AIIMS 2000 Physics Series LCR Circuit MCQ Question

In a series LCR circuit, the voltage across the resistance (R), capacitance (C), and inductance (L) can be represented as follows: when the circuit is at resonance, the voltages across R, C, and L are equal in magnitude but differ in phase. Each voltage is given as 10 V. When the capacitance is short-circuited, the capacitor is effectively removed from the circuit, and the total voltage across the inductor (L) will now be influenced by the remaining components. The voltage across the inductor can be calculated using the formula for the peak voltage in a series LCR circuit. The voltage across the inductor will become the vector sum of the voltages across the resistor and the inductor, which can be calculated as:

V_L = rac{V_R}{ ext{cos}( heta)}

Here, since the voltage across R is 10 V and the circuit is at resonance, the effective voltage across the inductor will be:

V_L = 10 imes rac{1}{ ext{cos}(45^{ ext{o}})} = 10 imes rac{1}{rac{1}{ ext{√2}}} = 10 ext{√2} ext{ V}

However, since the capacitor is short-circuited, the voltage across the inductor will be halved due to the phase difference, leading to:

V_L = rac{10 ext{√2}}{2} = 10 ext{V}

But since we are looking for the voltage across L after the capacitance is short-circuited, it will be:

$V_L = 10 / ext{√2} ext{ V}$

Thus, the correct answer is A) 10/√2 V.

The other options are incorrect because:

• B) 10 V does not consider the phase shift due to the short circuit.
• C) 10√2 V is the voltage across the inductor before the short circuit.
• D) 20 V is not possible as the voltage cannot exceed the supplied voltage in a series circuit.