Irodov 3.172: Cutting the Capacitance Drives Current Back

Problem Irodov 3.172

A source of constant emf $\mathcal{E}$ with negligible internal resistance, a resistance $R$ and a capacitor of capacitance $C$ are connected in series. At the moment $t = 0$ the capacitance is suddenly reduced by a factor $\eta$ . Find the current in the circuit as a function of the time $t$ .

The series circuit: the source E, the resistance R and the capacitor C. — The series circuit: the source $\mathcal{E}$ , the resistance $R$ and the capacitor $C$ .

Given

emf $\mathcal{E}$ , no internal resistance
$R$
$C$ , reduced to $C/\eta$ at $t = 0$
steady state before $t = 0$ (assumed)

Find

the current

I(t)

for

t > 0

Solution

Step 1The state just after the change

Before $t = 0$ the circuit is in a steady state (assumed): no current flows, so the capacitor carries the full emf and the charge

q_0 = C\mathcal{E}

The capacitance changes abruptly, but the charge on a plate can change only through the current, which is finite; so at $t = 0$ the charge is still $q_0$ while the capacitance is $C/\eta$ . The capacitor’s voltage jumps to

U_0 = \frac{q_0}{C/\eta} = \eta\,\mathcal{E} \tag{1}

which exceeds the emf for $\eta > 1$ .

Step 2The loop equation after the change

Just after t = 0: the capacitor, now C/η, keeps its charge, and the current I flows out of its positive plate through R and back into the source. — Just after $t = 0$ : the capacitor, now $C/\eta$ , keeps its charge, and the current $I$ flows out of its positive plate through $R$ and back into the source.

Take the current $I$ positive in the sense of the figure, out of the capacitor’s positive plate, through $R$ , into the positive terminal of the source, so that $dq/dt = -I$ . Going round the loop, the capacitor’s voltage exceeds the emf by the drop across $R$ :

\frac{\eta\,q}{C} - \mathcal{E} = R I \tag{2}

Step 3The current as a function of time

Differentiating (2) with respect to time, with $\mathcal{E}$ constant and $dq/dt = -I$ :

-\frac{\eta}{C}\,I = R\,\frac{dI}{dt}

The starting value of the current comes from (2) with $q = q_0$ , using (1):

I(0) = \frac{\eta\mathcal{E} - \mathcal{E}}{R} = \frac{(\eta - 1)\,\mathcal{E}}{R}

Separating the variables and integrating from $t = 0$ :

\begin{aligned} \int_{I(0)}^{I}\frac{dI'}{I'} &= -\int_0^t \frac{\eta\,dt'}{RC}\\[2pt] \ln\frac{I}{I(0)} &= -\frac{\eta t}{RC} \end{aligned}

an exponential decay with the time constant $RC/\eta$ :

I(t) = \frac{(\eta - 1)\,\mathcal{E}}{R}\,e^{-\eta t/RC}

(3)

For $\eta > 1$ the current flows against the emf: the capacitor drives charge back into the source until its voltage has fallen to $\mathcal{E}$ .

The current after t = 0: it is largest at t = 0 and decays exponentially with the time constant RC/η. — The current after $t = 0$ : it is largest at $t = 0$ and decays exponentially with the time constant $RC/\eta$ .

Check

The charge sent back is

\int_0^\infty I\,dt = \frac{(\eta - 1)\,\mathcal{E}}{R}\cdot\frac{RC}{\eta} = \frac{C\mathcal{E}(\eta - 1)}{\eta}.

It must be the old charge $C\mathcal{E}$ minus the new equilibrium charge $(C/\eta)\mathcal{E}$ , which is the same. With $\eta = 1$ nothing changes and (3) gives no current.

Remark

For $\eta < 1$ (the capacitance increased instead) the same formula holds with a negative current: the source then pushes extra charge onto the capacitor. Either way the capacitor alone cannot change its charge suddenly, which is why the jump appears in its voltage and the current carries the adjustment over the new time constant $RC/\eta$ .