A series LCR circuit has \(R=25\ \Omega\), \(L=0.5\ \mathrm{H}\) and \(C=50\ \mu\mathrm{F}\). It is joined to a variable-frequency 100 V (rms) ac source. The frequency is set so that the circuit is in resonance. The average power used in the circuit is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(400\ \mathrm{W}\)
At resonance \( X_L=X_C \). The impedance is then only \( R \), and the power factor is 1. The values of \( L \) and \( C \) decide the resonant frequency. They are not needed to find the power.
Given:
\( V=100\ \mathrm{V} \)
\( R=25\ \Omega \)
\( \begin{aligned} I &= \frac{V}{R} \\ &= \frac{100}{25} \\ &= 4\ \mathrm{A} \end{aligned} \)
\( \begin{aligned} P &= I^{2}R \\ &= 4^{2}\times25 \\ &= 400\ \mathrm{W} \end{aligned} \)
NCERT (2026-27):
Power in AC Circuit: the Power Factor – “Case (iv) Power dissipated at resonance in LCR circuit: At resonance Xc – XL= 0, and φ = 0. Therefore, cosφ = 1 and P = I2Z = I2 R. That is, maximum power is dissipated in a circuit (through R) at resonance.”
AC Voltage Applied to a Series LCR Circuit – “At resonant frequency, the current amplitude is maximum; im = vm/R.”