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(A) The current follows the form $I(t) = I_0 \sin(\omega t)$. Since the loop is for $0 \rightarrow T$,we have $\omega T = \pi$,so $\omega = \pi/T$.The

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$Q^2\pi^2R / 8T$

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(A) The current follows the form $I(t) = I_0 \sin(\omega t)$. Since the loop is for $0 \rightarrow T$,we have $\omega T = \pi$,so $\omega = \pi/T$.The total charge $Q$ is the integral of current over time:$Q = \int_0^T I_0 \sin(\frac{\pi}{T} t) dt = I_0 [-\frac{T}{\pi} \cos(\frac{\pi}{T} t)]_0^T = I_0 \frac{T}{\pi} (1 - (-1)) = \frac{2 I_0 T}{\pi}$.Solving for $I_0$,we get $I_0 = \frac{Q \pi}{2T}$.The heat generated $H$ is given by $\int_0^T I^2 R dt$:$H = R \int_0^T I_0^2 \sin^2(\frac{\pi}{T} t) dt = R I_0^2 \int_0^T \frac{1 - \cos(\frac{2\pi}{T} t)}{2} dt$.$H = \frac{R I_0^2}{2} [t - \frac{T}{2\pi} \sin(\frac{2\pi}{T} t)]_0^T = \frac{R I_0^2}{2} [T - 0] = \frac{R I_0^2 T}{2}$.Substituting $I_0 = \frac{Q \pi}{2T}$:$H = \frac{R T}{2} (\frac{Q \pi}{2T})^2 = \frac{R T Q^2 \pi^2}{2 \cdot 4 T^2} = \frac{Q^2 \pi^2 R}{8 T}$.

$A$ 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 sine curve in the range $0 \rightarrow \pi$. The total heat generated in the resistor is

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