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: Read off peak voltage and angular frequency.
From \(V = 135\sqrt{2}\sin 100t\), the amplitude is \(V_0 = 135\sqrt{2}\) V and \(\omega = 100\) rad/s.

Step 2: Impedance from the given reactances.
Here \(X_L = X_C = 4\,\Omega\), so the inductive and capacitive effects cancel. The only opposition left is the resistance, therefore \(Z = R = 45\,\Omega\). Because \(X_L = X_C\), the circuit satisfies the resonance condition and the current is maximum.

Step 3: Peak current, then rms.
\(I_0 = \dfrac{V_0}{Z} = \dfrac{135\sqrt{2}}{45} = 3\sqrt{2}\) A.
The ammeter records the rms value: \(I_{rms} = \dfrac{I_0}{\sqrt{2}} = \dfrac{3\sqrt{2}}{\sqrt{2}} = 3\) A.

Step 4: Voltage across L and across C.
\(V_L = I_{rms}X_L = 3 \times 4 = 12\) V (leads the current by \(90^\circ\)).
\(V_C = I_{rms}X_C = 3 \times 4 = 12\) V (lags the current by \(90^\circ\)).

Step 5: Net voltmeter reading.
Since \(V_L\) and \(V_C\) are equal in size but opposite in phase, their phasor sum is \(|V_L - V_C| = |12 - 12| = 0\) V. The voltmeter therefore reads \(0\) V, confirming resonance.

\[\boxed{\text{Ammeter} = 3\ \text{A},\ \text{Voltmeter} = 0\ \text{V},\ \text{resonance: yes}}\]