A series LCR circuit with `L=0.5` H and `R=10 Omega` is connected to an AC supply with rms voltage and frequency equal to 2000 V and `(150)/(pi)` Hz, respectively. The magnitude of the capacitance is varied so that current amplitude in the circuit becomes maximum. The rms voltage difference across the inductor is

Options

A. 3000 V
B. 2500 V
C. 2000 V
D. 2600 V

Answer

A

Explanation

Key Idea Current amplitude in series LCR circuit is maximum, if there is resonance i.e, `omega L=(1)/(omega C)`.
Given, `L=0.5H, R=10 Omega, V_(rms)=200 V ` and
`f=(150)/(pi) Hz`
Angular frequency,
`omega=2 pi f = 2pi xx (150)/(pi)= 300 `rad `s^(-1)`
`:.` For `I_("max")` in series LCR circuit,
`omega^(2)=(1)/(LC)`
`rArr C= (1)/(L omega^(2))`
`rArr C= (1)/(0.5xx300 xx300)=22.2xx10^(-6)F`
Current in the circuit,
`I_("max")=(V_("rms"))/(R)=(200)/(10)=20A`
`:.` rms voltage difference across the inductor `V_(L)`
and capacitor `V_(C)` are equal. Then
`V_(L)=I_("max")omegaL=20xx300xx0.5`
`rArr V_(L)=3000V`
Hence, the correct option is (a).