Question

The charge flowing through a resistor $R$ varies with time $t$ as $Q=3t-6t^2.$ The heat produced in $R$ till the current in it becomes zero is

• A. $\frac{3R}{4}$
• B. $\frac{3R}{2}$
• C. $\frac{4R}{2}$
• D. $\frac{9R}{2}$

Answer 1

The correct option is A $\frac{3R}{4}$

Given that the charge flowing through a resistor $R$ varies with time $t$ is: $Q=3t-6(t{)}^2$
Hence, the current $I$ will be: $I=\frac{dQ}{dt}=3-12t$
The heat produced in $R$ till the current in it becomes zero is
$H={\int }_0^{\frac{1}{4}}{I}^{2}Rdt$

$H={\int }_0^{\frac{1}{4}}(3-12t{)}^{2}Rdt$

$H=[-\frac{(3-12t{)}^3}{3\times 12}{]}_0^{\frac{1}{4}}R$ $=\frac{27R}{36}=\frac{3R}{4}$