JEE Mains · Physics · STD 12 -Alternating current
- JEE Mains 2015
- Hard
- MCQ
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
Pick an option to check your answer.
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 correct answer.
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