What is the value of inductance `L` for which the current is a maximum in series `LCR` circuit with `C=10 muF` and `omega=1000(rad)/s`?

Options

A. 100 mH
B. 1 mH
C. Cannot be calculated unless R is unknow
D. 10 mH

Answer

A

Explanation

Current in LCR series circuit,
`i=(V)/(sqrt(R^(2)+(X_(L)-X_(C))^(2)))`
where V is rms valus of current, R is resstance, `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,
`omegaL=(1)/(omegaC)`
or `L=(1)/(omega^(2)C)" "......(i)`
Given, `omega=1000s^(-1),C=10muF`
`=10xx10^(-6)F`
Hence, `L=(1)/((1000)^(2)xx10xx10^(-6))`
`=0.1H`
`=100mH`