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GATE 2015 EE - SET 1 - Question 11

Solution :

A = 10cm x 5 cm = 50 cm2

A = 20 x 10 -4 m2

B(t) = 0.25 sin\omega t , R = 0.452

P_{avg}=\int_{0}^{T} \frac{V^{2}}{R}

V_{rms} = -\frac{d\phi }{dt}

V = \frac{-d}{dt} (B x A)

V = \frac{-d}{dt}\left [ 0.25sin\omega T * 50 *10^{-4} \right ]

V = \frac{-d}{dt}\left [ sin(2\pi *50t) *50 *10 ^{-4} \right ]

V = 0.25 * 50 * 10^{-4}*2\pi *50*cos(\omega t)

V = \frac{-1 * 100\pi }{800} cos\omega T

V^{2}=\frac{100\pi }{800}^{2}cos^{2}\omega t

V^{2} = \frac{100\pi }{800}^{2}†\left [ \frac{1+cos2\omega t}{2}\right ]

P_{avg} = \int_{0}^{t} =\frac{100\pi }{800}^{2}*\frac{1}{0.4*2}\int_{0}^{t}(1+cos\omega t)

P = \frac{\pi }^{2}{64}*\frac{1}{0.4*2} = 0.193w

 

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