In the circuit shown in the figure, the ac source gives a voltage of $V = 10 \cos (500t)$. The ammeter reading will be:

$\frac{\sqrt{2}}{3}A$

The correct answer is Option (3) → $\frac{\sqrt{2}}{3}A$

• $V = 10 \cos(500t)$
• Peak voltage, $V_0 = 10 \text{ V}$
• Angular frequency, $\omega = 500 \text{ rad/s}$

Circuit components:

• Resistance $R = 10 \Omega$
• Inductor $L = 10 \text{ mH} = 10 \times 10^{-3} \text{ H}$
• Internal resistance of coil $= 5 \Omega$
• Capacitor $C = 400 \mu\text{F} = 400 \times 10^{-6} \text{ F}$

Inductive reactance:

$X_L = \omega L$

$X_L = 500 \times 10 \times 10^{-3} = 5 \Omega$

Capacitive reactance:

$X_C = \frac{1}{\omega C}$

$X_C = \frac{1}{500 \times 400 \times 10^{-6}} = 5 \Omega$

Since, $X_L = X_C$ , the circuit is in resonance, so net reactance becomes zero.

Total resistance in series: $R_{total} = 10 + 5 = 15 \Omega$

RMS voltage:$V_{rms} = \frac{V_0}{\sqrt{2}} = \frac{10}{\sqrt{2}}$

Current: $I_{rms} = \frac{V_{rms}}{R_{total}}$

$I_{rms} = \frac{10 / \sqrt{2}}{15}$

$I_{rms} = \frac{\sqrt{2}}{3} \text{ A}$