A) more resistance
B) less resistance
C) zero resistance
D) none of these
Correct Answer: B
Solution :
The peak value of current in a capacitor circuit is given by \[{{i}_{0}}=\frac{{{E}_{0}}}{1/\varepsilon c}\] From ohm's law, the quantity \[\frac{1}{\omega c}\] has dimensions of resistances, it represents the effective opposition of capacitor to flow of alternating current. It is capacitive resistance. \[{{X}_{C}}=\frac{1}{\omega c}\] Where \[\omega \] is angular frequency \[(\omega =2\pi f)\] \[\therefore \] \[{{X}_{C}}=\frac{1}{2\pi fc}\] Therefore, from above expression, we note that high frequency capacitor offers less resistance. Alternative Method: At a higher frequency, the charge on the capacitor plate will vary more rapidly, hence, \[\left( \frac{dq}{dt} \right)\] will be higher resulting in an increase in the instantaneous current \[\left( i=\frac{dq}{dt} \right)\]. Hence, lower capacitive resistance.
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