Irodov 3.326: Cut the Inductance, Jump the Current – Amit & Deepali Sinha Foundation

Problem Irodov 3.326

A source of constant emf $\mathcal{E}$ and a choke coil of inductance $L$ are connected in series in a closed circuit whose total resistance is $R$ . At the moment $t = 0$ the inductance of the coil is suddenly made $\eta$ times smaller. Find the current in the circuit as a function of the time $t$ . Hint: during a sudden change of the inductance the flux linkage of the coil stays the same.

Given

$\mathcal{E}$ constant
$L$
$R$ (whole circuit)
at $t = 0$ , $L \to L/\eta$ abruptly
flux linkage unchanged in the jump
steady current before $t = 0$ (assumed)

Find

I(t)

for

t > 0

Solution

Step 1The current before the jump

Before $t = 0$ the current is steady (assumed: the circuit has been closed long enough), so the coil has no emf and the whole of $\mathcal{E}$ drives the current through $R$ :

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

Step 2The current just after the jump

In the jump the flux linkage $LI$ keeps its value, so with the new inductance $L/\eta$ and the old current (1):

\begin{align*} \frac{L}{\eta}\,I(0) &= L I_0 \\ I(0) &= \eta I_0 = \frac{\eta\,\mathcal{E}}{R} \tag{2}\end{align*}

Step 3The relaxation that follows

For $t > 0$ the inductance is the constant $L/\eta$ , and the loop rule balances the source against the resistance and the coil’s emf:

\mathcal{E} - \frac{L}{\eta}\,\frac{dI}{dt} = RI

Writing it for the excess over the steady value, $I - \mathcal{E}/R$ , and separating the variables:

\frac{d\left(I - \mathcal{E}/R\right)}{I - \mathcal{E}/R} = -\frac{\eta R}{L}\,dt

Integrating from $t = 0$ , where by (2) the excess is $(\eta - 1)\mathcal{E}/R$ :

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

I = \frac{\mathcal{E}}{R}\left[1 + (\eta - 1)\,e^{-\eta R t/L}\right]

(3)

Step 4The graph

The current jumps from $\mathcal{E}/R$ to $\eta\mathcal{E}/R$ at $t = 0$ and then relaxes back to $\mathcal{E}/R$ exponentially, with the time constant $L/(\eta R)$ of the new circuit.

The current after the jump, drawn for η = 3: it starts at ηE/R and decays to E/R with the time constant L/η R. — The current after the jump, drawn for $\eta = 3$ : it starts at $\eta\mathcal{E}/R$ and decays to $\mathcal{E}/R$ with the time constant $L/\eta R$ .

Check

At $t = 0$ (3) gives $\eta\mathcal{E}/R$ , matching (2), and as $t \to \infty$ it gives $\mathcal{E}/R$ , the steady current, which does not depend on $L$ . With $\eta = 1$ nothing happens and (3) is the constant $\mathcal{E}/R$ , as it should be. Units: $Rt/L$ is $\Omega\cdot\text{s}/\text{H}$ , a pure number.

Remark

The coil’s magnetic energy $LI^2/2$ jumps from $LI_0^2/2$ to $(L/\eta)(\eta I_0)^2/2$ , an increase by the factor $\eta$ : whatever cuts the inductance (pulling out an iron core, say, against its attraction) must do that work, and the field energy released as the current relaxes goes into heat in $R$ . It is the same rule that makes a superconducting coil’s current rise when it is stretched, here followed by the decay that a resistance allows.