Irodov 3.327: The Shunt Doubles the Coil's Time – Amit & Deepali Sinha Foundation

Problem Irodov 3.327

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 coil of inductance $L$ 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 current through the inductance $L$ varies with time 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 the coil L, 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 the coil $L$ , the other in series with the source.

Given

$\mathcal{E}$
$R$ (both resistors)
$L$
$Sw$ closed at $t = 0$
source and coil resistances negligible (assumed)
no current in the coil at $t = 0$ (assumed)

Find

the coil current

I(t)

for

t > 0

Solution

Step 1Kirchhoff’s two rules after closing

Before the switch closes the coil is connected only across the parallel resistor, with no source in that loop, so any current in it has died away; we take the coil current zero at $t = 0$ , and the resistances of the source and the coil as negligible.

After the switch closes: I through the coil, I' through the parallel R, and their sum through the series R, each in the sense the source drives it. — After the switch closes: $I$ through the coil, $I'$ through the parallel $R$ , and their sum through the series $R$ , each in the sense the source drives it.

The parallel resistor and the coil share one voltage, the coil’s $L\,dI/dt$ , so the current in that resistor is

I' = \frac{L}{R}\,\frac{dI}{dt} \tag{1}

Going round the outer loop, the emf equals the drop across the series resistor, which carries $I + I'$ , plus the voltage across the coil:

\mathcal{E} = (I + I')R + L\,\frac{dI}{dt} \tag{2}

Step 2One equation for the coil current

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

\begin{aligned} \mathcal{E} &= RI + L\,\frac{dI}{dt} + L\,\frac{dI}{dt}\\[2pt] 2L\,\frac{dI}{dt} &= \mathcal{E} - RI \end{aligned}

\frac{dI}{\mathcal{E}/R - I} = \frac{R\,dt}{2L}

(3)

Step 3Integrating from the closing of the switch

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

\begin{aligned} \int_0^{I} \frac{di}{\mathcal{E}/R - i} &= \int_0^t \frac{R\,dt'}{2L}\\[2pt] -\ln\frac{\mathcal{E}/R - I}{\mathcal{E}/R} &= \frac{tR}{2L} \end{aligned}

Taking the exponential and solving for $I$ :

I(t) = \frac{\mathcal{E}}{R}\left(1 - e^{-tR/2L}\right)

(4)

The current flows down through the coil, in the sense the source drives it, at every $t > 0$ .

The coil current after the switch closes: a rise from zero toward E/R with time constant 2L/R. The mark sits where I has covered 1 - 1/e of the rise. — The coil current after the switch closes: a rise from zero toward $\mathcal{E}/R$ with time constant $2L/R$ . The mark sits where $I$ has covered $1 - 1/e$ of the rise.

Check

At $t = 0$ the formula gives $I = 0$ , and its slope is $dI/dt = \mathcal{E}/2L$ : at that instant the coil carries nothing, so the two resistors form a divider and the coil, in parallel with the vertical one, sees half the emf, $L\,dI/dt = \mathcal{E}/2$ . As $t \to \infty$ , $dI/dt \to 0$ , the coil is a plain wire that shorts the parallel resistor, and the whole current $\mathcal{E}/R$ flows through the series $R$ and the coil, as (4) gives. The exponent $tR/2L$ is dimensionless, since $L/R$ is a time.

Remark

Seen from the coil, the source with its two resistors is equivalent to a single source of emf $\mathcal{E}/2$ behind a resistance $R/2$ (Thevenin’s theorem), so the coil charges up as in the plainest RL circuit, with the time constant $L/(R/2) = 2L/R$ and the final current $(\mathcal{E}/2)/(R/2) = \mathcal{E}/R$ . With a capacitor in place of the coil the same reduction halves the time constant instead, to $(R/2)C$ : a capacitor’s time constant goes as the resistance it sees, an inductance’s as its inverse.