Question

An $AC$ voltage source described by $V=10\cos \left(\frac{\pi }{2}\right)t$ is connected to a $1\mu \text{F}$ capacitor as shown in figure. The key $K$ is closed at $t=0$. The time $(t>0)$ after which the magnitude of current $I$ reaches its maximum value for the first time is

• A. $1\text{s}$
• B. $2\text{s}$
• C. $3\text{s}$
• D. $4\text{s}$

Answer 1

The correct option is A $1\text{s}$

Given

$V=10\cos \left(\frac{\pi }{2}\right)t$ $\Rightarrow \omega =\frac{\pi }{2}$

In capacitor, current leads the voltage by $\frac{\pi }{2}$.

$\therefore$ current is given by

$I=I_0\cos (\frac{\pi }{2}t+\frac{\pi }{2})=-I_0\sin \left(\frac{\pi }{2}t\right)$


Current reaches to $I_{max}=I_0$ for the first time at

$t=\frac{T}{4}=\frac{2\pi }{\omega }\times \frac{1}{4}=\frac{\pi }{2\omega }$

As, $\omega =\frac{\pi }{2}$

$\therefore t=1\text{s}$

Hence, option $\left(A\right)$ is the correct answer.