Question

What will be the reading in the voltmeter and ammeter of the circuit shown ?

• A. 90V, 2A
• B. 0V, 2A
• C. 90V, 1A
• D. 0V, 1A

Answer 1

The Correct Option is B

1. Impedance of the circuit:

The impedance ($Z$) of a series RLC circuit is given by:

$Z = \sqrt{R^2 + (X_L - X_C)^2}$

In this case, $X_L = X_C = 4\,\Omega$, so the impedance is:

$Z = \sqrt{(45\,\Omega)^2 + (4\,\Omega - 4\,\Omega)^2} = \sqrt{(45\,\Omega)^2} = 45\,\Omega$

2. Current in the circuit:

The current ($I$) in the circuit is given by Ohm's law for AC circuits:

$I = \frac{V}{Z}$

where $V$ is the source voltage. Substituting the given values:

$I = \frac{90\,\text{V}}{45\,\Omega} = 2\,\text{A}$

Since the ammeter is connected in series with the circuit, it will read this current.

3. Voltage across the voltmeter:

The voltmeter is connected across the inductor and capacitor. Since $X_L$ and $X_C$ are equal, the voltage across the inductor ($V_L = IX_L$) and the voltage across the capacitor ($V_C = IX_C$) are equal in magnitude but opposite in phase. Therefore, the net voltage across the inductor and capacitor combination is:

$V_{LC} = V_L - V_C = IX_L - IX_C = I(X_L - X_C) = 2\,\text{A}(4\,\Omega - 4\,\Omega) = 0\,\text{V}$

Thus, the voltmeter will read 0 V.

The correct answer is (B) 0V, 2A.