In a series resonant R-L-C circuit, the voltage across R is 100 V and the value of `R=1000Omega`. The capacitance of the capacitor is `2xx10^(-6)` F, angular frequency of AC is 200 `"rad s"^(-1)`. Then the P.D. across the inductance coil is

Options

A. 100 V
B. 40 V
C. 250 V
D. 400 V

Answer

C

Explanation

Here, `V_(R)=100V,R=100Omega`
`C=2xx10^(-6)F,omega=200" rad s"^(-1)`
The current in the circuit is
`I=(V_(R))/(R)=(100V)/(1000Omega)=0.1A`
At resonance
`V_(L)=V_(C)=IX_(C)=(I)/(omegaC)=(0.1)/(200xx2xx10^(-6))=(1000)/(4)=250V`