# Electricity question potential difference

#### Jnblose

1. The problem statement, all variables and given/known data Givens: ΔV1 = 14 V, ΔV2 = 4.6 V, ΔV3 = 8.8 V, R = 3.0 )
2. Relevant equations

3. The attempt at a solution
I thought what I was doing was close lol.... apparently not. I decided to Combine 10 ohm and 3 ohm into 13 ohms in series with 14V and Combine 5 and 2 ohms into 7 ohms in series with 8.8 V. After that Im lost... IDK if thats even a good start hahaha.

#### Delphi51

Homework Helper
Oh, a complex one! Are you looking for the potential between a and b?

This is a network, not a simple circuit. You must use this theorem, whose name I forget:
"The sum of the potentials around a closed loop is zero."
It applies to the upper loop, the lower loop, and the loop around the outside of the whole thing. So, 3 possibilities. Sometimes you have to do more than one loop to find more than one unknown. In this problem you have unknown current and unknown Vab.

Begin by drawing an arrow to show which way the current might go in the loop. Either way is okay. Then if the current comes out of a source of potential from the + side, you take it to be positive.

#### Jnblose

Yes just Vab... this one is a beeeeeeeaaaaaaaassssssssttttttt.

#### Delphi51

Homework Helper
Oops - even more complicated that I thought at first because the current in the loops is not the same. You'll need I1 around the upper loop, I2 around the lower loop. The current through the middle part will be I1 + I2 (or minus depending on how you draw those current arrows). Three unknowns. Use all three loops.

I don't think you need it, but the other network theorem is
"The sum of the currents out of any point in the circuit is zero."

#### Kruum

The current through the middle part will be I1 + I2.
Actually no, if the picture is drawn correctly. Because the circuit is open between points a and b the middle part won't have a current at all, only potential.

I guess the easiest way to solve this is to calculate the effects of each voltage supply independently and sum them at the end.

Last edited:

#### mplayer

You can still go around the two inner loops using KVL to solve for I1 and I2...just use Vab within your calculation. Using the third outer loop wont help, it does not generate a linearly independent KVL equation. See if you can come up with a constraint equation for Vab to somehow relate it to the unknown currents.

#### Kruum

You guys are making this a lot harder than it actually is. ;) You can reduce the circuit into a very basic circuit. From there you can solve I, yep there's only one. And now it should be a walk in a park to determine the potential between a and b.

#### Delphi51

Homework Helper
Terribly sorry for misleading you, Jnblose!
I thought of the ab gap as a voltage source and totally blew this problem!

So we just have 14 - 8.8 Volts in series with 20 ohms for .26A of current?

#### Kruum

So we just have 14 - 8.8 Volts in series with 20 ohms for .26A of current?
You're forgetting the effect of V2. Even if there is no current in that wire, it doesn't mean the voltage source isn't working.

#### Delphi51

Homework Helper
But V2 can't contribute to the current because one side of it is not connected to anything.

#### Kruum

But V2 can't contribute to the current because one side of it is not connected to anything.
Yes it can. If current is the flow of electrons, it's easy to see why there is no current in an open circuit. But potential doesn't need anything to flow through. You could compare electric potential with gravitational potential.

#### Delphi51

Homework Helper
To me it is like the circuit around the outside stands alone. Someone has attached ONE terminal of a battery to it at one point, but that makes no difference to the current around that closed circuit - .26A. You could sum the voltages around that closed loop with a few IR's in it and get the same answer.

Of course V2 does affect the ab potential. I get 10.62 Volts for either the lower or upper portion of the loop, so 1.62 - 4.6 = 6.02 V from a to b.

#### Kruum

To me it is like the circuit around the outside stands alone. Someone has attached ONE terminal of a battery to it at one point, but that makes no difference to the current around that closed circuit - .26A. You could sum the voltages around that closed loop with a few IR's in it and get the same answer.

Of course V2 does affect the ab potential. I get 10.62 Volts for either the lower or upper portion of the loop, so 1.62 - 4.6 = 6.02 V from a to b.
Yep, you're calculations are right, but only because of the other supplies. Let's say the V3 would've been turned off or around, but the rest of the circuit remained the same, the current would have been different.

Last edited:

#### Delphi51

Homework Helper
Of course V1 and V3 affect the current. But V2 need not be considered in finding the current so we have just the simple series circuit. Thanks very much for catching my blunder in thinking we had a two or three network loop!

#### Kruum

Of course V1 and V3 affect the current. But V2 need not be considered in finding the current so we have just the simple series circuit.
That's actually not what I meant. I think we can agree on the fact that V2 creates a potential between a and b. If so, that potential is the same everywhere, since V2 doesn't create a current. So V2 creates that same potential near V1, which then affects the voltage created by V1. Now, in this case it does the exact same thing to V3. But due to the placement of the supplies they cancel each others out. So if the V3 was the other way around, your method vs. mine would give different answer.

I know I shouldn't argue with a teacher, but if I'm wrong here, I'm definitely going to remember that lesson forever. =D

### The Physics Forums Way

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving