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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 sine curve in the range 0 to <img alt="\pi" src="

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 sine curve in the range 0 to \pi. The total heat generated in the resistor is

Option: 1

Q^2 \pi^2 R / 8 T


Option: 2

2 Q^2 \pi^2 R / T


Option: 3

\mathrm{Q}^2 \pi^2 \mathrm{R} / \mathrm{T}


Option: 4

Q^2 \pi^2 R / 2 T


Answers (1)

best_answer

\text { Time period: } T_0=2 T, \omega=\frac{2 \pi}{T_0}=\frac{\pi}{T}

\begin{aligned} & I=I_0 \sin \omega t=I_0 \sin \left(\frac{\pi}{T} t\right) \\ & Q=\int_0^T I d t \Rightarrow Q=I_0 \int_0^T \sin \left(\frac{\pi}{T} t\right) d t \\ & =\frac{-I_0}{\pi / T}\left[ \cos \left(\frac{\pi}{T} t\right)\right]_0=\frac{2 T I_0}{\pi} \\ & \Rightarrow I_0=Q \pi / 2 T \end{aligned}

\text { Heat }=\int_0^T I^2 R d t=\int_0^T\left(\frac{\pi}{T} t\right) R d t=Q^2 \pi^2 R / 8 T

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sudhir.kumar

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