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Step-by-Step Solution
Step 1: Understand the Problem
We are given that charge flowing through a resistor R as a function of time is $Q(t) = a\,t - b\,t^2$, where $a$ and $b$ are positive constants. We need to find the total heat generated in the resistor over the period during which current flows.
Step 2: Express the Current as a Function of Time
The current $i(t)$ is obtained by differentiating the charge $Q(t)$ with respect to time:
$i(t) = \frac{dQ(t)}{dt} = \frac{d}{dt} \bigl(a\,t - b\,t^2\bigr) = a - 2b\,t.$
Step 3: Determine the Time Interval Over Which the Current Flows
The current flows as long as $i(t)$ is non-negative. Set $i(t) = 0$ to find when the current ceases to flow:
$ a - 2b\,t = 0 \quad \Rightarrow \quad t = \frac{a}{2b}.$
Thus, the current flows from $t = 0$ to $t = \frac{a}{2b}.$
Step 4: Apply Joule’s Law of Heating
The infinitesimal heat generated $dH$ in a resistor R carrying current $i(t)$ over a small time interval $dt$ is given by:
$dH = i^2(t)\,R\,dt.$
Hence, the total heat $H$ produced in the resistor over the time interval $[0, \frac{a}{2b}]$ is:
$H = \int_{0}^{\frac{a}{2b}} i^2(t)\,R\,dt.$
Substitute $i(t) = a - 2b\,t$ into the integral:
$H = \int_{0}^{\frac{a}{2b}} \bigl(a - 2b\,t\bigr)^2 R \, dt.$
Step 5: Evaluate the Integral
Expand $\bigl(a - 2b\,t\bigr)^2$:
$\bigl(a - 2b\,t\bigr)^2 = a^2 - 4ab\,t + 4b^2\,t^2.$
So the integral becomes:
$H = R \int_{0}^{\frac{a}{2b}} \left(a^2 - 4ab\,t + 4b^2\,t^2\right) dt.$
Integrate term by term:
$\int a^2 \, dt = a^2 t$
$\int (-4ab\,t)\,dt = -4ab \cdot \frac{t^2}{2} = -2ab\,t^2$
$\int 4b^2\,t^2 \, dt = 4b^2 \cdot \frac{t^3}{3} = \frac{4b^2\,t^3}{3}.$
Hence,
$\displaystyle H = R \left[ a^2 t - 2ab\,t^2 + \frac{4b^2\,t^3}{3} \right]_{0}^{\frac{a}{2b}}.$
Now substitute $t = \frac{a}{2b}$:
$t = \frac{a}{2b}, \quad t^2 = \frac{a^2}{4b^2}, \quad t^3 = \frac{a^3}{8b^3}.$
$\begin{aligned} H &= R \Biggl[ a^2 \left(\frac{a}{2b}\right) \;-\; 2ab \left(\frac{a^2}{4b^2}\right) \;+\; \frac{4b^2}{3} \left(\frac{a^3}{8b^3}\right) \Biggr] \\ &= R \Biggl[\frac{a^3}{2b} - \frac{2ab \cdot a^2}{4b^2} + \frac{4b^2 \cdot a^3}{3 \cdot 8b^3}\Biggr] \\ &= R \Biggl[\frac{a^3}{2b} - \frac{2a^3}{4b} + \frac{4 a^3}{24b}\Biggr] \\ &= R \Biggl[\frac{a^3}{2b} - \frac{a^3}{2b} + \frac{a^3}{6b}\Biggr] \\ &= \frac{a^3 R}{6b}. \end{aligned} $
Step 6: State the Final Result
Therefore, the total heat produced in the resistor R during the time the current flows is:
$\displaystyle H = \frac{a^3 R}{6b}.$
