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) → $\frac{\sqrt{2}}{3}A$

Given:

$V = 10 \cos(500t)$, so peak voltage $V_0 = 10 \ \text{V}$

RMS voltage: $V_{rms} = \frac{V_0}{\sqrt{2}} = \frac{10}{\sqrt{2}} = 5\sqrt{2} \ \text{V}$

Angular frequency: $\omega = 500 \ \text{rad/s}$

Inductance: $L = 10 \ \text{mH} = 10 \times 10^{-3} \ \text{H}$

Resistance of coil: $R_L = 5 \ \Omega$

Capacitance: $C = 400 \ \mu F = 400 \times 10^{-6} \ \text{F}$

Series resistor: $R = 10 \ \Omega$

Inductive reactance:

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

Capacitive reactance:

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

Since $X_L = X_C$, LC is at resonance, so net reactance cancels and only $R_L = 5 \ \Omega$ remains in that branch.

Thus, parallel combination: $10 \ \Omega$ in one branch and $5 \ \Omega$ in other branch.

Equivalent resistance:

$R_{eq} = \frac{10 \times 5}{10 + 5} = \frac{50}{15} = \frac{10}{3} \ \Omega$

Current in circuit:

$I_{rms} = \frac{V_{rms}}{R_{eq}} = \frac{5\sqrt{2}}{\frac{10}{3}} = \frac{15\sqrt{2}}{10} = \frac{\sqrt{2}}{3} \ \text{A}$

Answer: Ammeter reading = $\frac{\sqrt{2}}{3} \ \text{A}$