# Inductors in parallel. Link between inductance and current in a D.C. circuit

A problem from the FIITJEE review package: Or, paraphrased:

When two inductors in parallel connected to a battery with some internal resistance, what is the current through each of the inductors after achieving a steady state.

The one conclusion that I can make is that the potential across the inductors after achieving steady state is 0.

How do I link inductance with current? Is there any expression for resistance of an inductor?

Since this is a circuit theory question, it's possible the question author wants us to assume the inductors are ideal, that is they have zero series resistance.

If the inductors initially have 0 current through them1 as the switch is closed, and the voltage across them is $v(t)$, then the current through the first is $$i_1(t) = \int_0^t \frac{v(t)}{L_1}\rm{d}t$$ And across the second one $$i_2(t) = \int_0^t \frac{v(t)}{L_2}\rm{d}t$$

Because the integral is a linear operator, we can pull out the constant term and find

$$L_1 i_1(t) = L_2 i_2(t)$$

for any choice of $t$, including in the limit as $t\to\infty$.

Therefore in the steady state we find2

$$I_1 = \frac{L_2}{L_1}I_2$$

From this you can find the current through the individual inductors in the situation given. (Hint: the answer will be very similar to if they were two resistors with values $R_1$ and $R_2$)

Note 1: This is actually not a great assumption for the circuit as given, because when the switch is open there is no current flow through the resistor, and therefore no mechanism for the inductors to lose energy and decay to a 0-current state when the switch is open.

Note 2: In a real world circuit, however, the steady state behavior would be dominated by the equivalent series resistance of the two inductors, and their inductance would not come in to play. See rob's answer for more details.

My suspicion is that the author of your problem is making an incorrect assumption using the following chain of reasoning:

1. Most inductors are just wire coils
2. The inductance is proportional to the length of the coil.
3. The resistance of a wire is proportional to its length.
4. Therefore, inductors have $R\propto L$. (wrong!)

In that case you can solve the problem using the standard voltage divider technique. The relationship that governs inductance, current, and voltage difference is $\Delta\!V = -L\ \mathrm dI/\mathrm dt$, so once you have reached the steady state and $\mathrm dI/\mathrm dt=0$ you can put any $L$ you want in your description of the circuit without changing your analysis.

However, inductors don't generally have $R\propto L$. For instance, suppose I build two inductors: one by wrapping a long wire around a pencil, and the other by wrapping the same length of the same diameter/composition wire around an iron nail. The two inductors will have the same resistance, given by the length of the wire, and so the steady-state solution is for them to divide the current equally when connected in parallel. However, building up that current requires magnetizing the iron in one case, which requires more energy than magnetizing a pencil and therefore takes longer. The two devices have the same $R$ but different $L$, and none of the solutions in your problem is correct. You can also change $L$ by adjusting the inductor's geometry. An inductor made of narrow-diameter wire will have more resistance than one with the same geometry but made of heavy wire. There are a million ways that $L$ and $R$ can be decoupled from each other.

If you buy a real inductor, you should be able to get a data sheet that specifies its DC resistance (or perhaps an upper limit on its DC resistance).