# Induced EMF within a part of a parallel circuit?

1. Apr 5, 2015

### PhiowPhi

A Power supply is connected to a parallel circuit with resistance R, there are three wires connected in parallel where the voltages would be the same,however, the current would be dived based on the number of wires and the resistance is the same.

What would happen if wire (c) was placed in a changing magnetic field, and there is an induced EMF as diagram here:

With two cases, one the polarity is opposing the power supply and and the other case is the induced EMF's polarity supporting the power supply as the diagram above.

What is the resulting voltage? If the induced EMF is opposing the power supply then the EMF would be: $V_ps - V_e$, and if the voltage is supporting the power supply(as the second diagram) the voltage is added?

I could not predict the outcome when thinking about the theory.

2. Apr 5, 2015

### Svein

The way you have drawn it, there is no connection between the upper part of b and the upper part of a - c.

3. Apr 5, 2015

### Staff: Mentor

That "hump" in your wiring should be a connection of all 3 wires.

Perhaps it may be clearer were you to draw 3 resistors in those wires, then show the induced emf as though it is a battery added into that wire. See how that affects things.

4. Apr 5, 2015

### PhiowPhi

Induced EMF opposing the power supply:

Induced EMF supporting the power supply:

I could not be certain of the outcome, for the first diagram where the induced-EMF is opposing the power supply I assumed that the voltage would be 15, and from the second diagram I assumed it to be 25V, however, not sure what is correct. Also, could not figure out the voltage at the blue points(a,c).

5. Apr 5, 2015

### Svein

In the upper diagram, the "induced E" will almost short-circuit the power supply. The voltage between a and c will be forced to 5V.

The lower diagram isn't much better, since the PS will try to force +20V between a and c. As drawn, the "induced E" will definitely short-circuit the power supply.

6. Apr 5, 2015

### jim hardy

Svein is saying "doesn't that wire going from upper right 'c' down to 6 volt source have some resistance ? "

7. Apr 5, 2015

### PhiowPhi

Yes, it does.

8. Apr 5, 2015

### Svein

OK. Look at the upper diagram. Let us assume a very thin wire and assign it a resistance of 0.1Ω. The current through that wire will then be (20V - 5V)/0.1Ω = 150A. Something is definitely going to break!

9. Apr 5, 2015

### Staff: Mentor

Don't bother posting your revised circuit. I intended for you to sketch it on paper and examine it by yourself. 3 wires, 3 resistances, with that extra battery in series with one of them. What is going to be different then?

10. Apr 5, 2015

### PhiowPhi

Its kind of frustrating that with all the time I've invested into understanding circuit theory I can't come up with a proper answer.

11. Apr 5, 2015

### jim hardy

Looks like a good time label some currents and write Kirchoff's laws equations around the loops....

Nobody works these in their head until they've been at it for a semester or two.

12. Apr 5, 2015

### PhiowPhi

Sure I'll give it a try, but isn't this a short-circuit? As Svein stated it.

13. Apr 5, 2015

### jim hardy

looks to me like a 0.1 ohm resistor not a short circuit.

In beginner courses a short circuit is zero ohms.
When one moves on to real life "short circuit analysis", the resistance (and inductance) of the wiring comes into play.
That's why you'll never get infinite amps out of your wall socket, but you might get a thousand for a millisecond or two.

14. Apr 5, 2015

### PhiowPhi

But wouldn't the existence of the induced-EMF on wire (c), cause a short circuit? Or would it change the overall voltage... not so sure I'll play around with KVL&KCL, and will share my results and see where i'm wrong.

15. Apr 5, 2015

### jim hardy

How ?

KVL and KCL will not fail you. Admittedly it's hard work when we're new to it but
what does not kill us , strengthens us.

16. Apr 5, 2015

### PhiowPhi

Not sure, just a gut feeling.

Daily motto.

17. Apr 6, 2015

### PhiowPhi

I worked on a few attempts and here is my answer: Based on the wires resistance being the same(in fact for the whole circuit), the voltages for wire a/b would be 20V and 15V for wire c due to the induced-EMF opposing the power supply, and 25V when its non-opposing.

Here is the most confusing part, the current. I think that the current would divide out based on each wire's voltage. PS is at 20V/100V = 0.2A/3 = 0.06A
However, because the last wire has lower PD it should be getting much less then that value(Not sure how to find that even with KCL and KVL). The same assumption goes when the induced-emf is non-opposing 25V would require higher current than the other wires.

18. Apr 6, 2015

### PhiowPhi

Also, would the voltage still be 20V all around for both circuits. Im confused now with the induced-EMF changing the PD for the whole circuit? Or the single wire?

19. Apr 6, 2015

### jim hardy

Well, looks like you successfully applied KVL .... Hooray !

Say what ?

Focus on the three legs one at a time.
What is voltage across R1 ? What does Ohm's law say the current must be through R1 ?
Repeat for other two resistances.
Then sum currents by KCL.

This is how we work problems, one baby step at a time. Redraw between each step , gradually building up the number of known quantities and reducing the number of unknowns.

Everybody wants to leap straight to the answer. But it is no more possible to do that than to climb stairs one whole flight at a time.

20. Apr 6, 2015

### PhiowPhi

21. Apr 6, 2015

### jim hardy

You know, i suffer terribly from "fear of failure" and "fear of frustration" . It causes me to fluster about instead of digging into problems.
That is the weak side of my nature trying to avoid the emotional pain of making a mistake. Somehow that weak side of me thinks it'll preserve my vanity to not even try.

Awareness of that flaw in my makeup keeps me on the lookout for the behavior. Symptom is I'll throw up all kinds of reasons why i cannot solve the problem confronting me.
I have this saying that i tell myself when i find myself flustering about:
"The easiest way out is straight through. "

Pride when swallowed is quite nourishing.

Success consists in large part of having a big inventory of little mistakes from which we've learned.

So, plug & chug those formulas. It is a huge mistake to not teach our "gut feel" about them.

old jim

Last edited: Apr 6, 2015
22. Apr 7, 2015

### PhiowPhi

@jim hardy Could help clarify my misconception of how this is not a short-circuit?
If you look at my last diagrams, point (c) would be the voltage from the power supply across all the resistor and the induced voltage( when non-opposing) and should be remaining voltage across the resistors - the induced voltage(when opposing), why wouldn't a short circuit be a possibility here? Considering that there is a un-equal potential difference in a parallel circuit?

I haven't work on the numbers yet, wanted to understand all aspects then crunch in the numbers.

23. Apr 7, 2015

### Staff: Mentor

Are you drawing the crcuit with three equal resistances yet? The last you posted here still shows only two resistances.

You'll get nowhere fast if you continue pondering an incomplete/incorrect schematic.

I suggest that you don't bother mentioning the phrase "a short circuit" again, there isn't and won't be such a thing in this exercise!

24. Apr 7, 2015

### PhiowPhi

Yes, I am considering equal resistances. I'm my papers there is, haven't posted one here since you all get the idea.
My understanding of a short-circuit itself is wrong, and if anyone else says there will be one I'd think there would be(which is wrong since I must understand it first).

25. Apr 7, 2015

### jim hardy

This image ?

Let's think about leg c for a moment.

If that voltage source is a fixed one,
then ,
as you observed earlier , by KVL leg c has current of 25(or 15) volts divided by resistance of wire in c.
If that resistance is zero then you have a short circuit, for that's what a short circuit is at beginner's level.

But in post #7 you admitted the wire in leg c has resistance, and that's not a short circuit.

So draw a resistor in leg c.
If you want a short circuit, write next to it "0 ohms".
If you do not want a short circuit, write a number other than zero.

Furthermore

Even if c has zero resistance,
you defined your voltage source not as fixed but as induced by ΔΦ/ Δt .
which means magnetic induction,
and induction works both ways
so your leg c has counter emf equal to its inductance L multiplied by Δi/Δt.
which is not a short circuit either.
Lacking any resistance , current will start at zero and increase forever, which is in accordance with ohm's law.

Are you ready to buckle down and tackle this problem yet ?
KIrchoff, Faraday and Ohm are crying out solve it for you, if only you'll let them.

Last edited: Apr 7, 2015