An alternating voltage volt is applied to a purely resistive load of 50. The time taken for the current to rise from half of the peak value to the peak value is

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Solution

Q.
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 [2024]

1 5 ms

2 3.3 ms

3 7.2 ms

4 2.2 ms

Ans.
(2)

$V (t) = 220 \sin (100 π t)$

$I = \frac{V_{s}}{R} = \frac{220}{50} \sin (100 π t)$

$I = \frac{22}{5} \sin (100 π t)$

$\cos θ = \frac{\frac{I_{0}}{2}}{I_{0}} = \frac{1}{2} \Rightarrow θ = 60 ° = \frac{π}{3}$

Using phasor, $t = \frac{π / 3}{100 π}$

$t = \frac{1}{300} sec = \frac{10}{3} msec$

$t = 3.3 msec$

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