# Finding current through resistor as a function of time in the presence of 2 inductors

1 vote
80 views After long time switch is shifted from position 1 to position 2. Find $i$ through the resistor as a function of time.

Attempt:

I know that the total inductance would become $L+ 2L= 3L$. But after that I am unable to understand the mechanism that is happening. So basically, since "long" time has passed, the current is constant in circuit and equal to $\dfrac{\varepsilon} R$, also $\dfrac{d\phi}{dt}$ through L is 0.

Now, when switch is shifted, there's flux change through both the inductors. Both will develop polarity with positive part on the same side as the positive side of the battery

From KVL,
At any instant when current is $i$
$\varepsilon = 3L\dfrac{di}{dt}+ iR$

I am just confused about the limits of current during integration. To make the function we will take the limt=its from $i_o$ to $i$ but how to find $i_o$. That's my question. edited Jul 28, 2018
I think your difficulty is conceptual, since your teacher has already provided a solution and explained the reason for it (conservation of flux). So the initial current when the switch is moved, is $i_0=i_1/3=E/3R$ where $i_1$ is the current before the switch is moved, as derived in your Physics SE question. This is the lower limit of integration ($t=0, i=i_0$).

1 vote

Now, when switch is shifted, there's flux change through both the
inductors. Both will develop polarity with positive part on the same
side as the positive side of the battery

The problem is somewhat pathological in that the voltage across the inductors cannot be ordinary functions of time.

Why? The voltage across an inductor is proportional to the time derivative of the current through which means that in order for the inductor voltage to be defined for all time $t$, the inductor current must be continuous.

But, in this problem, the current through the inductors cannot be continuous at the time the switch is 'shifted' (thrown) to position 2.

Just before the switch is thrown, the resistor (and $L$ inductor) current is $i_R = \frac{\mathcal{E}}{R}$ while the current through the $2L$ inductor is zero.

Now, just after the switch is thrown, the current through the two inductors must, by KCL, be equal and so, one (or both) inductor current(s) must be discontinuous which implies that the inductor voltages do not exist at that time (the inductor voltages have a delta 'function' at the switching time).

One might try placing a large resistance $r \gg R$ in parallel with the $L$ inductor which will permit a continuous current and thus a well defined voltage across each inductor.

answered Aug 4, 2018 by (120 points)