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  • A total charge flows across a resistor during a time interval = in such a way that the current vs. Time graph for 0 to is like the loop of a sine curve in the range 0 to <img alt="\pi" src="https:/

A total charge flows across a resistor during a time interval = in such a way that the current vs. Time graph for 0 to is like the loop of a sine curve in the range 0 to \pi.
The total heat generated in the resistor is

Option: 1

\mathrm{\frac{\frac{}{}Q^2 \pi^2 R}{8 T}}


Option: 2

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


Option: 3

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


Option: 4

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


Answers (1)

best_answer

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



\begin{aligned} &\mathrm{ I=I_0 \sin \omega t=I_0 \sin \left(\frac{\pi}{T} t\right) }\\ &\mathrm{ Q=\int_0^T I d t \Rightarrow Q=I_0 \int_0^T \sin \left(\frac{\pi}{T} t\right) d t }\\ &\mathrm{ =\frac{-I_0}{\pi / T}\left[\cos \left(\frac{\pi}{T} t\right)\right]_0=\frac{2 T I_0}{\pi} }\\ &\mathrm{ =I_0=\frac{Q \pi }{2 T} }\\ &\mathrm{ \text { Heat }=\int_0^T I^2 R d t=\int_0^T \sin ^2\left(\frac{\pi}{T} t\right) R d t=\frac{Q^2 \pi^2 R}{8 T}} \end{aligned}

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Nehul

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