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For a series LCR circuit the power loss at resonance is : -
A
$\dfrac{V^2}{\left[\omega L - \dfrac{1}{\omega C}\right]}$
B
$I^2L\omega$
C
$I^2R$
D
$\dfrac{V^2}{C\omega}$
Detailed Solution
Average power in an AC circuit: $P = V_{rms}I_{rms}\cos\phi$
At resonance $X_L = X_C$, so the impedance Z = R and the phase angle $\phi = 0$ ($\cos\phi = 1$).
$P = VI = I^2R$ (since V = IR at resonance)
Power is dissipated only in the resistor; inductors and capacitors do not consume average power.
So the power loss at resonance is $I^2R$.
At resonance $X_L = X_C$, so the impedance Z = R and the phase angle $\phi = 0$ ($\cos\phi = 1$).
$P = VI = I^2R$ (since V = IR at resonance)
Power is dissipated only in the resistor; inductors and capacitors do not consume average power.
So the power loss at resonance is $I^2R$.