Under which of the following conditions the power delivered by the source of circuit will be maximum
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \omega L = (\omega C)^{-1} \)
In a series L-C-R circuit the power delivered by the source is \( P = V_{\mathrm{rms}}\,I_{\mathrm{rms}}\cos\phi \), which can be written as
\( \displaystyle P = \frac{V_{\mathrm{rms}}^{2}\,R}{Z^{2}} = \frac{V_{\mathrm{rms}}^{2}\,R}{R^{2} + (\omega L - 1/\omega C)^{2}} \)
For a given source and resistance, \( P \) is maximum when the impedance is least, \( Z = R \), which needs \( \omega L - 1/\omega C = 0 \). This is resonance, with power factor \( \cos\phi = 1 \):
\( \displaystyle \omega L = \frac{1}{\omega C} = (\omega C)^{-1} \)
Note: The official NTA Answer Key accepts both Option (3) and Option (4), because Option (4), printed as \( (\omega L)^{-1} = \omega C \), expresses the same resonance condition as Option (3). Option (4) has been changed to \( (\omega L)^{-1} = \omega C^{2} \), for the benefit of students and to ensure a single answer.
NCERT (2026-27):
Resonance – “… This frequency is called the resonant frequency:
\( \displaystyle X_c = X_L \quad \text{or} \quad \frac{1}{\omega_0 C} = \omega_0 L \quad \text{or} \quad \omega_0 = \frac{1}{\sqrt{LC}} \qquad (7.28) \)”
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 = I2R. That is, maximum power is dissipated in a circuit (through R) at resonance.”