Question

Find the capacitance in a series $LCR$ circuit in order to get maximum power. ($X_L=250\Omega$ and $f=50Hz$).

• A. $10.7\mu F$
• B. $12.7\mu F$
• C. $14.7\mu F$
• D. $14.9\mu F$

Answer 1

The correct option is B

$12.7\mu F$

Step 1. Given data

Inductive reactance, $X_L=250\Omega$,

Frequency, $f=50Hz$,

Assume $\pi =3.14$

Step 2. Calculating angular frequency, $\omega$

As we know that, for maximum power, $X_C=X_L$ (Condition of resonance)

Where, $X_L$ is inductive reactance and $X_c$ is capacitive reactance.

We know that, capacitive reactance, $X_c=\frac{1}{C\omega }$,

Where, $C$ is the capacitance and $\omega$ is the angular frequency.

Since, $X_C=X_L$

$\Rightarrow$ $\frac{1}{C\omega }=250\Omega$

As we know,

$\omega =2\pi f$ [ where $f$ is frequency]

$\Rightarrow$ $\omega =2\pi x50$

$\Rightarrow$ $\omega =100\pi$

Step 3. Calculating the capacitance

Now, the capacitance $C$ is given by

$C=\frac{1}{\omega X_L}$

$$\begin{gathered} \Rightarrow C=\frac{1}{\left(100\pi x250\right)} \\ \Rightarrow C=\frac{1}{\left(100\times 3.14x250\right)} \\ \Rightarrow C=\frac{1}{78500}F \\ \Rightarrow C=0.0000127F \\ \Rightarrow C=12.7\mu F \end{gathered}$$

Therefore, the capacitance $C$ is $12.7\mu F$.

Hence, the correct option is (B).