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. `10sqrt(2) V`
B. 10 V
C. 20 V
D. `10/(sqrt(2)) V`

Answer

D

Explanation

Before disconnecting C, the total voltage across the combination is
`V = I[R^(2)+(X_(L)-X_(C))^(2)]^(1//2) = IR = 10V ( :' X_(L) = X_(C))`
`:.` When C is removed, the current I' given by
`I' = V/((R^(2)+X_(L)^(2))^(1//2)) = 10/(sqrt(2)R) ( :' X_(L) = R)`
`:.` P.D. across `L = I'X_(L) = 10/(sqrt(2)R) xx R = 10/(sqrt2)V`.