Potential difference measures the energy each coulomb gives up in a component
Definition
Potential difference (p.d.)
The work done by a unit charge passing through a component.
V=WQ V = \frac{W}{Q} V=QW- As charge passes through a lamp it transfers energy to the filament, which heats up and glows.
- The p.d. across the lamp counts how much energy each coulomb delivers while it is inside.
- A p.d. of 1 V1\ \text{V}1 V across a component means each coulomb gives up 1 J1\ \text{J}1 J of energy in it.
- P.d. is always measured between two points, one on each side of the component, which is why it is also called the voltage across it.
- E.m.f. and p.d. are both an energy per coulomb, but the e.m.f. is the energy given out by the source while a p.d. is the energy delivered in one part of the circuit.
- In a simple loop the p.d.s across the components add up to the e.m.f. of the supply, because all the energy handed out has to be delivered somewhere.
A voltmeter goes in parallel, across the component
- A voltmeter compares two points, so it is connected in parallel with the component, with one lead on each side of it.
- It must have a very high resistance, so that almost no current is diverted through it and the circuit is left as it was.
- Connect the red terminal to the point nearer the positive of the supply, or an analogue needle will swing the wrong way.
- Pick the range before switching on, starting high and stepping down until the reading fills most of the scale.
- On an analogue voltmeter read the needle at eye level and square to the scale, to avoid parallax error, and check which of the printed scales matches the range in use.
- A digital voltmeter removes that reading error and usually offers an extra decimal place, giving 2.47 V2.47\ \text{V}2.47 V where an analogue meter would only give 2.5 V2.5\ \text{V}2.5 V.
- Reading 2.47 V2.47\ \text{V}2.47 V on a 000 to 3 V3\ \text{V}3 V range is far more precise than reading the same p.d. squashed into the bottom of a 000 to 30 V30\ \text{V}30 V range.
Calculating potential difference from work and charge
- The equation to recall is V=WQV = \dfrac{W}{Q}V=QW, where VVV is the p.d. in V\text{V}V, WWW is the work done in J\text{J}J and QQQ is the charge in C\text{C}C.
- Rearranged, the energy transferred inside the component is W=VQW = VQW=VQ.
- Substituting Q=ItQ = ItQ=It turns this into W=VItW = VItW=VIt, which is the form used for electrical energy calculations.
- Keep energy in joules, charge in coulombs and time in seconds so that the answer comes out in volts.
Common Mistake
- A voltmeter never goes in series with the component it is measuring, because its high resistance would almost stop the current.
- Do not write about the voltage "through" a component, since a p.d. is always across it.
- E.m.f. and p.d. share the volt as a unit but not their meaning, so keep the source and the component separate in your answer.
- Do not describe p.d. as the amount of current or the "push of the electrons"; it is energy per unit charge.
Exam technique
- On a circuit diagram, draw the voltmeter as a loop around the component, joined to the wire at the two points either side of it.
- When asked why a voltmeter has a high resistance, the mark is for it drawing almost no current, so the circuit it is measuring is not changed.
- Quote a reading to the precision the instrument actually gives, and always attach the unit.
- The definition mark needs the unit charge idea and the words passing through a component, so do not stop at "energy transferred".
Self review
- Define potential difference.
- State the unit of p.d. and explain what a p.d. of 1 V1\ \text{V}1 V means.
- How is a voltmeter connected, and why must its resistance be high?
- Give one advantage of a digital voltmeter over an analogue one.
- 30 J30\ \text{J}30 J is transferred as 5.0 C5.0\ \text{C}5.0 C passes through a heater. Calculate the p.d. across it.
