Irodov 3.328: Ideal Coils Share by Flux – Amit & Deepali Sinha Foundation

Problem Irodov 3.328

Two coils of inductances $L_1$ and $L_2$ are connected in parallel between two junctions. A source of emf $\mathcal{E}$ , a resistor $R$ and a switch $Sw$ , which is open, are connected in series between the same two junctions, the source’s positive terminal toward the left junction. The internal resistance of the source and the resistances of the coils are negligible. The switch is closed. Find the steady currents that then flow in the two coils.

The circuit with the switch Sw open: the coils L_1 and L_2 in parallel between the two junctions, and the source E (long + plate on the left) with R and Sw in series across them. — The circuit with the switch $Sw$ open: the coils $L_1$ and $L_2$ in parallel between the two junctions, and the source $\mathcal{E}$ (long $+$ plate on the left) with $R$ and $Sw$ in series across them.

Given

$\mathcal{E}$
$R$
$L_1$ , $L_2$
source and coil resistances negligible
no mutual inductance between the coils (assumed)
no current in the coils before closing (assumed)

Find

the steady currents

I_1

and

I_2

in the coils

Solution

Step 1The two coils share one voltage

After the switch closes: I_1 through L_1 and I_2 through L_2, in the sense the source drives them, joining into I_1 + I_2 through the switch and R. — After the switch closes: $I_1$ through $L_1$ and $I_2$ through $L_2$ , in the sense the source drives them, joining into $I_1 + I_2$ through the switch and $R$ .

Take $I_1$ and $I_2$ in the coils as in the figure. Both coils sit between the same two junctions, and each, having no resistance, has across it only its own emf; the coils are taken not to link each other’s flux (assumed). So at every instant after closing

L_1\,\frac{dI_1}{dt} = L_2\,\frac{dI_2}{dt}

Integrating from the moment of closing, when neither coil carries current (assumed), to a later time $t$ :

\int_0^{I_1} L_1\,dI_1' = \int_0^{I_2} L_2\,dI_2'

L_1 I_1 = L_2 I_2

(1)

Step 2The total current in the steady state

Once the currents stop changing, the coils have no emf and the junctions are joined by plain wire. The whole loop through the source then carries

I_1 + I_2 = \frac{\mathcal{E}}{R} \tag{2}

Step 3The two currents

Writing $I_2 = L_1 I_1/L_2$ from (1) and substituting into (2):

\begin{aligned} I_1\left(1 + \frac{L_1}{L_2}\right) &= \frac{\mathcal{E}}{R}\\[2pt] I_1\,\frac{L_1 + L_2}{L_2} &= \frac{\mathcal{E}}{R} \end{aligned}

Solving for $I_1$ , and then $I_2$ by the same steps with the indices exchanged:

\begin{gathered} I_1 = \frac{\mathcal{E}L_2}{R(L_1 + L_2)}\\[4pt] I_2 = \frac{\mathcal{E}L_1}{R(L_1 + L_2)} \end{gathered}

(3)

Both flow in the sense the source drives them, from the left junction to the right one, and the coil with the smaller inductance carries the larger current.

Check

The two numerators add to $\mathcal{E}(L_1 + L_2)$ , so the currents add to $\mathcal{E}/R$ , as (2) requires, and their ratio $I_1/I_2 = L_2/L_1$ is (1). For two equal coils each carries $\mathcal{E}/2R$ , as symmetry demands; as $L_1 \to 0$ the first coil is a bare wire across the second, and it takes the whole $\mathcal{E}/R$ while $L_2$ gets nothing.

Remark

The split is not set by any resistance, unlike two resistive branches: it is fixed from the first instant by the zero-resistance loop the two coils form, whose linked flux cannot change. The pair behaves as one inductance $L_1L_2/(L_1 + L_2)$ , as for parallel resistors. Give the coils small resistances $r_1$ and $r_2$ and the loop is no longer flux-conserving: the currents start in the ratio $L_2 : L_1$ and then drift, over the time $(L_1 + L_2)/(r_1 + r_2)$ , to the ratio $r_2 : r_1$ that resistances alone would give.