Question

Current in a resistor $R$ linearly decreases from some value to zero in time $\Delta t$. Total charge that crosses the cross-section is $q$. Choose the correct option(s).

• A. Heat generated in the resistance during this process is $\frac{4}{3}\frac{q^{2}R}{\Delta t}$.
• B. Heat generated in the resistance during this process is $\frac{2}{3}\frac{q^{2}R}{\Delta t}$.
• C. Maximum current is $i_0=\frac{2q}{\Delta t}$.
• D. Maximum current is $i_0=\frac{q}{\Delta t}$.

Answer 1

Answer: (A), (C)

Maximum current is $i_0=\frac{2q}{\Delta t}$.

The variation of current through the resistor is given below.


Also, given that charge flow is $q$.
As $q=\int idt$,
Area under $i–t$ curve,
$q=\frac{1}{2}\times \Delta t\times i_0$
$\Rightarrow i_0=\frac{2q}{\Delta t}\text{amp}$

Equation of $i$ is given by
$i-i_0=(t-0)\left(\frac{i_0-0}{0-\Delta t}\right)$
or $i=\frac{-2q}{\Delta t^2}t+\frac{2q}{\Delta t}$

As heat generated in resistance $R$,
$\Rightarrow {\int }_0^{H}dH={\int }_0^{\Delta t}{i}^{2}Rdt$
$\Rightarrow H=\frac{4q^2}{\Delta t^2}R{\int }_0^{\Delta t}[1-\frac{2t}{\Delta t}+\frac{t^2}{\Delta t^2}]dt$
$\Rightarrow H=\frac{4q^2}{\Delta t^2}R{[t-\frac{t^2}{\Delta t}+\frac{t^3}{3\Delta t^2}]}_0^{\Delta t}$
$\Rightarrow H=\frac{4q^2}{\Delta t^2}R[\Delta t-\Delta t+\frac{\Delta t}{3}]$
$\Rightarrow H=\frac{4}{3}\frac{q^{2}R}{\Delta t}$
Hence, options (A) and (C) are correct.