# Potential Difference Across Resistor in Parallel with Inductor

1. Feb 26, 2014

### LikesIntuition

If we have a resistor and inductor in parallel, how does the resistor maintain a potential difference equal to our emf? Am I correct in saying that for the inductor to maintain a voltage difference, there has to be some change in current, so that the current going through the resistor is also changing? If so, how is the voltage across the resistor maintained with changing current?

2. Feb 26, 2014

### sophiecentaur

What circuit are you describing (diagram??) and what are the initial conditions? Your question is not well enough specified for a proper answer yet.

3. Feb 26, 2014

### LikesIntuition

I don't really have a specific circuit in mind. How about just an emf connected to a resistor, which is connected in parallel to an inductor? And maybe a switch separating the emf from the rest of the circuit. So when the switch is open, nothing is happening. So I guess my question is, what happens to the potential difference across the resistor and inductor when the switch is closed?
My understanding is that the voltage across the individual resistor and inductor has to equal the emf. And for the inductor's back emf (which I currently understand to be where its voltage difference we are talking about comes from) to equal the emf, there has to be a change in the current that passes through it. But if that's happening, the voltage through the resistor would also need to be changing. So how could the resistor maintain a potential difference equal to the emf?

4. Feb 26, 2014

### KingCrimson

if you are talking about an inductor , a resistance-less inductor , in a dc circuit , then what you say doesn't make sense after the current settles down
just as the current settles down , there's no change anymore , the inductor becomes nothing more than a normal -perfect- conductor in our case since its resistance-less and no current would flow through the resistor
in real life there is no perfect conductor , the inductor itself has a certain amount of resistance , just as soon as the current settles down , the inductor becomes a resistor
* all of that applies only to D.C. circuits *

5. Feb 26, 2014

### willem2

The voltage across the resistor doesn't need to change. If you connect an ideal voltage source to an ideal inductor the current through the inductor will keep rising while the voltage across it remains constant, according to dI/dt = V/L.
The voltage across the resistor will be the voltage of the ideal voltage source, and since that voltage doesn't change, neither will the voltage across the riesistor, or the current through it.

6. Feb 26, 2014

### LikesIntuition

Oh that's true. Thanks, that clears things up I think!

7. Feb 26, 2014

### KingCrimson

btw just to make things clear
i think that if the back emf produced by the inductor is equal to that of the battery , no current would flow through the resistor
* correct me if i am wrong *

8. Feb 26, 2014

### LikesIntuition

I believe this is correct. Because at first, I don't think the resistor has any potential difference across it. Won't the back emf be highest when the switch is first closed? Then gradually more and more current will flow through the resistor. If this is true, then my original question was due to a lack of understanding of how these things work with each other...

9. Feb 27, 2014

### willem2

This is not correct. There's only back EMF when there's a change in current. With an ideal inductor and an ideal voltage source the circuit will never settle down, and the current keeps rising, while the back emf is equal to the battery voltage all the time.

The back emf is only highest if the switch is first closed, if you have a non-ideal voltage source or inductor, and the current will finally settle down. When the current is constant there is no more back emf.

If the inductor does have some resistance R_i, then the back emf will decrease until the current through the inductor is equal to V/R_i. Current through the resistor R will always be equal to V/R, because it only depends on the voltage of the voltage source, wich is ideal and won't change, whatever the current through the inductor is.

If the voltage source isn't ideal and does have some resistance R_v, than the current throuhg the inductor will settle at V/R_v. In this case the entire voltage is across the internal resistance of the voltage source, so there will be no voltage across the inductor or the resistor, and no current through the resistor.

10. Feb 27, 2014

### KingCrimson

wouldn't the current keep rising until it settles down ?

11. Feb 27, 2014

### willem2

With an ideal conductor and ideal voltage source, the current will never settle down.

With a non-ideal inductor and voltage source the current will settle down when the current is equal to V/(R_i + R_v). At this point the back-emf will be 0, so it is not equal to the emf of the battery.

12. Feb 27, 2014

### sophiecentaur

The current will be Zero, aamof. Not intuitive, I agree but you actually would need to introduce some series resistance into the Inductor to allow any current to flow at all.
The parallel Resistor is a red herring - the current will just (and always) be V/R.

Ideal cases need to be treated with care; you end up with dividing by zero or infinity and stuff like that.

13. Feb 27, 2014

### LikesIntuition

So are you saying that in an ideal circuit like this, there will be no current at all through the inductor unless it has a resistance? Or there will just be no current in the circuit?

14. Feb 27, 2014

### willem2

This doesn't make sense.

for the ideal case you get dI/dt = V/L, so I = (V/L) t

If you add a small series resistance R_i you'll get dI/dt = (V - I * R_i)/L.
The current is always smaller in this case, altough there's little difference as long as I * R_i is small compared to V

15. Feb 27, 2014

### Staff: Mentor

Careful, this is not correct. The voltage-current relationship for an inductor is $V=L \frac{dI}{dt}$. So a constant V across an ideal L means a constantly growing I. The current through an ideal inductor would continue to rise linearly until something breaks.

However, in practice, all real inductors have resistance, and many have a substantial amount of it, and so your point is well taken. It is not necessary as you stated, but it is certainly the usual case.

This is correct. The resistor in parallel has no bearing on the behavior of the inductor.

16. Feb 27, 2014

### sophiecentaur

As there is only a zero series resistance with the L, the time constant is L/R, which is infinite. Like I said, non-intuitive and, of course, unattainable.

17. Feb 27, 2014

### Staff: Mentor

The time constant is indeed infinite. That means that the inductor never reaches a steady state current even after an infinite amount of time. It most certainly does not mean that the current will be zero.

18. Feb 28, 2014

### sophiecentaur

That is not the definition of 'time constant'. T is the time to reach 1-1/e of steady state. With an exponential variation like that, no RL circuit ever reaches the steady state. You are finding it hard to accept, possibly, but consider what happens for a finite R. Then double the R and redouble and then keep doing it. The limit takes you to zero increase in current for All intervals you may choose.
Alternatively, consider what
1-(1/(e^10000000) gives you.

19. Feb 28, 2014

### Staff: Mentor

Correct. It never reaches the steady state, therefore the current must always be changing. If the current were always zero then it would be in its steady state instantaneously. That is a time constant of 0, not infinity. You are misunderstanding what an infinite time constant implies.

No, it has nothing to do with acceptance from either of us. The math is clear and unambiguous. The equation for a series RL circuit is:
$$v=iR+L\frac{di}{dt}$$
For constant v and for i(0)=0 this has the solution:
$$i=\frac{v}{R}\left( 1-e^{-\frac{R}{L}t} \right)$$
For R=0 the above equation becomes undefined: $i=\infty 0$. You are incorrectly assuming that $\infty 0 = 0$ but that is not legitimate. You have to take the limit instead. When you do that you find:
$$\lim_{R \to 0} i = \frac{v}{L}t$$

Furthermore, if you go back to the original differential equation and set R=0 in that equation then you have:
$$v=L\frac{di}{dt}$$
Which has the same solution
$$i = \frac{v}{L}t$$

So two different ways confirm the same answer and it is most emphatically not i=0.

Last edited: Feb 28, 2014
20. Feb 28, 2014

### sophiecentaur

You are correct that, " forR=0 the equation is undefined". You have to consider the limit as R approaches zero. You have not done that. In my analysis, which is perfectly valid, I consider the time constant L/R (that is correct) and apply that to the exponential form of the solution to yours. What you say about a non zero R is correct but the limit as R goes to zero is that the time approaches infinity and, moreover, the time differential approaches zero (=DC)

If it is as you say, you would have to give a value for this current. You cannot do that without having a non zero R value.

21. Feb 28, 2014

### willem2

But dalespam has just done that with his last equation. for R=0 one way of solving the differential equaton doesn't work, but solving it in another way works just fine.

22. Feb 28, 2014

### sophiecentaur

So your conclusion is that the current is a particular non zero value?

23. Feb 28, 2014

### sophiecentaur

Can you show how that expression
has the limit:
?
I would have to accept what you say if you can.

24. Feb 28, 2014

### Staff: Mentor

Certainly, simply do a Taylor series expansion:
$$i=\frac{v}{R}\left( 1-e^{-\frac{R}{L}t} \right)=\frac{v}{L}t-\frac{Rv}{2L^2}t^2+\frac{R^2v}{6L^3}t^3-\frac{R^3v}{24L^4}t^4+...$$

In the limit as R→0 all but the first term go to zero. Therefore:
$$\lim_{R \to 0} i = \frac{v}{L}t$$

25. Feb 28, 2014

### sophiecentaur

That's fine. Thank you - it makes sense