A 50 Hz AC circuit has a 10 mH inductor and a $2 \Omega$ resistor in series. The value of capacitance to be…
A 50 Hz AC circuit has a 10 mH inductor and a $2 \Omega$ resistor in series. The value of capacitance to be placed in series in the circuit to make the circuit power factor as unity is
$1.014 \times 10^{-6} \mathrm{~F}$
$1.014 \times 10^{-3} \mathrm{~F}$
$2.6 \times 10^{-3} \mathrm{~F}$
$4.125 \times 10^{-3} \mathrm{~F}$
Solution
For power factor, $\cos \phi=1$, the LCR circuit is in resonance.
$\begin{aligned}
& \therefore \quad \mathrm{w}=\frac{1}{\sqrt{\mathrm{LC}}} \Rightarrow 2 \pi \mathrm{f}=\frac{1}{\sqrt{\mathrm{LC}}} \\
& \Rightarrow 2 \pi \times 50=\frac{1}{\sqrt{10 \times 10^{-3} \times \mathrm{C}}} \\
& \therefore \quad \mathrm{C}=1.014 \times 10^{-3} \mathrm{~F}
\end{aligned}$