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A series LCR circuit with inductance 10 H10\text{ H}, capacitance 10 μF10\text{ }\mu\text{F}, resistance 50 Ω50\text{ }\Omega is connected to an ac source of voltage, V=200sin⁡(100πt) voltV = 200\sin(100\pi t)\text{ volt}. If the resonant frequency of the LCR circuit is f0f_0 and the frequency of the ac source is ff, then

1

f0=50 Hzf_0 = 50\text{ Hz}

2

f0=50π Hzf_0 = \frac{50}{\pi}\text{ Hz}

3

f0=50π Hzf_0 = \frac{50}{\pi}\text{ Hz}, f=50 Hzf = 50\text{ Hz}

4

f=100 Hzf = 100\text{ Hz}; f0=100π Hzf_0 = \frac{100}{\pi}\text{ Hz}

Step-by-Step Solution

Resonant frequency f0=12πLC=12π10×10×10−6=12π10−4=12π×10−2=1002π=50π Hzf_0 = \frac{1}{2\pi\sqrt{LC}} = \frac{1}{2\pi\sqrt{10 \times 10 \times 10^{-6}}} = \frac{1}{2\pi\sqrt{10^{-4}}} = \frac{1}{2\pi \times 10^{-2}} = \frac{100}{2\pi} = \frac{50}{\pi}\text{ Hz}. The source frequency ff is derived from ω=100π\omega = 100\pi, so 2πf=100π2\pi f = 100\pi, which gives f=50 Hzf = 50\text{ Hz}.

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