Question

In series LCR resonant circuit, R=800 \(\Omega\), C = 2 \(μ\)F and voltage across resistance is 200 V. The angular frequency is 250 rad/s. At resonance the voltage across the capacitance is

• A. \(250\) V
• B. \(500\) V
• C. \(1000\) V
• D. \(750\) V

Hint:

At resonance the current is the resistor voltage divided by R, and the capacitor voltage is the current times its reactance.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
In a series LCR circuit at resonance, the impedance is just \(R\). The current is set by the voltage across the resistor, and the capacitor voltage is \(I X_C\).

Step 2: Find the current.
\[ I = \frac{V_R}{R} = \frac{200}{800} = 0.25\text{ A} \]

Step 3: Find the reactance.
\[ X_C = \frac{1}{\omega C} = \frac{1}{250\times 2\times 10^{-6}} = \frac{1}{5\times 10^{-4}} = 2000\ \Omega \]

Step 4: Find the voltage.
\[ V_C = IX_C = 0.25\times 2000 = 500\text{ V} \]

Step 5: Check the options.
250 V, 750 V and 1000 V do not match \(0.25\times 2000\).

Final Answer:
The voltage across the capacitor is 500 V, option (B). \[ \boxed{500\text{ V}} \]