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

Options

A. (A) 10 V
B. (B) `10//(sqrt(2))V`
C. (C) `(10//3)V`
D. (D) `20 V`

Answer

B

Explanation

Here `R=X_(L) =X_(C ) (: "Voltage across them is same")`
When capacitor is short circuited
`I=(10)/((R^(2)+x_(L)^(2))^(1//2))=(10)/(sqrt(2)R)`
`:.` Potential drop across inductance `=IX_(L) =IR = 10 sqrt(2)V`.