An alternating voltage

Question:

An alternating voltage $v(t)=220 \sin 100 \pi t$ volt is applied to a purely resistive load of $50 \Omega$. The time taken for the current to rise from half of the peak value to the peak value is :

  1. (1) $5 \mathrm{~ms}$

  2. (2) $2.2 \mathrm{~ms}$

  3. (3) $7.2 \mathrm{~ms}$

  4. (4) $3.3 \mathrm{~ms}$


Correct Option: , 4

Solution:

(4) As $V(t)=220 \sin 100 \pi t$

so, $I(t)=\frac{220}{50} \sin 100 \pi \mathrm{t}$

i.e., $I=I_{m}=\sin (100 \pi \mathrm{t})$

For $I=I_{m}$

$t_{1}=\frac{\pi}{2} \times \frac{1}{100 \pi}=\frac{1}{200} \mathrm{sec}$

and for $I=\frac{I_{m}}{2}$

$\Rightarrow \frac{I_{m}}{2}=I_{m} \sin \left(100 \pi t_{2}\right)$

$\Rightarrow \frac{\pi}{6}=100 \pi t_{2} \Rightarrow t_{2}=\frac{1}{600} s$

$\therefore t_{\text {req }}=\frac{1}{200}-\frac{1}{600}=\frac{2}{600}=\frac{1}{300} s=3.3 \mathrm{~ms}$

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