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An amount of charge Q passes through a coil of resistance R. If the current in the coil decreases to zero at a uniform rate during time \(T\), then the amount of heat generated in the coil will be,
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Verified Answer
The correct answer is:
\(\frac{4 Q^2 R}{3 T}\)
Hint : 
Given, \(\frac{1}{2} \mathrm{I}_0 T=Q \Rightarrow \mathrm{I}_0=\frac{2 \mathrm{Q}}{\mathrm{T}}\)
Equation of \(\mathrm{I}(\mathrm{t}) \Rightarrow \frac{\mathrm{I}}{\mathrm{I}_0}+\frac{\mathrm{t}}{\mathrm{T}}=1\)
\(\begin{aligned}
& I=I_0\left(1-\frac{t}{T}\right)=\frac{2 Q}{T}\left(1-\frac{t}{T}\right) \\
& \text { Heat }=\int_0^T I^2 R d f \\
& =R \int_0^T \frac{4 Q^2}{T^2}\left(1-\frac{t}{T}\right)^2 \\
& =\frac{4 Q^2 R}{T^2}\left[\int_0^T d t+\frac{1}{T^2} \int_0^T t^2 d t-\frac{2}{T} \int_0^T t d t\right]=\frac{4 Q^2 R}{T^2}\left[\not T+\frac{T}{3}-\not T\right]=\frac{4 Q^2 R}{3 T}
\end{aligned}\)
Given, \(\frac{1}{2} \mathrm{I}_0 T=Q \Rightarrow \mathrm{I}_0=\frac{2 \mathrm{Q}}{\mathrm{T}}\)
Equation of \(\mathrm{I}(\mathrm{t}) \Rightarrow \frac{\mathrm{I}}{\mathrm{I}_0}+\frac{\mathrm{t}}{\mathrm{T}}=1\)
\(\begin{aligned}
& I=I_0\left(1-\frac{t}{T}\right)=\frac{2 Q}{T}\left(1-\frac{t}{T}\right) \\
& \text { Heat }=\int_0^T I^2 R d f \\
& =R \int_0^T \frac{4 Q^2}{T^2}\left(1-\frac{t}{T}\right)^2 \\
& =\frac{4 Q^2 R}{T^2}\left[\int_0^T d t+\frac{1}{T^2} \int_0^T t^2 d t-\frac{2}{T} \int_0^T t d t\right]=\frac{4 Q^2 R}{T^2}\left[\not T+\frac{T}{3}-\not T\right]=\frac{4 Q^2 R}{3 T}
\end{aligned}\)
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