Alternating Current Question 4
Question 4 - 25 January - Shift 1
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 $\pi^{2}=10$ )
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Answer: (100)
Solution:
Formula: Resonant Frequency
$f=\frac{1}{2 \pi \sqrt{LC}}$
$2000 Hz=\frac{m}{2 \pi \sqrt{L \times 62.5 \times 10^{-9}}}$
$L=\frac{1}{4 \pi^{2} \times 2000^{2} \times 62.5 \times 10^{-9}}=0.1 H=100 mH$
