Question

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

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

Answer 1

The correct option is A $\frac{a^{3}R}{6b}$

$I=\frac{dQ}{dt}=a-2bt$
Current becomes zero at $t=\frac{a}{2b}$
Heat $=R\int I^{2}dt=R\int (a^2+4b^2{t}^2-4abt)dt$
$=R{[a^{2}t+4b^2\frac{t^3}{3}-2ab{t}^2]}_0^{a/2b}$

$=R[a^2\frac{a}{2b}+\frac{4}{3}{b}^2\frac{a^3}{8b^3}-2ab\frac{a^2}{4b^2}]$

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