JEE Mains · Physics · STD 12 -7. Alternating current
An inductor of \(0.5\,mH\), a capacitor of \(20\,\mu\,F\) and resistance of \(20\,\Omega\) are connected in series with a \(220\,V\) ac source. If the current is in phase with the emf, the amplitude of current of the circuit is \(\sqrt{x} A\). The value of \(x\) is -
- A \(242\)
- B \(241\)
- C \(240\)
- D \(236\)
Answer & Solution
Correct Answer
(A) \(242\)
Step-by-step Solution
Detailed explanation
\(X _{ L }= X _{ C }\) So, \(Z=R=20\,\Omega\) \(i_{rms}=\frac{220}{20}=11\) \(i_{\max }=11 \sqrt{2}=\sqrt{242}\)
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