Irodov 3.188: Half the Emf, Twice as Fast – Amit & Deepali Sinha Foundation

Problem Irodov 3.188

A source of emf $\mathcal{E}$ with its positive terminal at the top drives a circuit through a switch $Sw$ , which is open. Beyond the switch, a resistor $R$ runs down to the bottom wire, and a capacitor $C$ is connected in parallel with that resistor. The bottom wire returns to the source’s negative terminal through a second resistor, also of resistance $R$ . The switch is closed at the moment $t = 0$ . Find how the voltage across the capacitor varies with the time $t$ after that.

The circuit before t = 0, with the switch Sw open. The source's long (+) plate is at the top; one R is in parallel with C, the other in series with the source. — The circuit before $t = 0$ , with the switch $Sw$ open. The source’s long ( $+$ ) plate is at the top; one $R$ is in parallel with $C$ , the other in series with the source.

Given

$\mathcal{E}$
$R$ (both resistors)
$C$
$Sw$ closed at $t = 0$
source resistance negligible (assumed)
$C$ uncharged at $t = 0$ (assumed)

Find

the capacitor voltage

V(t)

for

t > 0

Solution

Step 1Kirchhoff’s two rules after closing

Before the switch closes the capacitor is connected straight across a resistor, so any charge it held has drained away; we take it uncharged at $t = 0$ , and the source’s own resistance as negligible. Let $V$ be the capacitor voltage, the upper plate positive, and $I$ the current in the series resistor of the bottom wire. At the upper junction that current divides between the parallel resistor, which carries $V/R$ , and the capacitor, which takes $C\,dV/dt$ :

I = \frac{V}{R} + C\,\frac{dV}{dt} \tag{1}

Going round the outer loop, the emf equals the drop across the series resistor plus the voltage across the parallel pair:

\mathcal{E} = IR + V \tag{2}

Step 2One equation for

V

Substituting (1) into (2) eliminates $I$ :

\begin{aligned} \mathcal{E} &= V + RC\,\frac{dV}{dt} + V\\[2pt] RC\,\frac{dV}{dt} &= \mathcal{E} - 2V \end{aligned}

\frac{dV}{\mathcal{E}/2 - V} = \frac{2\,dt}{RC}

(3)

Step 3Integrating from the closing of the switch

Integrating (3) from $V = 0$ at $t = 0$ to $V$ at the time $t$ :

\begin{aligned} \int_0^{V} \frac{dV'}{\mathcal{E}/2 - V'} &= \int_0^t \frac{2\,dt'}{RC}\\[2pt] -\ln\frac{\mathcal{E}/2 - V}{\mathcal{E}/2} &= \frac{2t}{RC} \end{aligned}

Taking the exponential and solving for $V$ :

V(t) = \frac{\mathcal{E}}{2}\left(1 - e^{-2t/RC}\right)

(4)

The capacitor voltage after the switch closes: a rise from zero toward E/2 with time constant RC/2. The mark sits where V has covered 1 - 1/e of the rise. — The capacitor voltage after the switch closes: a rise from zero toward $\mathcal{E}/2$ with time constant $RC/2$ . The mark sits where $V$ has covered $1 - 1/e$ of the rise.

Check

At $t = 0$ the formula gives $V = 0$ , the uncharged start, and its slope is $dV/dt = \mathcal{E}/RC$ : at that instant the empty capacitor shorts the parallel resistor, so the whole current $\mathcal{E}/R$ flows into it through the series $R$ , and $I/C$ is exactly $\mathcal{E}/RC$ . As $t \to \infty$ , $V \to \mathcal{E}/2$ : with the capacitor full, the two equal resistors form a divider, and the parallel one carries half the emf. The exponent $2t/RC$ is dimensionless, since $RC$ is a time.

Remark

Seen from the capacitor, the source with its two resistors is equivalent to a single source of emf $\mathcal{E}/2$ behind a resistance $R/2$ , the two $R$ in parallel (Thevenin’s theorem). The capacitor therefore charges as in the plainest RC circuit, to $\mathcal{E}/2$ with the time constant $(R/2)\,C$ : the parallel resistor halves both the final voltage and the time it takes to get there.