Question

Find the readings of the voltmeter and ammeter in the given AC circuit. Is the circuit in the state of resonance?
Given: \( V = 135\sqrt{2}\,\sin 100t \) volt, \( R = 45\,\Omega \), \( X_L = 4\,\Omega \), \( X_C = 4\,\Omega \). The voltmeter is connected across the series L-C combination and the ammeter is in series with the circuit.

Hint:

Since \( X_L = X_C \), the reactances cancel, so \( Z = R \) (resonance). Find \( I = V_{rms}/R \) for the ammeter; the voltmeter across the L-C part reads \( I(X_L - X_C) = 0 \) because \( V_L \) and \( V_C \) cancel.

Answer 1

Step 1: rms value of the source voltage.
The source is \( V = 135\sqrt{2}\,\sin 100t \). Comparing with \( V = V_0 \sin \omega t \), the peak voltage is \( V_0 = 135\sqrt{2} \) V and \( \omega = 100 \) rad/s.
rms voltage \( V_{rms} = \dfrac{V_0}{\sqrt{2}} = \dfrac{135\sqrt{2}}{\sqrt{2}} = 135 \) V.

Step 2: Compare the reactances.
Inductive reactance \( X_L = 4\,\Omega \) and capacitive reactance \( X_C = 4\,\Omega \). Since \( X_L = X_C \), the net reactance \( X_L - X_C = 0 \).

Step 3: Impedance of the circuit.
Formula: \( Z = \sqrt{R^2 + (X_L - X_C)^2} \).
Substitution: \( Z = \sqrt{45^2 + 0^2} = 45\,\Omega \).

Step 4: Condition of resonance.
At resonance \( X_L = X_C \), which is exactly satisfied here. Hence the circuit IS in the state of resonance and the impedance is minimum (\( Z = R \)).

Step 5: Ammeter reading (rms current).
Formula: \( I_{rms} = \dfrac{V_{rms}}{Z} \).
Substitution: \( I_{rms} = \dfrac{135}{45} = 3 \) A.

Step 6: Voltmeter reading.
The voltmeter is across the series L-C combination.
\( V_L = I X_L = 3 \times 4 = 12 \) V and \( V_C = I X_C = 3 \times 4 = 12 \) V.
In a series circuit \( V_L \) and \( V_C \) are exactly \( 180^\circ \) out of phase, so the voltmeter reads \( V_{LC} = I(X_L - X_C) = 3(4-4) = 0 \) V. This zero reading across the L-C section is a signature of resonance.

\[\boxed{I_{ammeter} = 3\ \text{A}, \quad V_{voltmeter} = 0\ \text{V}, \quad \text{Yes, at resonance}}\]