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Confused! kindly explain, - Electromagnetic Induction and Alternating currents - JEE Main-3

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)

$3.3\; ms$

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

$w=100\: \pi$

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

$R=50\Omega$

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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