First slide

Phasors and Phasor Diagrams

Question

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

Moderate

A

10 V
B

$10 / \sqrt{2} V$
C

$10 \sqrt{2} V$
D

20 V

Solution

Here $R = X_{L} = X_{C} (∵$ voltage across them is same)
Total voltage in the circuit,
$V = I {[R^{2} + {(X_{L} - X_{C})}^{2}]}^{1 / 2}$
= IR = 10 volt
When capacitor is short circuited,
$I = \frac{10}{{(R^{2} + X_{L}^{2})}^{1 / 2}} = \frac{10}{\sqrt{2} R}$
$∴$ Potential drop across inductance
$= {LX}_{L} = IR = \frac{10}{\sqrt{2}} volt$

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