Step 1: Determine total impedance in the circuit
The given circuit consists of a resistor \( R = 30\Omega \) in series with a parallel LC circuit, where the inductor reactance is \( X_L = 25\Omega \) and the capacitor reactance is \( X_C = 25\Omega \).
Since \( X_L = X_C \), the net reactance of the LC branch is:
\[
X_{\text{net}} = X_L - X_C = 25\Omega - 25\Omega = 0\Omega.
\]
Thus, the LC branch acts as a short circuit, meaning the impedance of the parallel combination is zero. Step 2: Calculate total circuit current
The only effective impedance in the circuit is the resistor \( R = 30\Omega \), so the total impedance \( Z \) is:
\[
Z = R = 30\Omega.
\]
Using Ohm’s law, the current through the circuit is:
\[
I = \frac{V}{Z} = \frac{240V}{30\Omega} = 8A.
\]
Step 3: Determine the voltmeter reading
Since the LC branch is a short circuit, the voltage across the parallel LC circuit (voltmeter reading) is:
\[
V_{\text{meter}} = 0V.
\]
Final Answer: The ammeter reads \( 8A \) and the voltmeter reads \( 0V \), which matches option (1).
