current through branch 1: $|I_1| = \frac{V}{|Z_1|} = \frac{50}{25} = 2$ A.