Determining Total Heat Produced in Resistance R

This solution explains how to calculate the total heat produced in a resistance $R$ when the charge $Q$ flowing through it changes over time $t$ according to the given formula. We will find the moment the current becomes zero and then integrate the power dissipated over that time interval.

Step-by-Step Calculation of Heat Produced

The problem provides the relationship between charge $Q$ and time $t$ as $Q = At^2 - Bt^3$, where $A$ and $B$ are positive constants. We need to find the total heat produced until the current becomes zero.

1. Finding the Instantaneous Current $I(t)$

Current is defined as the rate of change of charge with respect to time. We can find the expression for the current $I(t)$ by differentiating the charge $Q(t)$ with respect to time $t$.

Given: $Q(t) = At^2 - Bt^3$

The current $I(t)$ is:

$ I(t) = \frac{dQ}{dt} = \frac{d}{dt}(At^2 - Bt^3) $

$ I(t) = 2At - 3Bt^2 $

2. Finding the Time When Current Becomes Zero

The current becomes zero at the beginning ($t=0$) and potentially at a later time. We need to find this non-zero time, let's call it $t_{zero}$. We set the expression for $I(t)$ to zero:

$ 2At - 3Bt^2 = 0 $

Factor out $t$:

$ t(2A - 3Bt) = 0 $

This equation gives two solutions for $t$: $t=0$ and $2A - 3Bt = 0$. We are interested in the time after the start, so we solve the second part:

$ 2A = 3Bt $

$ t_{zero} = \frac{2A}{3B} $

This is the time until which we need to calculate the heat produced.

3. Calculating the Total Heat Produced $H$

The heat $H$ produced in a resistance $R$ is calculated by integrating the power dissipated ($P = I^2 R$) over the time interval.

The formula for total heat is:

$ H = \int_0^{t_{zero}} P \, dt = \int_0^{t_{zero}} I(t)^2 R \, dt $

Substitute the expressions for $I(t)$ and $t_{zero}$:

$ H = R \int_0^{2A/(3B)} (2At - 3Bt^2)^2 \, dt $

First, expand the square term:

$ (2At - 3Bt^2)^2 = (2At)^2 - 2(2At)(3Bt^2) + (3Bt^2)^2 $

$ = 4A^2t^2 - 12ABt^3 + 9B^2t^4 $

Now substitute this back into the integral:

$ H = R \int_0^{2A/(3B)} (4A^2t^2 - 12ABt^3 + 9B^2t^4) \, dt $

Perform the integration:

$ H = R \left[ \frac{4A^2t^3}{3} - \frac{12ABt^4}{4} + \frac{9B^2t^5}{5} \right]_0^{2A/(3B)} $

$ H = R \left[ \frac{4A^2t^3}{3} - 3ABt^4 + \frac{9B^2t^5}{5} \right]_0^{2A/(3B)} $

Now, evaluate the expression at the upper limit $t = \frac{2A}{3B}$ (the value at $t=0$ is zero):

$ H = R \left( \frac{4A^2}{3} \left(\frac{2A}{3B}\right)^3 - 3AB \left(\frac{2A}{3B}\right)^4 + \frac{9B^2}{5} \left(\frac{2A}{3B}\right)^5 \right) $

Let's calculate the powers:

$ \left(\frac{2A}{3B}\right)^3 = \frac{8A^3}{27B^3} $

$ \left(\frac{2A}{3B}\right)^4 = \frac{16A^4}{81B^4} $

$ \left(\frac{2A}{3B}\right)^5 = \frac{32A^5}{243B^5} $

Substitute these back into the expression for $H$:

$ H = R \left( \frac{4A^2}{3} \cdot \frac{8A^3}{27B^3} - 3AB \cdot \frac{16A^4}{81B^4} + \frac{9B^2}{5} \cdot \frac{32A^5}{243B^5} \right) $

Simplify each term:

$ H = R \left( \frac{32A^5}{81B^3} - \frac{48A^5}{81B^3} + \frac{288A^5}{1215B^3} \right) $

To combine these terms, find a common denominator, which is $1215$. Note that $81 \times 15 = 1215$.

$ H = R \left( \frac{32 \times 15 A^5}{1215B^3} - \frac{48 \times 15 A^5}{1215B^3} + \frac{288 A^5}{1215B^3} \right) $

$ H = \frac{R A^5}{1215B^3} (32 \times 15 - 48 \times 15 + 288) $

$ H = \frac{R A^5}{1215B^3} (480 - 720 + 288) $

$ H = \frac{R A^5}{1215B^3} (48) $

Simplify the fraction $\frac{48}{1215}$. Both numbers are divisible by 3:

$ \frac{48}{3} = 16 $

$ \frac{1215}{3} = 405 $

So, the fraction simplifies to $\frac{16}{405}$.

$ H = \frac{16 R A^5}{405 B^3} $

Conclusion

The total heat produced in the resistance $R$ until the current becomes zero is $\frac{16A^5R}{405B^3}$. This corresponds to the first option.