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

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

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.