JEE PYQ: Alternating Current Question 26
Question 26 - 2020 (06 Sep Shift 2)
In a series LR circuit, power of 400 W is dissipated from a source of 250 V, 50 Hz. The power factor of the circuit is 0.8. In order to bring the power factor to unity, a capacitor of value $C$ as $\left(\frac{n}{3\pi}\right) \mu$F is added in series. The value of $n$ is ____.
Show Answer
Answer: 400
Solution
$R = 100\Omega$, $X_L = 75\Omega$. For unity power factor, $X_C = X_L = 75$. $C = \frac{1}{75 \times 2\pi \times 50} = \frac{400}{3\pi}\mu$F. So $n = 400$.
BETA
