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Question

Physics Question on Alternating current

What is the value of inductance L for which the current in a maximum in a series L-C-R circuit with C = 10 μ\muF and ω=1000s−1\omega = 1000 s^{ - 1} ?

A

100 mH

B

1 mH

C

Cannot be calculated unless R is known

D

10 mH

Answer

100 mH

Explanation

Solution

Current in L-C-R series circuit, i=VR2+(XL−XC)2i = \frac{ V}{ \sqrt{ R^2 + ( X_L - X_C)^2 }} For current to be maximum, denominator should be minimum which can be done, if XL−XCX_L - X_C This happens in resonance state of the circuit i, e, ωL=1ωC\omega L = \frac{ 1}{ \omega C } or L = 1ω2C\frac{ 1}{ \omega^2 C } \hspace15mm ..(i) Given, ω=1000 s−1, C=10 μF=10×10−6 F \omega = 1000\, s^{ - 1}, \, C = 10 \, \mu F = 10 \times 10^{ - 6} \, F Hence, L=1(1000)2×10×10−6L = \frac{ 1}{ (1000)^2 \times 10 \times 10^{ - 6}} = 0.1 H = 100 m H.