Answer:
B) leads the voltage by [tex]90[/tex] degree.
Explanation:
The capacitor having capacitance [tex]C[/tex] is connected with the AC power supply as shown in the figure.
Let the voltage of the power source be
[tex]V= V_0 \sin (\omega t)\; \cdots (i)[/tex],
where [tex]\omega[/tex] is the angular frequency, [tex]t[/tex] represents time and [tex]V_0[/tex] is the maximum voltage. Here, [tex]\omega[/tex] and [tex]V_0[/tex] is constants.
At any instant, let [tex]q[/tex] be the charge on the capacitor and as the potential difference across the capacitor is [tex]V[/tex], so [tex]q=CV[/tex].
[tex]\Rightarrow q= CV_0 \sin (\omega t)\; \cdots (ii)[/tex]
As the current, [tex]i[/tex] , in the circuit is the rate of change of charge, [tex]q[/tex], so
[tex]i=\frac {dq}{dt}[/tex]
[tex]\Rightarrow i= \frac{d}{dt}(CV_0 \sin (\omega t))[/tex] [from equation[tex](ii)[/tex]]
[tex]\Rightarrow i= C\omega V_0 \cos (\omega t)[/tex]
[tex]\Rightarrow i= C\omega V_0 \cos\left(\omega t+\frac{\pi}2\right)\;\cdots(iii)[/tex]
Now, from equations [tex](i)[/tex] and [tex](iii)[/tex], the phase difference between current and the voltage is [tex]\frac {\pi}{2}[/tex] where current is leading.
So, the current leads the voltage by [tex]\frac {\pi}{2}[/tex] or [tex]90[/tex] degree.
Hence, option [tex](b)[/tex] is correct.
