"Distinction between e.m.f. and potential difference (p.d.)",

TL;DR

E.m.f. (electromotive force) is the total energy provided per unit charge by a source, even when no current flows. Potential difference (p.d.) is the energy converted per unit charge between two points in a circuit when current is flowing. Think of e.m.f. as the battery's maximum push, and p.d. as the actual push it delivers to a component.

1. The Mental Model

Imagine a water pump (the source) pushing water through pipes (the circuit). E.m.f. is like the maximum pressure the pump can generate. P.d. is the actual pressure drop across a specific section of the pipe as water flows through it, doing work (e.g., turning a water wheel).

2. The Core Material

Understanding e.m.f. and p.d. is crucial for analyzing circuits. They are often confused because both are measured in volts, but they represent different aspects of electrical energy.

Electromotive Force (e.m.f.)

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E.m.f. is the work done per unit charge by the energy source (like a battery or generator) to move charge around a complete circuit. It represents the total electrical energy that the source can provide.
- Source of energy: E.m.f. originates within the source (e.g., chemical reactions in a battery).
- No current? No problem: A battery still has an e.m.f. even if it's not connected to anything (open circuit). In this case, the p.d. across its terminals is equal to its e.m.f.
- Internal resistance: Real sources have internal resistance ($r$). When current ($I$) flows, some energy is lost inside the source itself, causing a "voltage drop" ($Ir$) within the source.

Potential Difference (p.d.)

P.d., or voltage, is the work done per unit charge to move charge between two points in an electrical circuit. It represents the energy converted from electrical energy to other forms (like heat, light, or mechanical energy) as charge moves through a component (e.g., a resistor, bulb, or motor).
- Across a component: P.d. is always measured across external components in a closed circuit.
- Requires current: For a p.d. to exist across an external component, current must be flowing through it. If no current flows, there's no energy conversion, and thus no p.d. across that component.
- Terminal p.d.: When a current flows from a source, the p.d. measured across its terminals (known as terminal p.d., $V$) is less than the e.m.f. This is because of the voltage drop due to the internal resistance.
- $V = \text{e.m.f.} - Ir$
- If the external circuit resistance is $R$, then $V = IR$.
- So, $\text{e.m.f.} = IR + Ir = I(R+r)$. This shows that e.m.f. drives the current through both the external resistance and the internal resistance.

Here's a diagram to help visualize the relationship:

graph TD
    A["Energy Source (e.g., Battery)"] --> B["Provides e.m.f. (E)"]
    B --> C{{"Is Circuit Open?"}}
    C -- "Yes" --> D["Terminal p.d. (V) = E"]
    C -- "No, Circuit Closed" --> E["Current (I) Flows"]
    E --> F["Voltage Drop across Internal Resistance (Ir)"]
    E --> G["Voltage Drop across External Resistance (IR) = Terminal p.d. (V)"]
    F --> H["E = V + Ir"]
    G --> H

Key Differences Summarized:

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3. Worked Example

Let's say you have a battery with an e.m.f. of 12V and an internal resistance of 0.5 Ω.

Scenario 1: Open Circuit
If you measure the voltage across the battery terminals with a voltmeter when nothing is connected (open circuit), what will it read?
Since no current is flowing ($I=0$), the voltage drop across the internal resistance ($Ir$) is $0 \times 0.5 \text{ Ω} = 0\text{V}$.
Therefore, the terminal p.d. ($V$) will be equal to the e.m.f.:
$V = \text{e.m.f.} - Ir = 12\text{V} - 0\text{V} = 12\text{V}$.

Scenario 2: Closed Circuit
Now, you connect a resistor of 5.5 Ω across the battery. What is the current flowing and the p.d. across the resistor?
First, calculate the total resistance in the circuit:
Total Resistance ($R_{total}$) = External Resistance ($R$) + Internal Resistance ($r$)
$R_{total} = 5.5 \text{ Ω} + 0.5 \text{ Ω} = 6.0 \text{ Ω}$

Next, calculate the current ($I$) using Ohm's Law and the e.m.f.:
$I = \text{e.m.f.} / R_{total} = 12\text{V} / 6.0 \text{ Ω} = 2.0\text{A}$

Finally, calculate the p.d. across the external resistor (which is also the terminal p.d. of the battery):
$V = IR = 2.0\text{A} \times 5.5 \text{ Ω} = 11.0\text{V}$

Notice that the p.d. across the resistor (11.0V) is less than the battery's e.m.f. (12V). The "missing" voltage is the p.d. lost due to the internal resistance:
Voltage drop across internal resistance = $Ir = 2.0\text{A} \times 0.5 \text{ Ω} = 1.0\text{V}$
And indeed, $11.0\text{V} + 1.0\text{V} = 12.0\text{V}$ (e.m.f.).

4. Key Takeaways

• E.m.f. is the total potential energy supplied by a source, regardless of whether current is flowing.
• P.d. is the energy converted from electrical to other forms across a component when current is flowing.
• For an open circuit, the terminal p.d. of a source is equal to its e.m.f.
• For a closed circuit, the terminal p.d. of a source is less than its e.m.f. due to internal resistance.
• Both e.m.f. and p.d. are measured in Volts.
• E.m.f. drives current through the entire circuit (internal + external resistance).
• P.d. is measured across specific parts of the circuit where energy is being used.

Common Mistakes to Avoid

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• Confusing units: Always remember both are measured in Volts, but represent different concepts.
• Ignoring internal resistance: For real sources, internal resistance is always present and causes a difference between e.m.f. and terminal p.d.
• Applying p.d. to an open circuit: While e.m.f. can exist, p.d. across an external component requires current.
• Assuming terminal p.d. = e.m.f. in a closed circuit: This is only true if internal resistance is zero, which is an ideal case.

5. Now Try It

You have a power supply with an e.m.f. of 6.0V and an internal resistance of 0.2 Ω.
1. What would a voltmeter read if connected across the terminals of the power supply when it's not connected to anything else?
2. If you connect a light bulb with a resistance of 2.8 Ω to the power supply, calculate:
a. The total current flowing in the circuit.
b. The p.d. across the light bulb.
c. The voltage drop across the internal resistance.
d. Confirm that the sum of the p.d. across the bulb and the internal voltage drop equals the e.m.f.

Success means you can clearly differentiate the values you calculate for e.m.f., terminal p.d., and internal voltage drop, and show how they relate to each other.