Distinguish between electromotive force (e.m.f.) and potential difference (p.d.) in terms of energy considerations.
Key Concepts
Electromotive force (e.m.f.) – the work done per unit charge by a source in moving charge from its negative terminal to its positive terminal inside the source.
Potential difference (p.d.) – the work done per unit charge by external forces as charge moves between two points in a circuit, typically across a component.
Energy Perspective
The energy transferred to a charge \$q\$ when it moves through a source or a component is given by \$W = q \times V\$, where \$V\$ is the relevant voltage (e.m.f. or p.d.).
Inside the source (e.g., a battery) the chemical or other internal processes do work on the charge. The energy supplied is \$W_{\text{source}} = q \, \mathcal{E}\$, where \$\mathcal{E}\$ is the e.m.f.
Across an external component (e.g., a resistor) the electric field does work on the charge. The energy dissipated is \$W{\text{component}} = q \, V{\text{pd}}\$, where \$V_{\text{pd}}\$ is the potential difference across that component.
Comparison Table
Aspect
e.m.f. (\$\mathcal{E}\$)
Potential Difference (p.d.) \$V\$
Definition
Work done per unit charge by the source internally.
Work done per unit charge by external forces between two points.
Location
Across the terminals of the source (inside the source).
Across any circuit element (outside the source).
Sign convention
Positive from negative to positive terminal of the source.
Positive in the direction of current flow (from higher to lower potential).
Energy transfer
Supplies energy to the charge: \$W = q\mathcal{E}\$.
Consumes or releases energy: \$W = qV\$ (e.g., \$V\$ across a resistor is dissipated as heat).
Measurement
Measured with a voltmeter when no current flows (open‑circuit voltage).
Measured with a voltmeter when the circuit is operating (closed‑circuit voltage).
Practical Determination
In the laboratory the following steps are used to differentiate the two quantities.
Connect a voltmeter across the terminals of a battery without any external load. The reading is the e.m.f. \$\mathcal{E}\$ (open‑circuit voltage).
Connect a known resistor \$R\$ across the battery and measure the voltage again. The new reading is the terminal p.d. \$V = \mathcal{E} - I r\$, where \$r\$ is the internal resistance and \$I\$ the current.
Calculate the internal resistance using \$r = (\mathcal{E} - V)/I\$.
Worked Example
Consider a 12 V battery with an internal resistance of \$0.5\ \Omega\$ supplying a current of \$2\ \text{A}\$ to a resistor.
\$\$
\begin{aligned}
\mathcal{E} &= 12\ \text{V} \\
V_{\text{terminal}} &= \mathcal{E} - I r = 12 - (2)(0.5) = 11\ \text{V} \\
\text{Energy delivered to the external resistor per coulomb} &= V_{\text{terminal}} = 11\ \text{J C}^{-1}
\end{aligned}
\$\$
The difference \$12\ \text{V} - 11\ \text{V} = 1\ \text{V}\$ corresponds to the energy lost inside the battery due to its internal resistance.
Suggested Diagram
Suggested diagram: Simple circuit showing a battery (with e.m.f. \$\mathcal{E}\$ and internal resistance \$r\$) connected to an external resistor \$R\$, with a voltmeter across the terminals and another voltmeter across \$R\$.
Key Take‑aways
e.m.f. is the source’s ability to do work on charge inside the source; it is an open‑circuit quantity.
Potential difference is the work done on charge between any two points in the external circuit; it depends on the current flowing.
Energy considerations provide a clear physical distinction: \$W{\text{source}} = q\mathcal{E}\$ versus \$W{\text{component}} = qV\$.