GATE 2015 EE – SET 1 – Question 11
GATE 2015 EE – SET 1 – Question 11,Answer Key & Full Solutions to GATE 2017 Questions Papers for All branches .Full Solutions to GATE 2017 Questions Papers
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 GATE 2015 EE – SET 1 – Question 11
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Solution :
A = 10cm x 5 cm = 50 cm^{2}
A = 20 x 10^{ 4} m^{2}
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