The charge flowing through a resistance R var | vedclass

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(A) The charge flowing through the resistor is given by $Q = 2t - 8t^2$.The current $I$ is the rate of change of charge with respect to time:$I = \fra

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$\frac{R}{6} \ J$

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(A) The charge flowing through the resistor is given by $Q = 2t - 8t^2$.The current $I$ is the rate of change of charge with respect to time:$I = \frac{dQ}{dt} = \frac{d}{dt}(2t - 8t^2) = 2 - 16t$.The heat produced $H$ in a resistor $R$ over a time interval is given by the integral:$H = \int_{0}^{t} I^2 R \ dt$.Substituting the limits $0$ to $\frac{1}{8}$:$H = \int_{0}^{1/8} (2 - 16t)^2 R \ dt$$H = R \int_{0}^{1/8} (4 - 64t + 256t^2) \ dt$$H = R \left[ 4t - 32t^2 + \frac{256t^3}{3} \right]_{0}^{1/8}$Evaluating at the limits:$H = R \left[ 4(\frac{1}{8}) - 32(\frac{1}{8})^2 + \frac{256}{3}(\frac{1}{8})^3 \right]$$H = R \left[ \frac{1}{2} - 32(\frac{1}{64}) + \frac{256}{3}(\frac{1}{512}) \right]$$H = R \left[ \frac{1}{2} - \frac{1}{2} + \frac{1}{6} \right] = \frac{R}{6} \ J$.

The charge flowing through a resistance $R$ varies with time as $Q = 2t - 8t^2$. The total heat produced in the resistance is (for $0 \leq t \leq \frac{1}{8} \ s$)

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