Recall and use the equation for e.m.f. E = W / Q

4.2.3 Electromotive force and potential difference

Learning objective

Recall and use the equation for electromotive force (e.m.f.)

\$E = \frac{W}{Q}\$

Key definitions

• Electromotive force (e.m.f.) – the work done per unit charge by a source (such as a battery or generator) in moving charge around a complete circuit. It is a measure of the source’s ability to drive current.
• Potential difference (p.d.) – the work done per unit charge as charge moves between two points in a circuit. It is the energy transferred between those points.
• Work (W) – energy transferred to or from the charge, measured in joules (J).
• Charge (Q) – the quantity of electricity, measured in coulombs (C).

Relationship between e.m.f. and potential difference

The e.m.f. of a source is the potential difference across its terminals when no current flows (open‑circuit condition). When a current flows, the terminal p.d. is reduced by internal resistance:

\$\text{Terminal p.d.} = E - I r_{\text{int}}\$

where I is the current and r_int is the internal resistance of the source.

Units

Deriving the e.m.f. equation

1. By definition, e.m.f. is the work done per unit charge:
   \$E = \frac{W}{Q}\$
2. If a current I flows for a time t, the charge transferred is
   \$Q = I t\$
3. Substituting into the e.m.f. definition gives an alternative form:
   \$E = \frac{W}{I t}\$
4. Since power P = \frac{W}{t}, we can also write
   \$E = \frac{P}{I}\$

Sample calculation

Example: A battery does 120 J of work to move 0.5 C of charge from its negative to its positive terminal. Find the e.m.f. of the battery.

Solution:

1. Write the e.m.f. formula: \$E = \frac{W}{Q}\$
2. Substitute the given values: \$E = \frac{120\ \text{J}}{0.5\ \text{C}}\$
3. Calculate: \$E = 240\ \text{V}\$

Therefore the e.m.f. of the battery is 240 V.

Common misconceptions

• Confusing e.m.f. with the terminal potential difference when a current is flowing. Remember e.m.f. is measured under open‑circuit conditions.
• Treating the symbol E as energy rather than a voltage. In this context E represents electromotive force, measured in volts.
• Assuming that internal resistance does not affect the terminal p.d. In real cells, the terminal p.d. is always slightly less than the e.m.f. when current flows.

Practice questions

1. A galvanometer requires 2.5 J of work to move a charge of 0.01 C through it. What is the e.m.f. of the galvanometer?
2. A battery has an e.m.f. of 12 V and an internal resistance of 0.5 Ω. If a current of 4 A is drawn, calculate the terminal potential difference.
3. During an experiment a student measures that a source supplies a current of 0.2 A for 30 s while doing 15 J of work. Determine the e.m.f. of the source.

Answers to practice questions

1. \$E = \frac{W}{Q} = \frac{2.5\ \text{J}}{0.01\ \text{C}} = 250\ \text{V}\$
2. Terminal p.d. = \$E - I r_{\text{int}} = 12\ \text{V} - (4\ \text{A})(0.5\ \Omega) = 12\ \text{V} - 2\ \text{V} = 10\ \text{V}\$
3. First find the charge transferred: \$Q = I t = (0.2\ \text{A})(30\ \text{s}) = 6\ \text{C}\$
   Then \$E = \frac{W}{Q} = \frac{15\ \text{J}}{6\ \text{C}} = 2.5\ \text{V}\$