Question

In a series LR circuit, power of $400W$ is dissipated from a source of $250V,50Hz.$ 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 $\left(\frac{n}{3 \pi }\right)μF,$ then value of $n$ is ________.

Answer 1

Given in $1^{s t}$ case power factor of LR circuit
$cos\phi=0.8=\frac{R}{\sqrt{R^{2} + X_{L}^{2}}}$
where $\sqrt{R^{2} + X_{L}^{2}}=Z\ldots \left(1\right)$
$\therefore P=VIcos\phi$
$\Rightarrow 400=\left(250\right)^{2}\times \frac{R}{Z^{2}}$
$400=\left(250\right)^{2}\times \frac{0 . 8}{Z}$
$Z=125...\left(2\right)$
$R=\frac{\left(250\right)^{2} \times 0 . 8 \times 0 . 8}{400}\Rightarrow 100\Omega...\left(3\right)$
from $\left(1\right),\left(2\right)$ and $\left(3\right)$
$\left(100+ X _{ L }^{2}=(125\right.$
$X_{L}^{2}=15625-10000$
$X_{L}^{2}=5625$
$X_{L}=75...\left(4\right)$
In $2^{n d}$ case given,
Power factor $=1$ that means $X_{L}=X_{c}$ (Resonance condition)
$X_{L}=\frac{1}{\left(\omega \right)_{C}}\Rightarrow 75=\frac{1}{\left(2 \pi F\right) C}$
$C=\frac{1}{2 \pi \times F \times 75}$
$C=\frac{1}{2 \pi \times 50 \times 75}F\ldots \left(5\right)$
$C=\frac{n}{3 \pi }μF$ given
From $\left(5\right)\&\left(6\right)$
$\frac{1}{2 \pi \times 50 \times 75}=\frac{n \times 10^{- 6}}{3 \pi }$
$n=\frac{10^{6}}{7500}\Rightarrow \frac{3 \times 10^{4}}{75}\Rightarrow \frac{30000}{75}$
$n=400$