Question

In the circuit shown in the figure, neglecting the source resistance, the voltmeter and ammeter readings respectively are:

• A. \(0 V, 8 A\)
• B. \(150 V, 3 A\)
• C. \(150 V, 6 A\)
• D. \(0 V, 3 A\)

Hint:

For parallel LC circuits with \( X_L = X_C \), the circuit behaves as a short circuit, reducing the impedance of the LC branch to zero.

Answer 1

The Correct Option is A

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).