In a series LCR circuit, the voltage across the resistance, capacitance and inductance is 10 V each. If the capacitance is short circuited the voltage across the inductance will be

Options

A. 10 V
B. `10sqrt2V`
C. `10//sqrt2V`
D. 20 V

Answer

C

Explanation

As `V_(R)=V_(L)=V_(C)`
`R=X_(L)=X_(C)Z=R`
When capacitor is short circuited,
`Z=sqrt(R^(2)+X_(L)^(2))=sqrt(R^(2)+R^(2))=Rsqrt2`
New current `I'=V//Z=V//Rsqrt2=(10)/(Rsqrt2)`
Potential drop across inductance
`=I'X_(L)=I'R=(10xxR)/(Rsqrt2)=(10)/(sqrt2)" volt"`