JEE Mains · Physics · STD 12 -7. Alternating current
An \(LCR\) series circuit of capacitance \(62.5\,nF\) and resistance of \(50\,\Omega\). is connected to an \(A.C.\) source of frequency \(2.0\,kHz\). For maximum value of amplitude of current in circuit, the value of inductance is \(..........mH\). (take \(\left.\pi^2=10\right)\)
- A \(101\)
- B \(10\)
- C \(995\)
- D \(100\)
Answer & Solution
Correct Answer
(D) \(100\)
Step-by-step Solution
Detailed explanation
\(f =\frac{1}{2 \pi \sqrt{ LC }}\) \(2000\,Hz =\frac{1}{2 \pi \sqrt{ L \times 62.5 \times 10^{-9}}}\) \(L =\frac{1}{4 \pi^2 \times 2000^2 \times 62.5 \times 10^{-9}}=0.1\,H =100\,mH\)
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