What is the value of inductance `L` for which the current is a maximum in series `LCR` circuit with `C=10 muF` and `omega=1000s^(-1)`?

Options

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

Answer

A

Explanation

In the resonance condition, maximum current flows in the circuit.
Current in `LCR` 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 happens is resonance state of the circuit, i.e.,
`omegaL=1/(omegaC)`
or `L=1/(omega^(2)C)......(i)`
Given, `omega=1000s^(-1), C=10 muF=10xx10^(-6)F`
Hence, `L=1/((1000)^(2)xx10xx10^(-6))=0.1 H=100 mH`