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What is the value of inductance L for which the current is maximum in a series LCR circuit with \mathrm{C}=10 \mu \mathrm{F} and \omega=1000 \mathrm{~s}^{-1}?

Option: 1

100 mH


Option: 2

1 mH


Option: 3

10 mH


Option: 4

can not be calculated unless R is known


Answers (1)

best_answer

In resonance conditions, maximum current flows in the circuit. Hence, current in LCR series circuit,\mathrm{{I=\frac{V}{\sqrt{R^2+\left(X_L-X_C\right)^2}}}}
Where V is rms value of voltage, R is resistance, \mathrm{X_L } is inductive reactance and \mathrm{X_C} is capacitive reactance. For current to be maximum, denominator should be minimum which can be done, if \mathrm{\mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}}}
This happens in resonance state of the circuit i.e, \mathrm{\omega \mathrm{L}=\frac{1}{\omega \mathrm{C}}}
or \mathrm{\quad \mathrm{L}=\frac{1}{\omega^2 \mathrm{C}} (i)}
Given,\mathrm{ \omega=1000 \mathrm{~s}^{-1}, \mathrm{C}=10 \mu \mathrm{F}=10 \times 10^{-6} \mathrm{~F}}
Hence, \mathrm{\mathrm{L}=\frac{1}{(1000)^2 \times 10 \times 10^{-6}}=0.1 \mathrm{H}=100 \mathrm{mH}}

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manish

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