question_answer

A capacitor of 100 \[\mu F\]and resistance of 100\[\Omega \] are connected in series with an a. c source of 50 Hz. The phase angle between the current and source voltage is:

A)  \[{{\tan }^{-1}}(1/\pi )\]     

B)  \[{{\tan }^{-1}}(1/2\pi )\]

C)  \[{{\tan }^{-1}}(\pi /2)\]              

D)         \[{{\tan }^{-1}}(2\pi /2)\]

Correct Answer: A

Solution :

Let the angle between current and source voltage is\[\phi \] \[\therefore \]  \[\tan \phi =\left( \frac{1}{\omega CR} \right)\] or            \[\tan \phi {{\tan }^{-1}}\left( \frac{1}{\omega CR} \right)\] Hence:                                 \[\omega =2\pi f=2\pi \times 50=100\pi \,rad/\sec \]                                 \[\begin{align}   & C=100\mu F=100\times {{10}^{-6}}F=1\times {{10}^{-4}}F \\  & R=100\,\Omega  \\ \end{align}\]           Therefore, \[\phi ={{\tan }^{-1}}\left( \frac{1}{100\mu \times {{10}^{-4}}\times 100} \right)\]                                      \[={{\tan }^{-1}}\left( \frac{1}{\pi } \right)\]