question_answer

If a circuit made up of a resistance 10, and inductance 0.01H, and alternating emf 200V at 50Hz is connected, then the phase difference between the current and the emf in the circuit is:

A) \[{{\tan }^{-1}} \left( \pi  \right)\]          

B) \[{{\tan }^{-1}} \left( \frac{\pi }{2} \right)\]

C) \[{{\tan }^{-1}} \left( \frac{\pi }{4} \right)\]      

D) \[{{\tan }^{-1}} \left( \frac{\pi }{3} \right)\]

Correct Answer: A

Solution :

Capacitors \[{{\operatorname{C}}_{1}}\,\,and\,\,{{C}_{2}}\]are in series with \[{{C}_{3}}\] in parallel with them \[{{\operatorname{C}}_{1}}=\frac{{{K}_{1}}A{{\varepsilon }_{o}}\times 2}{2d}=\frac{{{k}_{1}}A{{\varepsilon }_{o}}}{\operatorname{d}}\] \[{{\operatorname{C}}_{2}}=\frac{{{K}_{2}}A\times 2{{\varepsilon }_{o}}}{2d}=\frac{{{k}_{2}}A{{\varepsilon }_{o}}}{\operatorname{d}}\] \[{{\operatorname{C}}_{1}}=\frac{{{k}_{1}}A{{\varepsilon }_{o}}}{2\operatorname{d}}\] \[{{\operatorname{C}}_{\operatorname{eq}}}={{\operatorname{C}}_{3}}+\frac{{{\operatorname{C}}_{1}}{{\operatorname{C}}_{2}}}{{{\operatorname{C}}_{1}}+{{\operatorname{C}}_{2}}}\] \[{{\operatorname{C}}_{\operatorname{eq}}}=\frac{A{{\varepsilon }_{o}}}{2d}=\left[ \frac{{{k}_{3}}}{2}+\frac{{{k}_{1}}{{k}_{2}}}{{{k}_{1}}+{{k}_{2}}} \right]\]