# In a series LR circuit, power of 400W is dissipated from a source of 250 V

Question:

In a series LR circuit, power of $400 \mathrm{~W}$ is dissipated from a source of $250 \mathrm{~V}, 50 \mathrm{~Hz}$. The power factor of the circuit is $0.8$. In order to bring the power factor to unity, a capacitor of value $\mathrm{C}$ is added in series to the $\mathrm{L}$ and $\mathrm{R}$. Taking the value

$\mathrm{C}$ as $\left(\frac{\mathrm{n}}{3 \pi}\right) \mu \mathrm{F}$, then value of $\mathrm{n}$ is

Solution:

$(400)$

Given: Power $P=400 \mathrm{~W}$, Voltage $V=250 \mathrm{~V}$

$P=V_{m} \cdot I_{\mathrm{rms}} \cdot \cos \phi$

$\Rightarrow 400=250 \times I_{\mathrm{ms}} \times 0.8 \Rightarrow I_{\mathrm{rms}}=2 \mathrm{~A}$

Using $P=I_{\mathrm{rms}}^{2} R$

$\left(I_{\mathrm{rms}}\right)^{2} \cdot R=P \Rightarrow 4 \times R=400$

$\Rightarrow R=100 \Omega$

Power factor is, $\cos \phi=\frac{R}{\sqrt{R^{2}+X_{L}^{2}}}$

$\Rightarrow 0.8=\frac{100}{\sqrt{100^{2}+X_{L}^{2}}} \Rightarrow 100^{2}+X_{L}^{2}=\left(\frac{100}{0.8}\right)^{2}$

$\Rightarrow X_{L}=\sqrt{-100^{2}+\left(\frac{100}{0.8}\right)^{2}} \Rightarrow X_{L}=75 \Omega$

When power factor is unity,

$X_{C}=X_{L}=75 \Rightarrow \frac{1}{\omega C}=75$

$\Rightarrow C=\frac{1}{75 \times 2 \pi \times 50}=\frac{1}{7500 \pi} F$

$=\left(\frac{10^{6}}{2500} \times \frac{1}{3 \pi}\right) \mu F=\frac{400}{3 \pi} \mu F$

$N=400$