Question

A capacitance of $\left(\frac{{10}^{-3}}{2\pi }\right)F$ and an inductance of $\left[\frac{100}{\pi }\right]mH$ and a resistance of $10\Omega$ are connected in series with an $AC$ voltage source of $220V,50Hz$. The phase angle of the circult is

• A. ${60}^{\circ }$
• B. ${30}^{\circ }$
• C. ${45}^{\circ }$
• D. ${90}^{\circ }$

Answer 1

The correct option is C ${45}^{\circ }$

Given - $C=\left(\frac{{10}^{-3}}{2\pi }\right)F;L=\left(\frac{100}{\pi }\right)mH=\frac{{10}^{-1}}{\pi }H;R=10\Omega ;n=50Hz$

The phase angle of a series $LCR$ circuit is given by ,

$\tan \varphi =\frac{\omega L-1/\omega C}{R}$ ..............eq1

Now , $\omega C=2\pi nC=2\pi \times 50\times \frac{{10}^{-3}}{2\pi }=1/20$

and $\omega L=2\pi nL=2\pi \times 50\times \frac{{10}^{-1}}{\pi }=10$

Putting these values in eq1 , we get

$\tan \varphi =\frac{10-20}{10}=1$

or $\tan \varphi =1=\tan 45$

or $\varphi ={45}^o$