Question

In the A.C. circuit given below, voltmeters $ V_1 $ and $ V_2 $ read 100 V each. Find the reading of the voltmeter $ V_3 $ and the ammeter A.

• A. \( 220 \, \text{V}, \, 2 \, \text{A} \)
• B. \( 110 \, \text{V}, \, 2 \, \text{A} \)
• C. \( 110 \, \text{V}, \, 4 \, \text{A} \)
• D. \( 220 \, \text{V}, \, 1 \, \text{A} \)

Hint:

In circuits with both capacitive and inductive reactance, the total impedance is calculated using \( Z = \sqrt{R^2 + (X_L - X_C)^2} \). Always check the correct formula for the reactances and impedance before solving.

Answer 1

The Correct Option is A

In an A.C. circuit involving inductance, capacitance, and resistance, we calculate the readings of the voltmeter and ammeter using the concept of impedance. The voltmeter \( V_1 \) and \( V_2 \) both measure the voltage across individual components of the circuit, while the ammeter measures the current in the circuit. 
Given: 
- \( L = 20 \, \text{mH} \) 
- \( C = 50 \, \mu \text{F} \) - \( R = 110 \, \Omega \) 
- Frequency \( f = 50 \, \text{Hz} \) 
The total impedance \( Z \) in the circuit can be calculated using: \[ Z = \sqrt{R^2 + \left( X_L - X_C \right)^2} \] Where: - \( X_L = 2\pi f L \) is the inductive reactance - \( X_C = \frac{1}{2\pi f C} \) is the capacitive reactance Now, calculate: - \( X_L = 2\pi \times 50 \times 20 \times 10^{-3} = 6.28 \, \Omega \) - \( X_C = \frac{1}{2\pi \times 50 \times 50 \times 10^{-6}} = 63.66 \, \Omega \) The total impedance \( Z \) is: \[ Z = \sqrt{110^2 + (6.28 - 63.66)^2} = \sqrt{110^2 + (-57.38)^2} = \sqrt{12100 + 3291.6} = 116.47 \, \Omega \] Now, calculate the total current \( I \) using Ohm's Law: \[ I = \frac{V_{\text{total}}}{Z} = \frac{220}{116.47} \approx 2 \, \text{A} \] 
Thus, the current through the circuit is 2 A. Now, the reading of \( V_3 \) is approximately the total voltage drop across the impedance, which is the applied voltage of \( 220 \, \text{V} \).