Question

Current in a resistor R uniformly 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)

The correct options are
A Heat generated in the resistance during this process is $\frac{4}{3}\frac{q^{2}R}{\Delta t}$
C Maximum current is $i_0=\frac{2q}{\Delta t}$


From the above graph current flows given by,
$i=-\frac{i_0}{\Delta t}t+i_0$.....(1)
on further solve eq (1)
$i=i_0(1-\frac{t}{\Delta t})$
Here, area under i-t curve gives the total no. of charges.
So, $i_0\frac{\Delta t}{2}=q$
$i_0=\frac{2q}{\Delta t}$
Where, $i_0$=maximum current
Hence option (C) matches correctly.

Total heat dissipation $\left(H\right)$ = $i_0^{2}R$

$dH=i^{2}Rdt=i_0^{2}R{(1-\frac{t}{\Delta t})}^{2}dt$

$H=i_0^{2}R{\int }_0^{\Delta t}(1+\frac{t^2}{\Delta t^2}-\frac{2t}{\Delta t})dt$

$=i_o^{2}R(t+\frac{t^3}{3\Delta t^2}-\frac{t^2}{\Delta t}){|}_0^{\Delta t}$

$=\frac{i_o^{2}R\Delta t}{3}$

On solving further solving we get

$H=\frac{4}{3}\frac{q^{2}R}{\Delta t}$

($\because i_o=\frac{2q}{\Delta t}$)