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WBJEE · Physics · Capacitance

A \(400 \Omega\) resistor, a \(250 \mathrm{mH}\) inductor and a \(2.5 \mu \mathrm{F}\) capacitor are connected in series with an AC source of peak voltage \(5 \mathrm{~V}\) and angular frequency \(2 \mathrm{rad/sec}\). What is the peak value of the electrostatic energy of the capacitor?

  1. A \(2 \mu \mathrm{J}\)
  2. B \(2.5 \mu \mathrm{J}\)
  3. C \(3.33 \mu \mathrm{J}\)
  4. D \(5 \mu \mathrm{J}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(5 \mu \mathrm{J}\)

Step-by-step Solution

Detailed explanation

Hint: The angular frequency is \(2 \mathrm{KHz}\). The unit given is incorrect Assuming it to be in radian / sec \(X_{L}=2 \times 10^{3} \times 250 \times 10^{-3}=500 \Omega\) \(\mathrm{X}_{\mathrm{C}}=\frac{1}{2.5 \times 10^{-6} \times 2 \times 10^{3}}=200 \Omega\)…