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Q.
The value of inductance $L$ for which the current is maximum in series LCR circuit with $C=10 \, μF$ and $\omega =1000 \, rad \, s^{- 1}$ .
NTA AbhyasNTA Abhyas 2020
Solution:
In resonance condition, maximum current flows in the circuit.
Current in the LCR circuit is given by
$i=\frac{V}{\sqrt{R^{2} + \left(X_{L} - X_{C}\right)^{2}}}$
For current to be maximum denominator should be minimum.
$\left(X_{L} - X_{C}\right)^{2}=0$
$\Rightarrow \, \, \, X_{L}=X_{C}$
$\Rightarrow \, \, \, \omega L=\frac{1}{\omega C}$
$\therefore \, \, \, L=\frac{1}{\omega ^{2} C}$
$=\frac{1}{\left(1000\right)^{2} \times 10 \times \left(10\right)^{- 6}}$
$=\frac{1}{10}H$
$\therefore \, \, \, L=0.1 \, H=100 \, mH$