An alternating voltage $v (t) = 220 \sin 100 π t$ volt is applied to a purely resistive load of 50 $Ω$ .The time taken for the current to rise from half of the peak value to the peak value is (in milli-second)

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answer is 3.33.

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Detailed Solution

As V(t)= $220 \sin 100 π t$
So, I(t)= $\frac{220}{50} \sin 100 π t$
i.e; I= $I_{m} \sin (100 π t)$
For I= $I_{m}$
$t_{1} = \frac{π}{2} \times \frac{1}{100 π} = \frac{1}{200} \sec$
And for I= $\frac{I_{m}}{2}$
$\Rightarrow \frac{I_{m}}{2} = I_{m} \sin (100 π t_{2})$

$\begin{array}{l} \Rightarrow t_{2} = \frac{1}{600} s \\ ∴ t_{r e q} = \frac{1}{200} - \frac{1}{600} = \frac{2}{600} = \frac{1}{300} s = 3.33 m s \end{array}$