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Question

(A) The total charge $q$ is the area under the $i-t$ graph. Since the current decreases uniformly from $I_0$ to $0$ in time $\Delta t$,the graph is a

Vedclass · Accepted Answer

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

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(A) The total charge $q$ is the area under the $i-t$ graph. Since the current decreases uniformly from $I_0$ to $0$ in time $\Delta t$,the graph is a triangle with base $\Delta t$ and height $I_0$.$q = \frac{1}{2} I_0 \Delta t \implies I_0 = \frac{2q}{\Delta t}$.The current as a function of time is $i(t) = I_0 \left(1 - \frac{t}{\Delta t}\right) = \frac{2q}{\Delta t} \left(1 - \frac{t}{\Delta t}\right)$.The heat generated $H$ is given by $H = \int_0^{\Delta t} i^2 R dt$.$H = R \int_0^{\Delta t} \left[ \frac{2q}{\Delta t} \left(1 - \frac{t}{\Delta t}\right) \right]^2 dt = R \frac{4q^2}{(\Delta t)^2} \int_0^{\Delta t} \left(1 - \frac{t}{\Delta t}\right)^2 dt$.Let $u = 1 - \frac{t}{\Delta t}$,then $du = -\frac{1}{\Delta t} dt$.$H = R \frac{4q^2}{(\Delta t)^2} (-\Delta t) \int_1^0 u^2 du = R \frac{4q^2}{\Delta t} \int_0^1 u^2 du = R \frac{4q^2}{\Delta t} \left[ \frac{u^3}{3} \right]_0^1 = \frac{4}{3} \frac{q^2 R}{\Delta t}$.

What amount of heat will be generated in a coil of resistance $R$ due to a charge $q$ passing through it if the current in the coil decreases to zero uniformly during a time interval $\Delta t$?

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