[thedoc (OP)]

Lily Peace  asked the Naked Scientists:
   
Hi Chris,

I'm doing A level physics, and am currently dragging my way through current and electricity.  I was looking at internal resistance and terminal p.d. and my book says that as the current increases, the terminal p.d. decreases.  I get how it works through V=E-Ir.

BUT then my friend pointed out that V=IR, so how can this exist at the same time, if the current is directly proportional to the voltage?  This problem has been plaguing us for months, if you can, please help!

Thanks

Lily

What do you think?

[graham.d]

Hi Lily. I am not sure of what the circuit is that you are looking at or where you are measuring the Voltages V and E. However, I will make a guess that you are somehow applying a voltage from some power supply or battery across a resistive load and measuring the current. If you do this then Ohm's Law will apply and V=IR where 'V' is the voltage across the terminals of the load, 'I' will be the measured current and 'R' is the resistance of the load. When you speak of "internal resistance" I am guessing you mean the resistance of the power supply which may include that of the ammeter measuring the current. For low currents this "internal resistance" of the power supply is usually negligible as it is much smaller than the resistance of the load. If the the current is high, however, the voltage applied to the load may not be exactly what you think because of the supply's internal resistance, any ammeter in series and the resistance of the connecting wires.

I don't know if this is exactly what you mean, however. You can often neglect these effects but if making accurate measurements in certain situations you may need to take precautions to eliminate the effects of internal resistance of supplies and ammeters (the resistance is ideally zero but is finite in practice) or of voltmeters (the resistance is ideally infinite but, of course, isn't in practice).

[evan_au]

It sounds like you are looking at the internal resistance of a battery (but it's similar to a DC power supply as described by Graham).

A battery is often modeled as:

• an "ideal" voltage source with a certain voltage E, and infinite current capacity
  
• in series with a resistor, which limits the current to a maximum value, and also causes a voltage droop at lower currents.
  

In reality these are not separate components within the battery casing, but represent multiple effects like the resistance of the battery terminals, and energy lost by the ions diffusing through the electrolyte.

Perhaps the conundrum you see is that:

• For an external load resistor, you connect a voltmeter's (say) negative terminal to the terminal of the battery, and measure the voltage across the resistor. You get Ohm's Law: V=IR
  
• For an internal source resistance, you connect a voltmeter's negative terminal to the negative terminal of the battery, and the positive terminal of the voltmeter to the positive terminal of the battery. Increasing the current drawn from the battery causes the voltage of the battery to decrease. In this case, the voltage seems change like -IR, the opposite of Ohm's Law above.

The problem here is that you cannot measure the current across the internal source resistance, because it is a distributed resistance inherent in the construction of the battery. Because you can't gain access to both terminals of this imaginary resistor, it looks like instead of voltage increasing with current, voltage decreases with current.

In electrical engineering, this is handled by Kirchoff's voltage law, which states that the sum of all voltages around a circuit is zero. This works for external load resistances, as well as the modeled internal resistance of the battery. The key is to ensure that you follow the around the circuit in a single direction.

[lunar11]

One trick is to simply think of the circuit with two resistors in series. One the internal resistance, r and the other the load resistance, R. The current I=emf/(R+r). Then to find the terminal pd, V=IR. The 'lost volts' would be Ir