Question

An alternating voltage $v\left(t\right)=220\sin 100\pi tvolt$ is applied to a purely resistive load of $5\Omega$. The time taken for the current to rise from half of the peak value to the peak value is

• A. $2.2ms$
• B. $5ms$
• C. $3.3ms$
• D. $7.2ms$

Answer 1

The correct option is C $3.3ms$

$V\left(t\right)=220\sin \left(100\pi t\right)volt$
$I\left(t\right)=\frac{220}{5}\sin \left(100\pi t\right)volt$
$=44sin\left(100\pi t\right)volt$


Comparing with $I=I_{0}sin\left(\theta \right)$,
$I=\frac{I_0}{2}$ when $\theta ={30}^0$
$I=I_0$ when $\theta ={90}^0$
$\therefore$ phase to be covered $\theta ={60}^0=\frac{\pi }{3}$
Time taken,
$t=\frac{\theta }{\omega }=\frac{\frac{\pi }{3}}{100\pi }=\frac{1}{300}\sec$
$=3.3ms$.