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 to the peak value to the peak value is :

  • Option 1)

    7.2\; ms

  • Option 2)

    2.2\; ms

  • Option 3)

    5\; ms

     

  • Option 4)

    3.3\; ms

Answers (1)

V=220\; \sin (100\: \pi\: t)

w=100\: \pi

T=\frac{2\pi }{w}=\frac{2\pi }{100\pi }=\frac{1}{50}sec.

R=50\Omega

Let time taken from half of peak value to the peak value =\Delta t

\Delta t=\frac{T}{6}=\frac{1}{50\times 6}=\frac{1}{300}sec

\Delta t=\frac{10}{3}ms=3.3 \; ms
 


Option 1)

7.2\; ms

Option 2)

2.2\; ms

Option 3)

5\; ms

 

Option 4)

3.3\; ms

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