Question

In a series LCR circuit, the voltage across R is \(100\) V, \(R = 1\,\text{K}\,\Omega\) and \(C = 2\,μ\text{F}\). The angular frequency \(ω\) is \(200\) rad s\(^{-1}\). At resonance, the voltage across '\(L\)' is

• A. \(150\) V
• B. \(200\) V
• C. \(250\) V
• D. \(300\) V

Hint:

At resonance X_L = X_C, so V_L = I X_C with I = V_R / R.

Answer 1

The Correct Option is C

Step 1: Quality factor idea:
At resonance $V_L = V_C = QV_R$ where $Q = \frac{X_L}{R}$.

Step 2: Compute Q:
$X_C = \frac{1}{\omega C} = 2500\ \Omega$, $R = 1000\ \Omega$, so $Q = 2.5$.

Step 3: Voltage:
$V_L = Q\,V_R = 2.5\times100 = 250$ V.

Final Answer:
V across L is 250 V, option (C). \[ \boxed{250\text{ V}} \]