In a series LCR circuit \( R = 6\ \Omega \), \( L = 25 \) mH and \( C = 1.6\ \mu \)F. The applied voltage is represented as \( V = 240\sin\omega t \) (volts), where \( \omega \) is variable. In the state of resonance, consider the following statements.
A. Angular frequency is \( 5\times10^{3} \) rad/s
B. Power factor is \( 0.5 \)
C. Current leads voltage by \( \frac{\pi}{2} \).
D. The power dissipated is \( 9.6 \) kW
E. Current in the circuit is \( 20 \) A
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → A and D only
Note: The given answer is as per the official NTA Answer Key. However, none of the given options appears to be correct, because the 9.6 kW in statement D is the peak power of the circuit, while the power dissipated in an ac circuit is the average power, which is 4.8 kW here, as explained below.
At resonance, \( X_{L} = X_{C} \), so \( Z = R \): the current is in phase with the voltage and the power factor is 1. In \( V = 240\sin\omega t \), 240 V is the peak value \( V_{0} \).
(A) Angular frequency is \( 5\times10^{3} \) rad/s — Correct.
\( \displaystyle \omega_{0} = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{(25\times10^{-3}\,\mathrm{H})(1.6\times10^{-6}\,\mathrm{F})}} = \frac{1}{\sqrt{4\times10^{-8}}}\,\mathrm{rad\,s^{-1}} \)
\( \omega_{0} = 5\times10^{3}\,\mathrm{rad\,s^{-1}} \)
(B) Power factor is \( 0.5 \) — Incorrect. At resonance \( Z = R \), so \( \cos\phi = R/Z = 1 \).
(C) Current leads voltage by \( \frac{\pi}{2} \). — Incorrect. At resonance the current is in phase with the voltage.
(D) The power dissipated is \( 9.6 \) kW — Correct as per NTA; however, it is not correct as a general statement. The power dissipated is the average power:
\( \displaystyle P = \frac{V_{\mathrm{rms}}^{2}}{R} = \frac{V_{0}^{2}}{2R} = \frac{(240\,\mathrm{V})^{2}}{2\times6\,\Omega} \)
\( P = 4800\,\mathrm{W} = 4.8\,\mathrm{kW} \)
9.6 kW is \( V_{0}^{2}/R \), the peak power; it equals the power dissipated only if 240 V is taken as the rms value.
\( \displaystyle \frac{V_{0}^{2}}{R} = \frac{(240\,\mathrm{V})^{2}}{6\,\Omega} = 9600\,\mathrm{W} = 9.6\,\mathrm{kW} \)
(E) Current in the circuit is \( 20 \) A — Incorrect. The peak current is 40 A and the rms current about 28.3 A.
\( \displaystyle I_{0} = \frac{V_{0}}{R} = \frac{240\,\mathrm{V}}{6\,\Omega} = 40\,\mathrm{A},\quad I_{\mathrm{rms}} = \frac{40\,\mathrm{A}}{\sqrt{2}} \approx 28.3\,\mathrm{A} \)
So strictly only A is correct; NTA’s Option (3) takes 240 V as the rms value.