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Alternating Current
Series LCR Resonance
2 questions from 2 papers, each with its answer.
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All Questions
2023 1 question
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In a series LCR circuit, the inductance $L$ is $10\ mH$, capacitance $C$ is $1\ \mu F$ and resistance $R$ is $100\ \Omega$. The frequency at which resonance occurs is:
At resonance $\omega=\frac{1}{\sqrt{LC}}=10^4\ rad/s$, so $f=1.59\ kHz$.
For resonance frequency: $\omega L=\frac{1}{\omega C}$ $\Rightarrow\omega=\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{10\times10^{-3}\times1\times10^{-6}}}$ $=\frac{1}{\sqrt{10^{-8}}}=10^4\ rad/sec$ $f=\left(\frac{\omega}{2\pi}\right)=\frac{10^4}{2\pi}=1.59\ kHz$
2022 1 question
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A series LCR circuit with inductance 10 H, capacitance $10\ \mu F$, resistance $50\ \Omega$ is connected to an ac source of voltage, $V=200\sin(100t)$ volt. If the resonant frequency of the LCR circuit is $\nu_0$ and the frequency of the ac source is $\nu$, then :$V = 200\sin(100t)$, so $\omega = 100$ rad/s Source frequency $\nu = \frac{\omega}{2\pi} = \frac{100}{2\pi} = \frac{50}{\pi}$ Hz Resonant frequency $\nu_0 = \frac{1}{2\pi\sqrt{LC}} = \frac{1}{2\pi}\sqrt{\frac{1}{10 \times 10 \times 10^{-6}}} = \frac{50}{\pi}$ Hz So $\nu_0 = \nu = \frac{50}{\pi}$ Hz