question_answer

An AC supply of 100 V is applied to a capacitor of capacitance 20 \[\mu \text{F}.\]If the current in the circuit is 0.628 A, the frequency of AC. must be

A)  50 Hz                                   

B)  60 Hz

C)  25 Hz                                   

D)  40 Hz

Correct Answer: A

Solution :

                 Capacitive reactance \[{{X}_{C}}\] is given by \[{{X}_{C}}=\frac{1}{\omega C}\] where \[\omega \]  is angular frequency \[\omega =2\pi \,f\]                 \[{{V}_{C}}=100\,V\]                 \[\therefore \]  \[{{X}_{C}}=\frac{1}{2\,\pi fC}\]                 ? (i) Also, \[{{X}_{C}}\] is given by                 \[{{X}_{C}}=\frac{{{V}_{C}}}{I}\]                                ... (ii)                 \[\therefore \] From Eqs. (i) and (ii), we get                 \[\frac{1}{2\,\pi fC}=\frac{{{V}_{C}}}{I}\] \[\Rightarrow \]               \[f=\frac{I}{2\pi {{V}_{c}}C}\] Given,   \[I=0.628A\],      \[{{V}_{C}}=100\,V\], \[\Rightarrow \]               \[f=\frac{0.628}{2\times 3.14\times 20\times {{10}^{-6}}\times 100}\] \[\Rightarrow \]               \[f=50\,\,Hz\]