Question

A lossy capacitor $C_x$ rated for operation at 5 kV, 50 Hz is represented by an equivalent circuit with and ideal capacitor $C_P$ in parallel with a resistor$R_P.$ The value of $C_P$ is found to be $0.102\mu F$ and the value of $R_P=1.25\Omega$. Then the power loss and $tan\delta$ of the lossy capacitor operating at the rated voltage, respectively, are

• A. 10 W and 0.0002
• B. 10 W and 0.0025
• C. 20 W and 0.025
• D. 20 W and 0.04

Answer 1

The correct option is C 20 W and 0.025

$C_P=0.102\mu F$
$R_P=1.25M\Omega$
$Powerloss=\frac{V^2}{R_P}=\frac{(5\times {10}^3{)}^2}{1.25\times {10}^6}=\frac{25}{1.25}$
$=20W$
$tan\delta =\frac{1}{\omega C_P{R}_P}$
$=\frac{1}{2\times \pi \times 0.102\times {10}^{-6}\times 1.25\times {10}^6}$

$=0.025$