distinguish between e.m.f. and potential difference (p.d.) in terms of energy considerations

Practical Circuits: e.m.f. vs Potential Difference

What is Electromotive Force (e.m.f.)?

⚡️ Think of a battery as a water pump that pushes water (electrons) through a pipe (circuit). The pump’s power is the e.m.f. – the energy supplied per unit charge to start the flow.

In physics, e.m.f. is the energy per unit charge that a source (battery, generator) provides to move charges from one terminal to the other, regardless of the circuit’s internal resistance.

Mathematically, for a simple battery: \$\mathcal{E} = V{\text{source}}\$ where \$\mathcal{E}\$ is the e.m.f. and \$V{\text{source}}\$ is the voltage the source can deliver when no current flows.

What is Potential Difference (p.d.)?

💡 The potential difference, or voltage, is the energy per unit charge that a charge experiences as it moves from one point to another in a circuit. It is the “pressure” that drives electrons through the resistive elements.

Unlike e.m.f., p.d. depends on the current flowing and the resistances in the circuit: \$V = IR\$ where \$I\$ is the current and \$R\$ the resistance.

In a closed loop, the sum of all potential differences (including drops across resistors and rises across the battery) must equal the e.m.f. of the source.

Energy Considerations

  • e.m.f. supplies energy: \$W_{\text{emf}} = \mathcal{E}\,Q\$ where \$Q\$ is the charge moved.

  • Potential difference causes energy transfer: \$W_{\text{pd}} = V\,Q = I\,R\,Q\$.

  • Energy lost in a resistor: \$W_{\text{loss}} = I^2 R t\$ (heat).

  • In a steady state, energy supplied by e.m.f. equals energy lost in resistances: \$\mathcal{E} I = I^2 R_{\text{total}}\$.

Comparison Table

Aspecte.m.f.Potential Difference
DefinitionEnergy per unit charge supplied by a source.Energy per unit charge between two points in a circuit.
Depends onSource characteristics (chemical, mechanical, etc.).Current and resistance in the path.
UnitsVolts (V)Volts (V)
Role in circuitPushes charges around the loop.Determines current flow through elements.

Examination Tips

  1. When asked to calculate the e.m.f., use the source’s open‑circuit voltage.

  2. For potential difference across a resistor, apply Ohm’s law: \$V = IR\$.

  3. Remember that in a closed loop, the algebraic sum of all potential differences equals the e.m.f. (Kirchhoff’s Voltage Law).

  4. Use energy arguments: energy supplied by e.m.f. = energy dissipated in resistors.

  5. Check units: volts for both e.m.f. and p.d., but interpret them differently.

Quick Practice Question

A 12 V battery has an internal resistance of 0.5 Ω. If a 3 Ω resistor is connected across the battery, what is the current and the potential difference across the resistor?

🔍 Solution: Total resistance \$R_{\text{total}} = 0.5 + 3 = 3.5\,\Omega\$.

Current \$I = \dfrac{\mathcal{E}}{R_{\text{total}}} = \dfrac{12}{3.5} \approx 3.43\,\text{A}\$.

Potential difference across the 3 Ω resistor \$V_R = I R = 3.43 \times 3 \approx 10.3\,\text{V}\$.

Notice that the e.m.f. (12 V) is larger than the p.d. across the external resistor because some voltage is dropped across the internal resistance.