# Internal resistence: chicken and egg misconcpetion

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1. Oct 2, 2014

### davidbenari

I feel like have a chicken and egg type of problem regarding internal resistences because of the following problem:

The voltage across a battery with internal resistence is given by the equation
EMF-Ir=V
Given the voltage across the battery then the voltage in the circuit is
V=IR

So, if I have more current then my voltage decreases, if I have less voltage then my current decrease because of V=IR.

So what's happening here?

If I have an increase in current, then it immediately drops?

Thanks.

2. Oct 2, 2014

### Simon Bridge

... given the voltage across the battery (which is, properly, the voltage across the load) - but the voltage across the battery depends on the size of the load. The fact the voltage is "given" means some earlier steps have already been completed.

To treat the circuit properly, you need to know the EMF and internal resistance. Given those, you can work out the current in the circuit and, thus the voltage dropped across the load.

3. Oct 3, 2014

### CWatters

Correct.

If the load draws more current the battery terminal voltage V decreases due to the internal resistance of the battery.

If the battery terminal voltage V decreases then the current flowing in the load will decrease due to Ohms law (assuming the load obeys ohms law)..

These aren't inconsistent. You can plot a graph of I vs V for both of the above. Where the lines cross will be a valid solution to both.

4. Oct 3, 2014

### xAxis

The I in the battery is not the same I across the circuit.

5. Oct 3, 2014

### sophiecentaur

Not at all. As you lower the resistance of the load, the current will increase until an equilibrium situation is reached. Nothing has to drop back down again because it will not 'overshoot'. (Unless you include reactive elements in your circuit; but let's sort out the straightforward case first).

6. Oct 3, 2014

### CWatters

I think you will find it is.

7. Oct 3, 2014

### davidbenari

I'm going to be irrelevant here, but in a circuit, if we assume a constant current, then the electric field inside the ideal conductors (not the resistors) is zero right? Otherwise charges would accelerate.

Thanks.

8. Oct 3, 2014

### Staff: Mentor

Right. That's what makes them ideal.

Yes, but it's more helpful to consider that if the electric field is not zero then there must be some potential difference between the two ends of the conductor, therefore some non-zero resistance... And then it's not an ideal conductor.

9. Oct 4, 2014

### sophiecentaur

You are describing what can happen in a beam of electrons. But that isn't a metal conductor.

If there is no change in potential then no work is done and no acceleration. This is all very basic stuff but it does need thinking about from time to time, to avoid reaching some strange conclusions about the nature of things.

10. Oct 4, 2014

### Staff: Mentor

Combine the two equations:
EMF - I r = I R
EMF = I (R+r)

There is nothing surprising or "chicken and egg" here. The current is simply determined by the total resistance, which is the sum of the internal and external resistances. The open circuit voltage is equal to the EMF, and the short circuit current is equal to EMF/r.