In a series LCR circuit, the voltage across the resistor, capacitor and inductor is 10 V each. If the capacitor is short-circuited, the voltage across the inductor will be

Question

In a series LCR circuit, the voltage across the resistor, capacitor and inductor is 10 V each. If the capacitor is short-circuited, the voltage across the inductor will be (A) 10V (B) (10/√2)V (C) (10/3)V (D) 20V

Tardigrade Admin · Accepted Answer

Here, R=XL=XC , because voltage across them is the same. When the capacitor is short circuited, I=(10/(R2 + XL2)(1)2)=(10/√2 R) Hence, potential drop across inductor =IXL=IR=(10/√2)V

Q. In a series $L CR$ circuit, the voltage across the resistor, capacitor and inductor is $10 V$ each. If the capacitor is short-circuited, the voltage across the inductor will be

$10 V$

$\frac{10}{2} V$

$\frac{10}{3} V$

$20 V$

Here, $R = X_{L} = X_{C}$ , because voltage across them is the same.
When the capacitor is short circuited,
$I = \frac{10}{( R ^{2} + X _{L}^{2} ) ^{\frac{1}{2}}} = \frac{10}{2 R}$
Hence, potential drop across inductor $= I X_{L} = I R = \frac{10}{2} V$

Q. In a series LCR circuit, the voltage across the resistor, capacitor and inductor is 10V each. If the capacitor is short-circuited, the voltage across the inductor will be

10V

2​10​V

310​V

20V