MHT CET · Physics · Capacitance

A resistor of \(50 \Omega\), inductor of self inductance \(\left(\frac{3}{\pi^2}\right) \mathrm{H}\) and a capacitor of unknown capacity are connected in series to an a.c. source of 100 V and 50 Hz . When the voltage and current are in phase, the value of capacitance is (nearly)

A \(0.66 \times 10^{-4} \mathrm{~F}\)
B \(0.33 \times 10^{-4} \mathrm{~F}\)
C \(0.66 \times 10^{-2} \mathrm{~F}\)
D \(0.33 \times 10^{-2} \mathrm{~F}\)

Verified Solution

Answer & Solution

Correct Answer

(B) \(0.33 \times 10^{-4} \mathrm{~F}\)

Step-by-step Solution

Detailed explanation

Given: \(\mathrm{L}=\frac{3}{\pi^2} \mathrm{H}\) and \(\mathrm{f}=50 \mathrm{~Hz}\)
As voltage and current are in phase, the circuit is a resonant circuit.
\(\begin{aligned}
& \therefore \quad \mathrm{X}_{\mathrm{C}}=\mathrm{X}_{\mathrm{L}} \\
& \frac{1}{\omega \mathrm{C}}=\omega \mathrm{L} \\
& \mathrm{C}=\frac{1}{\omega^2 \mathrm{~L}}=\frac{1}{4 \pi^2 \mathrm{f}^2 \mathrm{~L}} \\
&=\frac{1}{4 \pi^2 \times(50)^2 \times \frac{3}{\pi^2}}=0.33 \times 10^{-4} \mathrm{~F}
\end{aligned}\)

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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