####   Syllabus Sections:-

3. Technical Aspects

Potential Difference and Electromotive Force.

3a.1 Understand the difference between potential difference (p.d.) and electromotive force (e.m.f.).

#### The abbreviation "emf" stands for electromotive force, which is measured in Volts and is the driving force or a pressure to propel electrons around an electrical circuit,

If the probes of a voltmeter were placed on the terminals of a battery whilst it is not connected to a circuit i.e. no current is being drawn the measured potential difference in Volts between the two points would under these conditions equate to the "emf" Volts of the battery.

Potential Difference is also measured in Volts and is the name given to the difference in voltage that may exist between two different points in an electrical circuit.

So what is the significance of all of this ?

The "emf" is the FORCE pushing the electrons round a circuit, from the source whether it is a battery, generator or power supply and as stated above is measured in volts.

The potential difference, is not a force, but is a voltage difference that exists between any two points in a circuit not at the same potential

Understand the concept of source resistance (impedance) and voltage drop due to current flow.

The term impedance in relation to a DC source may be confusing but it is an historical term which some physicists still like to use to describe the resistance to a flow of electrons inside a source of electrons.

Some of the total energy used in any circuit will be consumed forcing electrons through this internal resistance. So looking at the circuit above you will observe that we have a single resistor R connected across a battery of emf E.

"r" is shown as the internal resistance of the battery. The voltmeter, whilst it is connected to the battery terminals is actually measuring the potential difference across R and not the emf of the battery E because of the voltage drop across the internal resistance r when current is flowing.

So the voltage at the terminals now, with current flowing, will be lower than the battery emf due to the voltage lost across the battery internal resistance. As more current is drawn so the battery internal voltage drop will increase. Shorting the battery (a dangerous practice...do not do this) will cause all of the emf to be lost across this internal resistance. Therefore the internal resistance can now be seen as the limiting factor as to the maximum current that can actually be drawn from the battery!

#### So to sum up :-

1. The electromotive force (EMF) is the driving force pushing electrons around an electrical circuit.

2. That there will be an internal voltage drop in any source due to the current flow through the source's internal resistance.

3. The internal resistance of the electron source is also known as the source resistance (or source impedance)

#### Still mystified well look at this another way :-

A car battery is a good source of emf. In an ideal source of emf it will maintain a constant potential difference between its terminals, independent of the current, I, through it or the resistance, R, across it.

The formula for an ideal source of emf is V= I x R, which is known as Ohm's Law which you were introduced to in the Foundation licence as a magic triangle.

The potential difference across a real source in a circuit, however, is not equal to the ideal emf. The reason is that the flow of electrons passing through the materials of the battery encounters an internal resistance r and experiences a drop in potential difference equal to I x r (Ir).

Thus the EMF of the battery is given by

# Vemf = IR + Ir

Where Vemf  is the emf of the battery measured out of circuit,  I is the current in the external circuit, R is the load in the external circuit and r is the internal resistance of the battery.

If we transpose the equation  Vemf = IR + Ir
(by first subtract IR from both sides )  Vemf - IR = Ir
( then divide by each side by  I )  (Vemf - IR)/I = r

but as we want to know the internal resistance we can write it out as

# r = (Vemf - IR)/I  