2
$\begingroup$

The Question is:-

Find the current distribution in the inductors at steady state (i.e. assuming $t\rightarrow \infty$ after the switch is closed and circuit established and consider the inductors without any resistance and ideal).

enter image description here

My Efforts
Since both of the are in parallel, the equations for current at any time $t$ after establishing the circuit is given by ($i_1$=corrent in $5mH$ inductor and $i_2$ in the other, and the total current $i$):- $$i(t)=i_1(t)+i_2(t)=4(1-e^{-\frac{5t}{L_{net}}})$$ where $L_{net}=\frac{10}{3}mH$. But I do not know how this current divides into two in the inductances at any time $t$ or in the steady state. Had the inductors not been ideal, i.e. if they had different resistances, the current could divide in the ration of their resistances in the final steady state, but since they are ideal, this method won't work. How can I solve this? the answers given are $\frac83A$ in the $50mH$ inductor and $\frac43A$ in the other.

$\endgroup$
1
$\begingroup$

Forget the distraction of the steady state specification for a moment. Since the inductors are parallel connected, it follows that:

$$L_1 \dfrac{di_1}{dt} = L_2 \dfrac{di_2}{dt}$$

Integrating both side with respect to time and assuming zero initial conditions yields:

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

Thus:

$$\dfrac{i_1(t)}{i_2(t)} = \dfrac{L_2}{L_1} = 2$$

And so,

$$i_1 = 2 \cdot i_2$$

for all $ t \ge 0$.


The initial condition is true only for ideal (zero resistance) inductors, right?

If there is series resistance (non-ideal inductors) then the analysis proceeds as follows:

$$L_1 \dfrac{di_1}{dt} + R_1 \cdot i_1 = L_2 \dfrac{di_2}{dt} + R_2 \cdot i_2$$

In steady state we have:

$$ \dfrac{di_1}{dt} = \dfrac{di_2}{dt} = 0$$

Thus:

$$R_1 \cdot i_1(\infty) = R_2 \cdot i_2(\infty) \rightarrow i_1(\infty) = \dfrac{R_2}{R_1} i_2(\infty)$$

$\endgroup$
  • $\begingroup$ The initial condition is true only for ideal (zero resistance) inductors, right? Otherwise, the current distribution must be according to only the resistances of the branches and no inductance term comes into consideration for $t\rightarrow \infty$? $\endgroup$ – Satwik Pasani Oct 17 '13 at 15:07
  • $\begingroup$ @SatwikPasani, see the update to my answer $\endgroup$ – Alfred Centauri Oct 17 '13 at 15:31
1
$\begingroup$

How is the voltage drop across an inductor related to its current and inductance? It is given as:

\begin{equation*}V = L\frac{di}{dt}\end{equation*}

Now since the two inductors are connected in parallel you could relate their voltages and hence the rate of change of currents across them. Use that equation along with the equation for the sum of currents that you have obtained. You have two equations and two unknowns, $i_1$ & $i_2$, so you should be able to solve them! So you'll get them as functions of time and then you can put the limit that $t \rightarrow \infty$.

$\endgroup$
  • $\begingroup$ With changing magnetic flux, as is true for this case, the fields are non-conservative. Therefore, how can we assign a unique value to the potential drop when the drop is path dependent? $\endgroup$ – Satwik Pasani Oct 17 '13 at 15:02
  • $\begingroup$ As the net-electric field in an ideal inductor must be zero ($E=\rho J$), we obtain that the sum of the conservative and non-conservative electric fields is zero. Using the potential for the conservative electric field we obtain the above equation relating $V$ and $i$. $\endgroup$ – cutculus Oct 18 '13 at 15:18
1
$\begingroup$

The current voltage relation of an inductor is given by:

enter image description here

In steady state scenario, the time derivative is equal to zero. Hence the inductor acts as a short circuit. Now if you replaced the inductors in your circuit with short circuits, the question becomes meaningless as it asks about current division between two short circuits.

The only way to overcome this is to take a finite resistance into consideration. Although the question says the inductors are ideal, I assume the question means there is no bulk resistance.

Since one of them has double the value of the other, you can think of it as two 5 mH inductors in series. Hence the 10 mH inductor is going to have half the current according to current divider rule:

enter image description here

The total current is 4 A as you computed it. The current in the other branch is 2/3 the total current

$\endgroup$
1
$\begingroup$

You are basically on the right track. Actually I think this question is a bit unfair because the inductors being ideal (having no resistance) is unrealistic and therefore leads to a unrealistic solution. Your equation for the total current is correct. Note that in steady state, the current is just 4 A. The sum of the currents in the two inductors must therefore equal 4 A.

With absolutely perfect inductors starting at 0 current and with the same voltage applied to them, the current thru each will be inversely proportional to its inductance. The 5 mH inductor will therefore always have twice the current thru it that the 10 mH inductor will. Simply divide up the 4 A steady state current by that ratio to get your answer.

Again, this is a totally unrealistic question. This solution wouldn't work if the DC resistance of either inductor weren't exactly 0. In a real situation with non-zero resistance, the inductance wouldn't matter at all in the steady state, and each inductor would carry current inversely proportional to it resistance.

$\endgroup$
  • $\begingroup$ +1 and thank you for a good answer. But don't the answers I cite sum upto $4A$ as is required while considering the voltage of the battery and net resistance? $\endgroup$ – Satwik Pasani Oct 17 '13 at 15:16
  • 1
    $\begingroup$ @Satwik: Oops, yes. I misread the little numbers. They are rather small on my screen. Sorry for the confusion. $\endgroup$ – Olin Lathrop Oct 17 '13 at 17:21
  • 1
    $\begingroup$ It really is an evil problem. The inductors are resistance-less (in reality, they would be a pair of superconducting coils, bridged by superconducting shunts) and so the problem is heavily dependent on the persistent current flowing within the pair of inductors before the switch is closed. The persistent current could be anything, so the problem isn't even fully specified. $\endgroup$ – Nanite Oct 17 '13 at 19:29
1
$\begingroup$

Well there is NO switch in the circuit. And there is no information about the order in which the components are assembled. So the answer is indeterminate. The 2:1 current answer is only correct if the two inductors are connected in parallel before any Voltage is applied. If either of the two inductors, is the last element to be connected, then ALL of the current would be flowing in the other inductor, and there never would be any Voltage across the inductors to change the currents, so the second inductor connected, would have zero current.

$\endgroup$
  • $\begingroup$ Thank you for the answer. Can we show, what you just said (zero current for the inductor connected at last) mathematically or in a formal way so as to convince my more-conventional teacher? $\endgroup$ – Satwik Pasani Oct 18 '13 at 2:46
  • $\begingroup$ Well Satwik, when the circuit with one inductor is assembled, you get the initial inductive transient, with a time constant L/R, which decays to the steady state with 4 Amps flowing. (you waited exactly as long as the teacher waited, before he gave you the problem to look at), so we know the inductor Voltage IS zero (he says so). Then you connect the second inductor across this zero Voltage; so with zero Voltage neither inductor current can change; ergo 4 Amps, and zero Amps. QED. You can describe all that in your own words. He left out the switch, so he screwed up the problem. $\endgroup$ – user26165 Oct 18 '13 at 20:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.