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What is the value of inductance $L$ for which the current is a maximum in a series $L C R$ circuit with $C=10 \mu \mathrm{F}$ and $\omega=1000 \mathrm{~s}^{-1}$ ?
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Verified Answer
The correct answer is:
$100 \mathrm{mH}$
Current in $L C R$ series circuit,
$i=\frac{V}{\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}}$
where $V$ is rms value of current, $R$ is resistance, $X_{L}$ is inductive reactance and $X_{C}$ is capacitive reactance.
For current to be maximum, denominator should be minimum which can be done, if
$$
X_{L}=X_{C}
$$
This happens in resonance state of the circuit ie,
$$
\omega L=\frac{1}{\omega C}
$$
or $\quad L=\frac{1}{\omega^{2} C}$ ...(i)
$\begin{aligned} \text { Given, } \omega &=1000 \mathrm{~s}^{-1}, C=10 \mu \mathrm{F} \\ &=10 \times 10^{-6} \mathrm{~F} \\ \text { Hence, } L &=\frac{1}{(1000)^{2} \times 10 \times 10^{-6}} \\ &=0.1 \mathrm{H} \\ &=100 \mathrm{mH} \end{aligned}$
$i=\frac{V}{\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}}$
where $V$ is rms value of current, $R$ is resistance, $X_{L}$ is inductive reactance and $X_{C}$ is capacitive reactance.
For current to be maximum, denominator should be minimum which can be done, if
$$
X_{L}=X_{C}
$$
This happens in resonance state of the circuit ie,
$$
\omega L=\frac{1}{\omega C}
$$
or $\quad L=\frac{1}{\omega^{2} C}$ ...(i)
$\begin{aligned} \text { Given, } \omega &=1000 \mathrm{~s}^{-1}, C=10 \mu \mathrm{F} \\ &=10 \times 10^{-6} \mathrm{~F} \\ \text { Hence, } L &=\frac{1}{(1000)^{2} \times 10 \times 10^{-6}} \\ &=0.1 \mathrm{H} \\ &=100 \mathrm{mH} \end{aligned}$
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