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 known
D. 10 mH

Answer

A

Explanation

Current in L-C-R series circuit,
`i=(V)/(sqrt(R^(2)+(X_(L)-X_(C))^(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 happen in resonance state of the circuit i.e.
`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`