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

Question

In a series L C R circuit the voltage across resistance, capacitance and inductance is 10 V each. If the capacitance is short circuited, the voltage across the inductance will be (A) (10/√2)V (B) 10 V (C) 10√2V (D) 20 V

Tardigrade Admin · Accepted Answer

Since voltage across L, C and R is same, equal to 10 V, therefore R = X_L = X_C.
Also total voltage in the circuit is
I [R² + (X_L - X_C )²]^1/2 = I R = 10 V.
When capacitor is short circuited current in the circuit

I = \frac{10}{( R ^{2} + X _{L}^{2} ) ^{1/2}} = \frac{10}{2 R} .

Potential drop across inductance

= I X_{L} = I R = 10/ 2

V.

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

$\frac{10}{2} V$

$10 V$

$102 V$

$20 V$

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

2​10​V

10V

102​V

20V

Since voltage across L, C and R is same, equal to 10 V, therefore R = XL = XC. Also total voltage in the circuit is I [R2 + (XL - XC )2]1/2 = I R = 10 V. When capacitor is short circuited current in the circuit I=(R2+XL2​)1/210​=2​R10​. Potential drop across inductance =IXL​=IR=10/2​ V.

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

$\frac{10}{2} V$

$10 V$

$102 V$

$20 V$