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A total charge Q flows across a resistor R during a time interval T in such a way that the current vs. time graph for 0 to T is like the loop of a sin curve in the range 0 to $π$ . The total heat generated in the resistor is

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$Q^{2} π^{2} R / 8 T$

$2 Q^{2} π^{2} R / T$

$2 Q^{2} π^{2} R / 4 T$

$Q^{2} π^{2} R / 4 T$

answer is D.

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Detailed Solution

$I = I_{0} \sin ω t$

$Q = \int Idt = I_{0} \int_{0}^{T / 2} \sin ω tdt = I_{0} {[\frac{- \cos ω t}{ω}]}_{0}^{T / 2}$

$= \frac{I_{0}}{(2 π / T)} [- \cos ω t]_{0}^{T / 2} = \frac{I_{0} T}{π}$

$\Rightarrow I_{0} = \frac{Q π}{T} \dots . (1)$

Total heat $H = \int I^{2} Rdt = R \int {(I_{0} \sin ω t)}^{2} dt$

$= I_{0}^{2} R \int_{0}^{T / 2} (\frac{1 - \cos 2 ω t}{2}) dt$

$= \frac{I_{0}^{2} R}{2} {[t - \frac{\sin 2 ω t}{2 ω}]}_{0}^{T / 2}$

$= \frac{I_{0}^{2} R}{2} [T / 2 - 0 - 0 + 0]$

$= \frac{I_{0}^{2} RT}{4} = \frac{Q^{2} π^{2}}{T^{2}} \times \frac{RT}{4} \Rightarrow H = \frac{Q^{2} π^{2} R}{4 T} .$

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