$A$ total charge $Q$ flows across a resistor $R$ during a time interval $= T$ in such a way that the current vs. time graph for $0 \rightarrow T$ is like the loop of a sin curve in the range $0 \rightarrow \pi$ . The total heat generated in the resistor is
Diffcult
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Time period: $T_0=2 T, \omega=\frac{2 \pi}{T_0}=\frac{\pi}{T}$
$I=I_0 \sin \omega t=I_0 \sin \left(\frac{\pi}{T} t\right)$
$Q =\int \limits_0^{ T } Idt \Rightarrow Q = I _0 \int\limits_0^{ T } \sin \left(\frac{\pi}{ T } t \right) d t$
$=\frac{-I_0}{\pi T}\left[\cos \left(\frac{\pi}{T} t \right)\right]_0=\frac{2 TI _0}{\pi}$
$\Rightarrow \quad I _0= Q \pi / 2 T$
$\text { Heat }=\int_0^T I^2 R d t=\int_0^T I_0^2 \sin ^2\left(\frac{\pi}{T} t\right) R d t=\frac{Q^2 \pi^2 R}{8 T}$
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