Irodov 3.189: Same Charge, Different Heat – Amit & Deepali Sinha Foundation

Problem Irodov 3.189

A total charge $q$ passes through a coil of resistance $R$ . Find the heat generated in the coil if the current

falls to zero at a uniform rate over a time interval $\Delta t$
falls toward zero, halving its value every time interval $\Delta t$ .

The heat generated in the coil is Joule’s law summed over the whole flow, and the charge is the integral of the current:

Q = \int R I^2\,dt, \qquad q = \int I\,dt \tag{1}

In each case the second integral fixes the initial current $I_0$ from the given charge, and the first then gives the heat.

Given

$R$
total charge $q$
(a) linear fall to zero in $\Delta t$
(b) halving time $\Delta t$

Find

the heat

Q

in each case

Solution

Step 1(a) A current falling linearly

The two current laws for one and the same charge q under each curve (the tail of (b) runs on past the frame), I against t: (a) a straight fall to zero at Δ t, (b) a halving every Δ t. — The two current laws for one and the same charge $q$ under each curve (the tail of (b) runs on past the frame), $I$ against $t$ : (a) a straight fall to zero at $\Delta t$ , (b) a halving every $\Delta t$ .

The current runs down from $I_0$ to zero along a straight line:

I = I_0\left(1 - \frac{t}{\Delta t}\right), \qquad 0 \le t \le \Delta t

The charge in (1) is the area of a triangle, which fixes $I_0$ :

q = \int_0^{\Delta t} I_0\left(1 - \frac{t}{\Delta t}\right)dt = \frac{I_0\,\Delta t}{2}

I_0 = \frac{2q}{\Delta t}

(2)

The heat from (1), with the substitution $u = 1 - t/\Delta t$ :

Q = R I_0^2 \int_0^{\Delta t}\left(1 - \frac{t}{\Delta t}\right)^2 dt = R I_0^2\,\Delta t\int_0^1 u^2\,du = \frac{R I_0^2\,\Delta t}{3}

Substituting (2):

Q_a = \frac{4}{3}\,\frac{q^2R}{\Delta t}

(3)

Step 2(b) A current halving at equal intervals

Halving every $\Delta t$ is an exponential fall with the decay rate $\alpha = (\ln 2)/\Delta t$ :

I = I_0\,2^{-t/\Delta t} = I_0\,e^{-\alpha t}, \qquad t \ge 0

The current never quite reaches zero, but the charge it carries is finite:

q = \int_0^\infty I_0\,e^{-\alpha t}\,dt = \frac{I_0}{\alpha}

I_0 = \alpha q

(4)

The square of the current falls at twice the rate, so the heat from (1) is

Q = R I_0^2\int_0^\infty e^{-2\alpha t}\,dt = \frac{R I_0^2}{2\alpha}

Substituting (4):

Q = \frac{R(\alpha q)^2}{2\alpha} = \frac{R q^2\alpha}{2}

and with $\alpha = (\ln 2)/\Delta t$ :

Q_b = \frac{\ln 2}{2}\,\frac{q^2R}{\Delta t}

(5)

Check

Both answers have the units of $q^2R/\Delta t$ , that is $\mathrm{A^2\,s^2}\,\Omega/\mathrm{s} = \mathrm{A^2}\,\Omega\,\mathrm{s}$ , an energy. A comparison: the same charge carried by a steady current $q/\Delta t$ for the time $\Delta t$ would give $q^2R/\Delta t$ , and by the inequality $\int I^2\,dt \ge (\int I\,dt)^2/T$ over any interval $T$ , a flow lasting only $\Delta t$ must give at least that much. Case (a) does, with the factor $4/3 > 1$ . Case (b) gives less, $0.35$ , because its flow is not confined to $\Delta t$ : it trails on, and half its charge passes after the time $\Delta t$ .

Remark

For a fixed charge the heat goes as $1/\Delta t$ : the same charge pushed through faster means a larger current for a shorter time, and since the heat goes as the square of the current, the larger current wins. Law (b) is exactly the discharge of a capacitor $C$ through the coil, with the decay rate $\alpha = 1/RC$ ; then $Q = Rq^2\alpha/2$ becomes $q^2/2C$ , the energy the capacitor stored, whatever the resistance.