Irodov 3.335: A Third of the Stored Energy – Amit & Deepali Sinha Foundation

Problem Irodov 3.335

A coil of inductance $L = 2.0$ $\mu$ H and resistance $R = 1.0$ $\Omega$ is connected across a source of constant emf $\mathcal{E} = 3.0$ V through a switch $Sw$ , which is closed. A resistor $R_0 = 2.0$ $\Omega$ is connected in parallel with the coil, directly across the same two junctions. The internal resistance of the source is negligible. The switch is then opened. Find the heat generated in the coil after that.

The circuit before the switch is opened: the coil L with its own resistance R, the resistor R_0 across the same junctions, and the source branch with the switch Sw closed. — The circuit before the switch is opened: the coil $L$ with its own resistance $R$ , the resistor $R_0$ across the same junctions, and the source branch with the switch $Sw$ closed.

Given

$L = 2.0$ $\mu$ H
$R = 1.0$ $\Omega$ (the coil’s own)
$\mathcal{E} = 3.0$ V
$R_0 = 2.0$ $\Omega$
source resistance negligible
steady state before opening (assumed)

Find

the heat

Q

generated in the coil after

Sw

is opened

Solution

Step 1The coil’s current before the switch opens

With the switch closed long enough for the steady state (assumed), the current no longer changes and the coil’s emf $-L\,dI/dt$ is zero. The coil is then simply its resistance $R$ , connected straight across the source, whose own resistance is negligible:

I_0 = \frac{\mathcal{E}}{R} \tag{1}

The resistor $R_0$ carries $\mathcal{E}/R_0$ at the same time, but that current has no inertia: the moment the source is cut off, $R_0$ instead carries the coil’s current.

Step 2The decay through the coil and

R_0

After the switch opens: the coil's current I keeps its sense and closes through R_0, which now carries it the opposite way to before. — After the switch opens: the coil’s current $I$ keeps its sense and closes through $R_0$ , which now carries it the opposite way to before.

Once the switch opens, the source branch carries no current, and the coil and $R_0$ form a single closed loop. The current in an inductance cannot jump, so the loop starts with $I_0$ and the coil’s emf drives it through both resistances:

-L\,\frac{dI}{dt} = (R + R_0)\,I

Integrating from $I_0$ at the moment of opening, $t = 0$ :

\begin{aligned} \int_{I_0}^{I}\frac{dI'}{I'} &= -\frac{R + R_0}{L}\int_0^t dt'\\[2pt] \ln\frac{I}{I_0} &= -\frac{(R + R_0)\,t}{L} \end{aligned}

I(t) = I_0\,e^{-(R + R_0)t/L}

(2)

Step 3The heat in the coil

The coil’s own resistance $R$ turns $I^2R\,dt$ into heat in each interval $dt$ . Summing over the whole decay with (2):

\begin{aligned} Q &= \int_0^\infty I^2 R\,dt = I_0^2 R\int_0^\infty e^{-2(R + R_0)t/L}\,dt\\[2pt] &= I_0^2 R\,\frac{L}{2(R + R_0)} = \frac{\mathcal{E}^2}{R^2}\,R\,\frac{L}{2(R + R_0)} \end{aligned}

the last step substituting $I_0$ from (1). Numerically,

Q = \frac{(2.0\times 10^{-6}\text{ H})(3.0\text{ V})^2}{2\,(1.0\ \Omega)(1.0\ \Omega + 2.0\ \Omega)}

Q = \frac{L\mathcal{E}^2}{2R(R + R_0)} = 3.0\ \mu\text{J}

(3)

Check

Energy accounting. When the switch opens the coil holds $LI_0^2/2 = L\mathcal{E}^2/(2R^2) = 9.0\ \mu$ J, and all of it ends as heat in the loop. The same current flows through $R$ and $R_0$ at every instant, so the heat divides in the ratio $R : R_0 = 1 : 2$ , and the coil’s share is one third of the stored energy, which is the value in (3). Units: $L\mathcal{E}^2/R^2$ is $\text{H}\cdot\text{A}^2 = \text{J}$ .

Remark

At the instant of opening the current through $R_0$ reverses and jumps from $\mathcal{E}/R_0 = 1.5$ A to $\mathcal{E}/R = 3.0$ A, so the voltage across the coil’s terminals leaps to $R_0\mathcal{E}/R = 6.0$ V, twice the emf. A larger shunt resistance takes a larger share of the energy but raises that spike; with no shunt at all the current has nowhere to go and the stored energy is spent in a spark across the opening contacts, which is why inductive loads are switched with a resistor or a diode across them.