Question 37: A series LCR circuit containing 5.0 H inductor, 80 \mu \mathrm{~F} capacitor and 40 \Omega resistor is connected to 230 V variable frequency ac source. The angular frequencies of the source at which power transferred to the circuit is half the power at the resonant angular frequency are likely to be :
(1) 25 \mathrm{rad} / \mathrm{s} and 75 \mathrm{rad} / \mathrm{s}
(2) 50 \mathrm{rad} / \mathrm{s} and 25 \mathrm{rad} / \mathrm{s}
(3) 46 \mathrm{rad} / \mathrm{s} and 54 \mathrm{rad} / \mathrm{s}
(4) 42 \mathrm{rad} / \mathrm{s} and 58 \mathrm{rad} / \mathrm{s}
Answer: Option (3)
Explanation:
For a series LCR circuit, the resonant angular frequency is given by
\omega_0 = \frac{1}{\sqrt{LC}}Here, L = 5.0 \, \mathrm{H} and C = 80 \times 10^{-6} \, \mathrm{F}.
\omega_0 = \frac{1}{\sqrt{5 \times 80 \times 10^{-6}}} = \frac{1}{\sqrt{4 \times 10^{-4}}} = 50 \, \mathrm{rad/s}The power in a series LCR circuit becomes half of the maximum power at frequencies \omega_1 and \omega_2 given by
\omega_1 = \omega_0 - \frac{R}{2L} \omega_2 = \omega_0 + \frac{R}{2L}Substituting the given values,
\frac{R}{2L} = \frac{40}{2 \times 5} = 4 \, \mathrm{rad/s}Therefore,
\omega_1 = 50 - 4 = 46 \, \mathrm{rad/s} \omega_2 = 50 + 4 = 54 \, \mathrm{rad/s}Hence, the angular frequencies at which the power is half of the maximum power are
46 \, \mathrm{rad/s} and 54 \, \mathrm{rad/s},
which corresponds to option (3).