# Electric Potential in circuits

1. May 8, 2013

### Enterprise D

Hi,

I'm just starting to study circuits, and there's a thing I don't understand about electric potential.

I was under the impression that as a charge moves through the circuit from the high potential end of the battery to the low potential end, it lost electric potential energy which became kinetic energy.

However, I found this graph:

In circuit B, the diagram states that from A to B the electric potential stays at 6V. Surely the charge would lose some electric potential and gain some kinetic energy?

Can anyone explain this?

Thanks!

2. May 8, 2013

### Staff: Mentor

There is no gain in kinetic energy. You are confusing this with a mass moving in a gravitational field (where when it falls it loses PE and gains KE).

There is very little electrical potential loss in the wire because its resistance is so low. The resistance of the light bulbs is high, so that is where the bulk of the electrical potential (voltage) is dropped.

The current around the loop is constant, so all the charges move at the same speed the whole time they are flowing around the loop.

3. May 8, 2013

### nebbione

In circuit analysis, we usually suppose that resistance of a wire is '0', which is usually a good approximation for most electronics circuits, so if I have a certain potential 'A' in a point of a wire, if on the same wire, i don't go through a dipole (electronics component) i 'll have the same potential (which is an approximation, a good approximation), because the resistance of the wire is neglectable

4. May 8, 2013

### Enterprise D

If the charge doesn't lose electric potential, why is it at 0V when it reaches the low potential terminal? Does a charge not lose electric potential as it moves through an electric field?

5. May 8, 2013

### nebbione

Because all the potential gets lost on the lamp... let's restart form the begin.
In electromagnetism we talk more appropriately about difference of potential between two points of a circuit, (even if we usually simply say "potential").
So if we are talking about the potential in a point of a circuit it's always preferable to specify relative to which other point.
By default if we do not specify this relative potential and simply say... "At this point of the circuit i have a potential of 3.5V", even if we didn't specify to what this potential is relative to, we are referring to ground.
The ground by definition is a node of a circuit which has potential = 0.

That's why it's always a good idea to draw where the ground is in a circuit.

In your circuit the battery has a potential of 5V, this means that from the + and the - there are 5V, so (V+) - (V-)=0.

Moreover, if you write a KVL (Kirchoff Voltage Law) who says that the potential in an electrical net is conserved (holy law of the conservation of energy) you can see that V(battery)-V(lamp)=0 and i took positive sign for the battery and negative for the lamp which is a passive component.

6. May 8, 2013

### Introyble

What??? Check again, I thought Ben Franklin experimented with the idea and proved (think Franklin proved).... Anyway, don't the electrons actually flow from negitive back to positive????

Last edited by a moderator: May 8, 2013
7. May 8, 2013

### Staff: Mentor

Sure. When we talk about current flow, we mentally picture positive charges flowing from positive to negative. But in reality the electrons are going the other direction.

8. May 8, 2013

### Introyble

True. I was being a little facetious

9. May 8, 2013

### Staff: Mentor

Doh!

10. May 8, 2013

### Enterprise D

So if there were no lamp, would each charge have 5V of electric potential even when it reached the low-potential end of the circuit?

11. May 9, 2013

### davenn

if there was no lamp, just a wire between the terminals of the battery, there would be a short circuit between the terminals resulting in a zero potential difference and a high current flow

you can only have a potential difference across a load. In the case of the battery not connected to anything and you measure the voltage across the 2 terminals, you are measuring the potential difference across the internal resistance of the battery (or other power supply)

Dave

12. May 9, 2013

### davenn

Keep in mind we dont talk too much about potential difference as such in a circuit... what we are more interested in is voltage drop across various elements in a circuit

Take your circuit B, 6V battery and 2 lamps. IF the lamps are identical ie. have the same resistance, then the voltage drop across each of them will be equal measured at 3V....
putting a voltmeter across lamp 1, points B and C you will measure 3V
putting a voltmeter across lamp 2, points D and E you will again measure 3V

the addition of the 2 voltage drops across each lamp will equal the total voltage supplied by the battery = 6V

Dave

13. May 9, 2013

### Enterprise D

I think what's confusing me is the fact that I read that electric potential is a location-dependant quantity. So, for example, a positive test charge near a positively charged source will move away from the source, losing electric potential energy and gaining kinetic energy.

I don't understand how an electron moving through a wire can maintain the same electric potential (a location-dependant quantity) when it moves through the wire from one location to the next.

Thanks again!

14. May 9, 2013

### Staff: Mentor

It doesn't, because a real wire has a small but nonzero resistance. Consider a copper wire, 1.00 m long and 0.500 mm in diameter (circular cross section). It has a resistance of about 0.0856 Ω.

Use two of these wires to connect a 10.0 Ω resistor to a 10.0 V battery. Let the positive terminal of the battery be at 10.0 V and the negative terminal at 0.00 V. Treat the two wires and resistor as three resistances in series. You'll find that the "+" end of the resistor is at about 9.916 V and the "-" end is at about 0.084 V.

As you proceed around the circuit from the "+" terminal of the battery, the potential starts at 10.00 V, decreases linearly along the wire to 9.916 V as you enter the resistor, drops to 0.084V as you go through the resistor, and finally deceases linearly to 0 at the "-" terminal of the battery.

Last edited: May 9, 2013
15. May 10, 2013

### Enterprise D

Hello again! Sorry to be a pain, but I'm still having trouble understanding.

I understand now, thanks to your answers, how voltage changes throughout the circuit, but I don't understand why the following sentence is true:

"Two points in an electric circuit that are connected by an ideal conductor without resistance and not within a changing magnetic field, have a voltage of zero."

Take these two situations for example:

and

In the first situation, an electron moving from the upper box to the lower box loses electric potential. Across any two points in its trajectory there will be a certain voltage.

In the second situation, suddenly across two points before the bulb there is no voltage (I'm assuming it's an ideal conductor). How is this possible?

Thanks again.

16. May 10, 2013

### pumila

The voltage between two points is the product of the current times the resistance. The resistance of an ideal conductor is zero, so no matter how much current flows the voltage is zero.

If you have a perfect conductor between the battery and the lamp, then the lamp, then a perfect conductor back to the battery, there will be no voltage difference along either conductor (both ends will be at the same voltage in the circuit), and so the full battery voltage will appear across the lamp. The resistance of the lamp will then be what limits the current.

If you remove the lamp so that you have a perfect conductor across the battery, then for an ideal battery the current becomes infinite. However, even though you can get perfect conductors there is no such thing as a perfect battery and so the current is limited in this case by the internal limitations of the short-circuited battery. Both battery terminals will then be at the same voltage.

17. May 10, 2013

### Enterprise D

But what if I see the voltage as the amount of potential energy the electron loses as it moves towards the positive terminal of the battery? Why does it lose this energy in the first example but not in the second?