Question

A 400 $\Omega$ resistor, a 250 $mH$ inductor and a $2.5\mu F$ capacitor are connected in series with an $AC$ source of peak voltage 5 $V$ and angular frequency $2kHz$. What is the peak value of the electrostatic energy of the capacitor?

• A. $2\mu J$
• B. $2.5\mu J$
• C. $3.33\mu J$
• D. $5\mu J$

Answer 1

The correct option is D $5\mu J$

From given,
$X_C=\frac{1}{\omega C}=200\Omega$
$X_L=\omega L=500\Omega$

$Z=\sqrt{R^2+{(X_L-X_C)}^2}=\sqrt{{400}^2+{300}^2}=500\Omega$

$V_C=i{X}_C=\frac{5}{500}\times 200=2V{U}_C=\frac{1}{2}C{V}_C^2=\frac{1}{2}\times 2.5\times {10}^{-6}\times (2{)}^2=5\mu J$