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 $\frac{b^{3}R}{6a}$

• A. True
• B. False

Answer 1

The correct option is B False

Given: $Q=at-b{t}^2$

$\Rightarrow i=\frac{dQ}{dt}=a-2bt$

Therefore, current will be zero at $t=\frac{a}{2b}$

We know that: $dH=i^{2}Rdt$

$\Rightarrow H={\int }_0^{a/2b}(a-2bt{)}^{2}Rdt$

$\Rightarrow H=\left[\frac{(a-2b(\frac{a}{2b}){)}^{3}R}{-3\times 2b}\right]$$-\left[\frac{(a-2b(0){)}^{3}R}{-3\times 2b}\right]$

$\Rightarrow H=\frac{a^{3}R}{6b}$