1. Given data

• Voltage: $V = 250\,\text{V}$
• Frequency: $f = 50\,\text{Hz}$
• Power dissipated: $P = 400\,\text{W}$
• Power factor: $\cos\phi = 0.8$

We have a series LR circuit.

2. Find current in the circuit

Using the AC power relation:

$P = VI\cos\phi$

So,

$400 = 250 \cdot I \cdot 0.8$

$400 = 200I$

$I = 2\,\text{A}$

3. Find the impedance of the LR circuit

$Z = \frac{V}{I} = \frac{250}{2} = 125\,\Omega$

4. Find resistance

Since power is dissipated only in resistance,

$P = I^2R$

$400 = (2)^2 R = 4R$

$R = 100\,\Omega$

5. Find inductive reactance

For a series LR circuit,

$Z^2 = R^2 + X_L^2$

$125^2 = 100^2 + X_L^2$

$15625 = 10000 + X_L^2$

$X_L^2 = 5625$

$X_L = 75\,\Omega$

6. Condition for unity power factor after adding capacitor in series

To make power factor unity in a series $LRC$ circuit, net reactance must be zero:

$X_L = X_C$

So,

$X_C = 75\,\Omega$

Now,

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

where

$\omega = 2\pi f = 2\pi \cdot 50 = 100\pi$

Hence,

$\frac{1}{100\pi C} = 75$

$C = \frac{1}{75 \cdot 100\pi}$

$C = \frac{1}{7500\pi}\,\text{F}$

Convert into microfarads:

$C = \frac{10^6}{7500\pi}\,\mu\text{F}$

$C = \frac{1000}{7.5\pi}\,\mu\text{F}$

$C = \frac{400}{3\pi}\,\mu\text{F}$

Given,

$C = \left(\frac{n}{3\pi}\right)\mu\text{F}$

Comparing,

$n = 400$

7. Comparison with stored answer

Stored correct answer: $400$

Our derived answer also is $400$.