Actual electric potential at terminals of battery

Question

I understand that a 9V battery will produce an electrical potential difference/voltage of 9v across the positive and negative terminal, with the positive terminal having the higher potential by definition.

I am aware that the actual electric potential at the positive terminal may not be 9V or 0V for the negative terminal but the difference in the electric potential is 9v.

Wikipedia tells me that the electric potential of a point is the amount of energy needed to move a unit positive charge from a reference point, usually infinity, to that point. I do not understand how this relates to circuits.

My question is really what is meant by the actual electric potential at the terminals of each battery and is it possible to measure?

thanks

Answer 1

Potential is always measured relative to some reference point. This can be the earth, the moon, the car chassis, the negative or positive terminal of the power source or even to an AC signal. For most practical applications we don't use infinity as a reference but rather something much more local.

^{simulate this circuit – Schematic created using CircuitLab}

Figures 1 to 8. Various measurement and reference schemes for a 9 V battery and voltmeter.

1. The voltmeter measures the battery voltage as there is a complete circuit.
2. In this case there is a circuit through local earth. The voltmeter shows 9 V.
3. This represents a chassis connection. Again 9 V.
4. In this case some arbitrary point in the circuit has been taken as ground. In most battery powered equipment this will be the DC negative supply. One recent question on this site showed an old transistor radio circuit with the battery positive as the "ground".
5. We have the option of connecting the circuit ground to earth. This prevents floating of the device power supply and might be used for safety, to avoid audio hum, etc., depending on the application.
6. Without a reference for the voltmeter it will read 0 V. Note that a very sensitive digital meter will show random readings due to stray electric fields. Putting a medium resistance (say 100k) across the meter terminals will cause the reading to collapse to zero.
7. Inverting the battery we now have a positive ground. The voltmeter will read -9 V.
8. Inverting the meter so that its positive lead is grounded will also result in a -9 V reading.

For most practical electronics you just need to work out the potential between points. When debugging it is very often most convenient to attach the negative probe to the circuit ground and take all readings with reference to that point.

Answer 2

Wikipedia tells me that the electric potential of a point is the amount of energy needed to move a unit positive charge from a reference point, usually infinity, to that point. I do not understand how this relates to circuits.

It doesn't relate to circuits. It's a definition of potential, but not a practical one.

While each terminal of a 9v battery does have a potential with respect to a reference point at infinity, it's not a stable or useful potential, as it takes so little charge to change it. For instance, a 9v battery may have a capacitance to infinity of a few pFs. Adding just a few nC of charge to it, as you might easily do by walking across a carpet with it, would change the potential by thousands of volts.

Connecting it to another body using a resistance of 100Mohm (about the highest value of resistor whose accuracy won't be trashed by surface contamination if you touch it) would equalise the potentials with a time constant of mS. You touching it (body resistance 100k) would get it to the same potential as you with a time constant of \$\mu S\$.

It is possible to measure the potential to ground of an isolated battery with care, and a very, very high input impedance meter (often called an electrometer). It's not very useful though.

Answer 3

The units of volts (a joule per coulomb); moving 1 coulomb of charge across 1 volt of electric potential requires 1 joule of work.

From Q=CV and a battery with a given mAh capacity in 1h

It means you can compute the equivalent C of any battery if it starts at Vi and is depleted at Vf where final hipothetical energy ( if it were not depleted of charge) using.

E[Joules] =1/2C(Vi² -Vf)²

This should be approx the same energy=power * time accumulated for each smal increment of time as voltage and current changes with instantaneous power =V(t) * I (t)

In practice we use mAh *Vavg = mWh then /3.6 ks = Joules