In a series LR circuit, power of $$400\,\text{W}$$ is dissipated from a source of $$250\,\text{V}$$, $$50\,\text{Hz}$$. The power factor of the circuit is $$0.8$$. In order to bring the power factor to unity, a capacitor of value C is added in series to the L and R. Taking the value of C as $$\left(\frac{n}{3\pi}\right)\,\mu F$$, then value of n is___

We have an a.c. source of rms voltage $$V = 250\; \text{V}$$ and frequency $$f = 50\; \text{Hz}$$ feeding a series combination of a resistor $$R$$ and an inductor $$L$$.

The average power consumed in any a.c. circuit is given by the formula

$$P = V I \cos\phi = I^{2} R = \dfrac{V^{2} R}{R^{2} + X_{L}^{2}},$$

where $$\phi$$ is the phase angle and $$X_{L} = \omega L = 2\pi f L$$ is the inductive reactance. For the data given we know directly:

$$P = 400\; \text{W}, \qquad V = 250\; \text{V}, \qquad \cos\phi = 0.8.$$

First we calculate the current. From the definition of power factor,

$$\cos\phi = \dfrac{R}{Z},$$

where $$Z$$ is the magnitude of the series impedance. Therefore

$$R = Z\cos\phi = 0.8\,Z.$$

The power in terms of impedance is obtained by substituting $$I = V/Z$$ in the expression $$P = I^{2}R$$:

$$P = \left(\dfrac{V}{Z}\right)^{2} R = \dfrac{V^{2}}{Z^{2}}\,(0.8\,Z) = 0.8\,\dfrac{V^{2}}{Z}.$$

Re-arranging gives

$$Z = 0.8\,\dfrac{V^{2}}{P}.$$

Substituting the numerical values,

$$Z = 0.8\,\dfrac{(250)^{2}}{400} = 0.8\,\dfrac{62500}{400} = 0.8 \times 156.25 = 125\;\Omega.$$

Using $$R = 0.8\,Z$$, we find

$$R = 0.8 \times 125 = 100\;\Omega.$$

The inductive reactance is obtained from

$$Z^{2} = R^{2} + X_{L}^{2} \;\;\Longrightarrow\;\; X_{L} = \sqrt{Z^{2} - R^{2}} = \sqrt{125^{2} - 100^{2}} = \sqrt{15625 - 10000} = \sqrt{5625} = 75\;\Omega.$$

Now a capacitor is added in series to make the overall power factor unity. Unity power factor means the net reactance is zero, so the capacitive reactance must exactly cancel the inductive reactance:

$$X_{C} = X_{L} = 75\;\Omega.$$

For a capacitor, reactance is given by the formula

$$X_{C} = \dfrac{1}{\omega C}, \qquad \omega = 2\pi f = 2\pi \times 50 = 100\pi\;\text{rad/s}.$$

Setting $$X_{C} = 75\;\Omega$$, we get

$$75 = \dfrac{1}{\omega C} \;\;\Longrightarrow\;\; C = \dfrac{1}{\omega \times 75} = \dfrac{1}{100\pi \times 75} = \dfrac{1}{7500\pi}\;\text{F}.$$

To express the capacitance in micro-farads, we multiply by $$10^{6}$$:

$$C = \dfrac{10^{6}}{7500\pi}\;\mu\text{F}.$$

The problem statement writes the capacitance in the form

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

To match this with our value, equate the two numerators:

$$\dfrac{10^{6}}{7500} = \dfrac{n}{3} \;\;\Longrightarrow\;\; n = 3 \times \dfrac{10^{6}}{7500} = \dfrac{3000000}{7500} = 400.$$

So, the answer is $$400$$.