GATE 2015 EE – SET 1 – Question 11

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

GATE 2015 EE - SET 1 - Question 11

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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