Question

The charge flowing through a resistance $R$ varies with time t as $ Q = at - bt^2$ , where $a$ and $b$ are positive constants, The total heat produced in $R$ is :

• A. $\frac{a^3 R}{3b}$
• B. $\frac{a^3 R}{2b}$
• C. $\frac{a^3 R}{b}$
• D. $\frac{a^3 R}{6b}$

Answer 1

The Correct Option is D

To determine the total heat produced in the resistor $R$, we need to follow these steps:

1. First, recognize that the charge $Q$ passing through the resistor over time is given by:

Q = at - bt^2

2. To find current $I(t)$, differentiate $Q$ with respect to time $t$:

I(t) = \frac{dQ}{dt} = a - 2bt

3. The power dissipated across the resistor, which is equivalent to the rate of heat production, is given by $P(t) = I^2(t)R$:

P(t) = (a - 2bt)^2 R

4. Now, the total heat produced over time is the integral of $P(t)$ with respect to time. Therefore, we need to integrate the expression:

\int P(t) \, dt = \int (a - 2bt)^2 R \, dt

5. Executing the integration:

\int (a - 2bt)^2 \, dt = \int (a^2 - 4abt + 4b^2t^2) \, dt

= a^2t - 4ab \frac{t^2}{2} + 4b^2 \frac{t^3}{3}

= a^2t - 2abt^2 + \frac{4b^2t^3}{3}

6. We determine the limits from $t = 0$ to when I(t) = 0 (i.e., when the current flow stops). Solving a - 2bt = 0 gives the upper limit t = \frac{a}{2b}.
7. Substitute these limits into the integral to find the total heat H:

H = \left[ a^2t - 2abt^2 + \frac{4b^2t^3}{3} \right]_0^{\frac{a}{2b}} \cdot R

= \left[ a^2 \frac{a}{2b} - 2ab \left( \frac{a}{2b} \right)^2 + \frac{4b^2}{3} \left( \frac{a}{2b} \right)^3 \right] \cdot R

= \left[ \frac{a^3}{2b} - \frac{a^3}{2b} + \frac{a^3}{6b} \right] \cdot R

= \frac{a^3}{6b} \cdot R

7. Thus, the total heat produced in the resistor is: \frac{a^3R}{6b}.