Question

Current passing through a coil of resistance R decreases uniformly as shown in the figure. If total amount of charge q is passing through the coil during a time interval $t_0$ then what will be the total amount of heat generated through it ?

• A. $\frac{3}{4}\frac{q^{2}R}{t_0}$
• B. $\frac{4}{3}\frac{q^{2}R}{t_0}$
• C. $\frac{q^2}{t_{0}R}$
• D. $\frac{q^2{R}^2}{t_0^2}$

Answer 1

The correct option is B $\frac{4}{3}\frac{q^{2}R}{t_0}$

Equation of this graph:-

$\frac{i}{i_0}+\frac{t}{t_0}=1$ (i)

$\int dq=\int idt\int {dq={\int }_0^{t_0}{i}_0(1-\frac{t}{t_0})}^{dt}q={[i_{0}t-\frac{i_0}{t_0}\frac{t^2}{2}]}_0^{t_0}q=i_0{t}_0-\frac{i_0{t}_0}{2}q=\frac{i_0{t}_0}{2}$

Heat generated $=\int 1^{2}Rdt={\int }_0^{t_0}{(1-\frac{t}{t_0})}^2{t}_0^{2}dt={\int }_0^{t_0}[i_0^{2}t+\frac{t^3{i}_0^2}{3t_0^2}-\frac{t^2{i}_0^2}{t_0}]=i_0^2{t}_0+\frac{i_0^2{t}_0}{3}-i_0^2{t}_0$

Heat generated $=(\frac{i_0^2{t}_0}{3}\times R)J{i}_0=\frac{2q}{t_0}$

Heat generated $=\frac{4\times q^2}{t_0^2\times 3}{t}_0\times R=\frac{4}{3}\frac{q^{2}R}{t_0}Joules$