MHT CET · Physics · Alternating Current
An a.c. e.m.f. of peak value 230 V and frequency 50 Hz is connected to a circuit with \(\mathrm{R}=11.5 \Omega, \mathrm{~L}=2.5 \mathrm{H}\) and a capacitor all in series. The value of capacitance is ' \(C\) ' for the current in the circuit to be maximum. The value of ' \(C\) ' and maximum current are respectively ( \(\pi^2=10\) )
- A \(4 \mu \mathrm{~F}, \quad 20 \mathrm{~A}\)
- B \(5 \mu \mathrm{~F}, \quad 10 \mathrm{~A}\)
- C \(2 \mu \mathrm{~F}, \quad 20 \mathrm{~A}\)
- D \(8 \mu \mathrm{~F}, 12 \mathrm{~A}\)
Answer & Solution
Correct Answer
(A) \(4 \mu \mathrm{~F}, \quad 20 \mathrm{~A}\)
Step-by-step Solution
Detailed explanation
\( \omega = 2 \pi f = 2 \pi (50) = 100 \pi \text{ rad/s} \) \( C = \frac{1}{\omega^2 L} = \frac{1}{(100 \pi)^2 (2.5)} = \frac{1}{10000 \times 10 \times 2.5} = \frac{1}{250000} = 4 \times 10^{-6} \text{ F} = 4 \, \mu \text{F} \)
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