Question:
In a series LCR circuit connected to an ac source of voltage $V$ and frequency $\omega$, the power loss at resonance is:
In a series LCR circuit connected to an ac source of voltage $V$ and frequency $\omega$, the power loss at resonance is:
Show Hint
At resonance the reactances cancel, so Z equals R and the current is V/R. Only the resistor uses power, giving V squared over R.
Updated On: Oct 21, 2026
- $I^2\omega L$
- $V^2/(\omega C)$
- $V^2/R$
- $V^2/\left(\omega L - \dfrac{1}{\omega C}\right)$
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept:
In an ac circuit the average power is used up only in the resistance. The coil and capacitor store energy and give it back, so they use up no average power.
Step 2: Key Formula or Approach:
Average power: \(P=V_{rms}I_{rms}\cos\phi\), where \(\phi\) is the phase angle between the voltage and the current.
At resonance, \(\omega L=\dfrac{1}{\omega C}\). So the reactances cancel and the impedance is only \(R\).
Step 3: Detailed Explanation:
At resonance \(Z=R\), so the current is \(I=\dfrac{V}{R}\) and the phase angle is \(\phi=0\), which gives \(\cos\phi=1\).
\[ P=VI\cos\phi=V\cdot\frac{V}{R}\cdot1=\frac{V^{2}}{R} \]
This is also \(I^{2}R\).
Step 4: Why the other options are wrong:
(A) \(I^{2}\omega L\) is the reactive quantity of the coil. It is not power loss, since an ideal coil does not use up energy.
(B) \(V^{2}/\omega C\) has the capacitor reactance in it, and the capacitor also does not use up energy.
(D) \(V^{2}/(\omega L-1/\omega C)\) would become infinite at resonance since the denominator is zero. That is not possible.
Final Answer:
The power loss at resonance is \(V^{2}/R\), option (C). \[ \boxed{P=\frac{V^{2}}{R}} \]
In an ac circuit the average power is used up only in the resistance. The coil and capacitor store energy and give it back, so they use up no average power.
Step 2: Key Formula or Approach:
Average power: \(P=V_{rms}I_{rms}\cos\phi\), where \(\phi\) is the phase angle between the voltage and the current.
At resonance, \(\omega L=\dfrac{1}{\omega C}\). So the reactances cancel and the impedance is only \(R\).
Step 3: Detailed Explanation:
At resonance \(Z=R\), so the current is \(I=\dfrac{V}{R}\) and the phase angle is \(\phi=0\), which gives \(\cos\phi=1\).
\[ P=VI\cos\phi=V\cdot\frac{V}{R}\cdot1=\frac{V^{2}}{R} \]
This is also \(I^{2}R\).
Step 4: Why the other options are wrong:
(A) \(I^{2}\omega L\) is the reactive quantity of the coil. It is not power loss, since an ideal coil does not use up energy.
(B) \(V^{2}/\omega C\) has the capacitor reactance in it, and the capacitor also does not use up energy.
(D) \(V^{2}/(\omega L-1/\omega C)\) would become infinite at resonance since the denominator is zero. That is not possible.
Final Answer:
The power loss at resonance is \(V^{2}/R\), option (C). \[ \boxed{P=\frac{V^{2}}{R}} \]
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