Question

A series $LCR$ circuit driven by $300\text{V}$ at a frequency of $50\text{Hz}$ contains a resistance $R=3k\Omega$, an inductor of inductive reactance $X_L=250\pi \Omega$ and an unknown capacitor. The value of capacitance, to maximize the average power, should be:$\text{Take}({\pi }^2\approx 10)$

• A. $25\mu F$
• B. $40\mu F$
• C. $4\mu F$
• D. $400\mu F$

Answer 1

The correct option is C $4\mu F$

$\omega =2\pi f=100\pi \text{rad/s}$

For maximizing average power, $Z=R\Rightarrow X_L=X_C$

$\Rightarrow \omega L=\frac{1}{\omega C}$

$\Rightarrow 250\pi =\frac{1}{100\pi C}$

$\therefore C=\frac{1}{25\times {\pi }^2\times {10}^3}=\frac{{10}^{-4}}{250}=4\times {10}^{-6}\text{F}=4\mu F$

Hence, $\left(D\right)$ is the correct answer.