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: Understanding the Concept:
At resonance of a series LCR circuit, the inductive reactance equals the capacitive reactance, \(X_L = X_C\). The impedance is just \(R\), and the current is \(I = \frac{V_R}{R}\).

Step 2: Find the current:
\(I = \frac{100}{1000} = 0.1\) A.

Step 3: Find the reactance:
\[ X_C = \frac{1}{\omega C} = \frac{1}{200\times2\times10^{-6}} = \frac{1}{4\times10^{-4}} = 2500\ \Omega \]
At resonance \(X_L = X_C = 2500\ \Omega\).

Step 4: Voltage across L:
\[ V_L = IX_L = 0.1\times2500 = 250\text{ V} \]

Step 5: Why the other options are wrong.
150 V, 200 V and 300 V would require \(X_L\) of 1500, 2000 or 3000 \(\Omega\), which do not match \(\frac{1}{\omega C} = 2500\ \Omega\).

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