JEE Mains · Physics · STD 12 -7. Alternating current

For the \(LCR\) circuit, shown here, the current is observed to lead the applied voltage. An additional capacitor \(C '\) , when joined with the capacitor \(C\) present in the circuit , makes the power factor of the circuit unity. The capacitor \(C'\), must have been connected in

A series with \(C\) and has a magnitude \(\frac{C}{{\left( {{\omega ^2}LC - 1} \right)}}\)
B series with \(C\) and has a magnitude \(\,\frac{{\left( {1 - {\omega ^2}LC} \right)}}{{{\omega ^2}L}}\)
C parallel with \(C\) and has a magnitude \(\,\frac{{\left( {1 - {\omega ^2}LC} \right)}}{{{\omega ^2}L}}\)
D parallel with \(C\) and has a magnitude \(\frac{C}{{\left( {{\omega ^2}LC - 1} \right)}}\)

Verified Solution

Answer & Solution

Correct Answer

(C) parallel with \(C\) and has a magnitude \(\,\frac{{\left( {1 - {\omega ^2}LC} \right)}}{{{\omega ^2}L}}\)

Step-by-step Solution

Detailed explanation

Power factor \(\cos \phi=\frac{R}{\sqrt{R^{2}+\left[\omega L-\frac{1}{\omega\left(C+C^{\prime}\right)}\right]^{2}}}=1\) On solving we get, \(\omega L=\frac{1}{\omega\left(C+C^{\prime}\right)}\) \(C' = \frac{{1 - {\omega ^2}LC}}{{{\omega ^2}L}}\) Hence option \((c)\) is the…

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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