| An ac voltage source \[V\,=\,{{V}_{0}}\,\sin \,\,\omega t\] is connected across resistance R and capacitance C as shown in figure. It is given that \[R\,=\,\frac{1}{\omega \,C}\]. The peak current is \[{{I}_{0}}\]. If the angular frequency of the voltage source is changed to \[\frac{\omega }{\sqrt{3}}\] then the new peak current in the circuit is : |
 |
A) \[\frac{{{I}_{0}}}{2}\]
B) \[\frac{{{I}_{0}}}{\sqrt{2}}\]
C) \[\frac{{{I}_{0}}}{\sqrt{3}}\]
D) \[\frac{{{I}_{0}}}{3}\]
Correct Answer: B
Solution :
| The peak value of the current is\[{{I}_{0}}=\frac{{{V}_{0}}}{\sqrt{{{R}^{2}}+\frac{1}{{{\omega }^{2}}{{C}^{2}}}}}=\frac{{{V}_{0}}}{\sqrt{2}\,R}\]when the angular frequency is changed to\[\frac{\omega }{\sqrt{3}}\] |
| The new peak value is \[{{I}_{0}}'=\frac{{{V}_{0}}}{\sqrt{{{R}^{2}}+\frac{3}{{{\omega }^{2}}{{C}^{2}}}}}=\frac{{{V}_{0}}}{\sqrt{4{{R}^{2}}}}=\frac{{{V}_{0}}}{2R}\] |
| \[\therefore \] \[{{I}_{0}}'=\frac{{{I}_{0}}}{\sqrt{2}}\] |
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