Alternating Current

JEE Main level, single correct, medium

45 of 190

Question 45 of 190

An alternating voltage V(t)=220sin⁡(100πt)V(t)=220\sin \left(100\pi t\right) volt is applied to a purely resistive load of 50Ω50\Omega . The time taken for the current to rise from half of the peak value to the peak value is:

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Answer: (D) 3.33ms3.33ms

i=VR=22050sin⁡⁡(100πt)i=\dfrac{V}{R}=\dfrac{220}{50}\sin ⁡\left(100\pi t\right)
i=imaxsin⁡⁡(100πt)\text{i}={i}_{max}\sin ⁡\left(100\pi t\right)
For i=imax2:\text{i}=\dfrac{{i}_{max}}{2}:
imax2=imaxsin⁡⁡(100πt)\dfrac{{i}_{max}}{2}={i}_{max}\sin ⁡\left(100\pi t\right)
sin⁡⁡100πt=12⇒100πt=π6\sin ⁡100\pi t=\dfrac{1}{2} \Rightarrow 100\pi t=\dfrac{\pi }{6}
t=1600st=\dfrac{1}{600}s
For i=imax:i={i}_{max}:
imax=imaxsin⁡⁡(100πt){i}_{max}={i}_{max}\sin ⁡\left(100\pi t\right)
sin⁡⁡100πt=1⇒100πt=π2\sin ⁡100\pi t=1\Rightarrow 100\pi t=\dfrac{\pi }{2}
⇒t=1200s\Rightarrow t=\dfrac{1}{200}s
so time for imax2toimax\dfrac{{i}_{max}}{2}to{i}_{max} is
t(imax)−t(imax⁡2)=1200−1600=1300=3.33ms{t}_{({i}_{max})}-{t}_{\left(\dfrac{{i}_{\max }}{2}\right)}=\dfrac{1}{200}-\dfrac{1}{600}=\dfrac{1}{300}=3.33m{s}^{}

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