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

Joule's Law of Heating Calculation 

We are given the following equation for \( Q \):

\(Q = at - bt^2\)

The current \( i \) is given by:

\(i = a - 2bt\)

For \( i = 0 \), we solve for \( t \):

\(t = \frac{a}{2b}\)

Using Joule's Law of Heating:

According to Joule's law of heating, the infinitesimal heat \( dH \) is given by:

\(dH = i^2 R dt\)

To find the total heat \( H \), we integrate from \( 0 \) to \( \frac{a}{2b} \):

\(H = \int_0^{\frac{a}{2b}} (a - 2bt)^2 R dt\)

Expanding the integrand and integrating:

\(H = \frac{(a - 2bt)^3 R}{-6b} \Bigg|_0^{\frac{a}{2b}}\)

Substituting the limits:

\(H = \frac{a^3 R}{6b}\)

Conclusion:

The total heat \( H \) is given by \( \frac{a^3 R}{6b} \).