Question

An inductor $200mH$, capacitor $500\mu F$, resistor $10\Omega$ are connected in series with a $100V$, variable frequency ac source.
Calculate :
(i) frequency at which power factor of circuit its unity
(ii) current amplitude at this frequency
(iii) Q-factor.

Answer 1

Solution:-

Given:-$L=200\times {10}^{-3}\mu$

$C=500\times {10}^{-6}$F

$R=10\Omega$ V = 100V

i) For unity power factor

Z=R

$X_C=X_L\Rightarrow \frac{1}{\omega C}=\omega L$

$\omega =\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{200\times {10}^{-3}\times 500\times {10}^{-}6}}$

$\omega =100Hz$

ii)Current amplitude at this frequency will be

I=$\frac{V}{R}=\frac{100}{10}=10A$

iii)Q-factor = $\frac{1}{R}\sqrt{\frac{L}{C}}=\frac{1}{10}\sqrt{\frac{200\times {10}^{-3}}{500\times 10-6}}=2$

Q=2