$V=200\sin (400\, t)$ volt a.c. source is connected to a series combination of an inductance of 20 mH and a resistance of 6 Ω. What will be the rms current in the a.c. circuit?

14.14 A

The correct answer is Option (3) → 14.14 A

Given:

Voltage: $v = 200 \sin(400t)\ \text{V}$

Resistance: $R = 6\ \Omega$

Inductance: $L = 20\ \text{mH} = 0.02\ \text{H}$

Angular frequency: $\omega = 400\ \text{rad/s}$

Inductive reactance: $X_L = \omega L = 400 \cdot 0.02 = 8\ \Omega$

Impedance of series R-L circuit: $Z = \sqrt{R^2 + X_L^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10\ \Omega$

Peak voltage: $V_0 = 200\ \text{V}$

RMS voltage: $V_{\text{rms}} = \frac{V_0}{\sqrt{2}} = \frac{200}{\sqrt{2}} \approx 141.42\ \text{V}$

RMS current: $I_{\text{rms}} = \frac{V_{\text{rms}}}{Z} = \frac{141.42}{10} \approx 14.14\ \text{A}$

∴ RMS current in the circuit = 14.14 A