Irodov 3.205: Half the Energy Is Always Lost – Amit & Deepali Sinha Foundation

Problem Irodov 3.205

Two capacitors, each of capacitance $C$ , a resistor $R$ and a switch $Sw$ are connected in series in a single loop. With the switch open, one of the capacitors is charged to a voltage $V_0$ . At the moment $t = 0$ the switch is closed. Find:

the current $I$ in the loop as a function of the time $t$
the heat released, computed from the current $I(t)$

The circuit: the resistor R, two capacitors C and the switch Sw in one series loop, the switch open. — The circuit: the resistor $R$ , two capacitors $C$ and the switch $Sw$ in one series loop, the switch open.

Given

$C$ for each capacitor
$R$
one capacitor at $V_0$
the other initially uncharged (assumed)
$Sw$ closed at $t = 0$

Find

(a)

I(t)

, (b) the heat

Q

from

I(t)

Solution

Step 1The loop equation

The moment the switch closes: the current I leaves the positive plate of the charged capacitor, runs through R and charges the other one. — The moment the switch closes: the current $I$ leaves the positive plate of the charged capacitor, runs through $R$ and charges the other one.

Take the left capacitor as the charged one; the other starts uncharged (assumed). Let $q$ be the charge that has passed round the loop by the time $t$ , so the charged capacitor holds $CV_0 - q$ and the other holds $q$ , and $I = dq/dt$ . Going round the loop, the voltage of the first capacitor drives the current against $IR$ and the voltage of the second:

\frac{CV_0 - q}{C} = IR + \frac{q}{C} \tag{1}

Step 2(a) The current

Differentiating (1) in time, with $dq/dt = I$ , removes $q$ and leaves an equation for $I$ alone:

\begin{aligned} -\frac{I}{C} &= R\,\frac{dI}{dt} + \frac{I}{C}\\ \frac{dI}{I} &= -\frac{2\,dt}{RC} \end{aligned}

At $t = 0$ no charge has moved yet, so (1) with $q = 0$ fixes the starting current:

I(0) = \frac{V_0}{R}

Integrating from that value:

I = \frac{V_0}{R}\,e^{-2t/RC}

(2)

The time constant is $RC/2$ : the two capacitors in series act as one capacitance $C/2$ discharging through $R$ .

The current starts at its largest value and decays with the time constant RC/2. — The current starts at its largest value and decays with the time constant $RC/2$ .

Step 3(b) The heat, from the current

The heat is the Joule power $I^2R$ integrated over the whole transient, with (2):

\begin{aligned} Q = \int_0^\infty I^2R\,dt &= \frac{V_0^2}{R}\int_0^\infty e^{-4t/RC}\,dt\\ &= \frac{V_0^2}{R}\cdot\frac{RC}{4} \end{aligned}

Q = \frac{1}{4}\,CV_0^2

(3)

Check

Energy balance. At the start only one capacitor holds energy, $CV_0^2/2$ . At the end the current has stopped, so by (1) the two voltages are equal; the charge $CV_0$ is shared equally and each capacitor sits at $V_0/2$ . The loss of field energy is then exactly (3):

\frac{1}{2}CV_0^2 - 2\cdot\frac{1}{2}C\left(\frac{V_0}{2}\right)^2 = \frac{1}{4}CV_0^2

Remark

$R$ has dropped out of the heat: a small resistance makes a large current for a short time, and the product is the same. Half the stored energy is lost whatever the resistance, the price of sharing charge between equal capacitors, just as a perfectly inelastic collision of equal masses loses half the kinetic energy.