"Definition of electromotive force (e.m.f.)",

TL;DR

Electromotive force (e.m.f.) isn't a force; it's the energy provided per unit charge by a source to move charge around a circuit. It's measured in volts and represents the "push" a power source gives to electrons. Unlike potential difference, e.m.f. accounts for energy used within the source itself.

1. The Mental Model

Think of e.m.f. as the total "oomph" a battery or generator has before it even pushes current through anything outside of itself. It's the maximum energy boost each unit of charge gets from the source.

2. The Core Material

You've probably heard of voltage, and e.m.f. is very closely related, but there's a crucial distinction.

What e.m.f. really is

e.m.f. stands for electromotive force, but that name is a bit misleading. It's not actually a force in the mechanical sense (like pushing a box). Instead, it's the energy supplied by the source per unit electric charge to drive charge around a complete circuit.

Imagine you have a pump lifting water to a higher level. The e.m.f. is like the total energy the pump gives to each liter of water, including any energy lost due to friction inside the pump itself.

Mathematically, it's defined as:
$e.m.f. = \frac{\text{Energy supplied (W)}}{\text{Charge moved (Q)}}$

Since energy is measured in Joules (J) and charge in Coulombs (C), the unit for e.m.f. is Joules per Coulomb (J/C), which is equivalent to a Volt (V).

e.m.f. vs. Potential Difference (P.D.)

This is where it gets important.
* e.m.f. (ε): This is the total potential difference produced by the source when no current is flowing in the external circuit (i.e., on open circuit). It includes the energy lost within the source due to its internal resistance. It's the maximum "push" the source can provide.
* Potential Difference (V): This is the energy converted per unit charge in the external circuit. It's the "voltage" you'd measure across a component or across the terminals of the source when current is flowing. It's always less than the e.m.f. when current flows because some energy is "lost" inside the source due to its internal resistance.

Think of it this way:
A battery has an e.m.f. of 1.5 V. This is the voltage across its terminals when nothing is connected.
When you connect a light bulb, current flows. Now, if you measure the voltage across the battery's terminals, it might be slightly less, say 1.4 V. This 1.4 V is the terminal potential difference. The "missing" 0.1 V is the voltage drop across the battery's own internal resistance.

Here's a diagram to help visualize the flow:

graph TD
    A["Energy Source (e.g., Battery)"] --> B{"Provides total energy per charge (e.m.f.)"};
    B --> C["Some energy lost within source (Internal Resistance)"];
    C --> D["Remaining energy available externally (Terminal P.D.)"];
    D --> E["Energy used by external components (e.g., Light bulb)"];
    E --> A;

Internal Resistance

Photo by Zacharias Korsalka on Pexels

Every real-world power source (like a battery or a generator) has some internal resistance (r). This resistance is inside the source itself and causes some energy to be dissipated as heat when current flows.

Due to internal resistance:
* The voltage available to the external circuit (terminal P.D., V) is always less than the e.m.f. (ε) when current (I) is flowing.
* The "lost volts" or voltage drop across the internal resistance is given by Ir.

So, the relationship between e.m.f., terminal P.D., current, and internal resistance is:
ε = V + Ir
where:
* ε = electromotive force (e.m.f.)
* V = terminal potential difference (voltage across external circuit)
* I = current flowing
* r = internal resistance of the source

This equation is fundamental for understanding real-world circuits.

3. Worked Example

Let's say you have a 1.5 V battery. This is its e.m.f. (ε).
When you connect it to a circuit and a current of 0.2 A flows, you measure the voltage across the battery's terminals to be 1.4 V. This is the terminal P.D. (V).

Can you find the internal resistance (r) of the battery?

Using the formula:
ε = V + Ir

We know:
ε = 1.5 V
V = 1.4 V
I = 0.2 A

Substitute the values:
1.5 V = 1.4 V + (0.2 A * r)

Subtract 1.4 V from both sides:
1.5 V - 1.4 V = 0.2 A * r
0.1 V = 0.2 A * r

Now, solve for r:
r = $\frac{0.1 \text{ V}}{0.2 \text{ A}}$
r = 0.5 Ω

So, the internal resistance of this battery is 0.5 Ohms. This means that when 0.2 A flows, 0.1 V is "dropped" across the battery's own internal resistance, leaving 1.4 V for the rest of the circuit.

4. Key Takeaways

• e.m.f. is the total energy per unit charge supplied by a source, measured in Volts.
• It's the maximum potential difference a source can provide when no current is drawn (open circuit).
• Potential difference (P.D.) is the energy per unit charge converted in the external circuit or across terminals when current is flowing.
• Internal resistance within a source causes some energy to be lost, making the terminal P.D. less than the e.m.f. when current flows.
• The relationship is ε = V + Ir, where Ir represents the "lost volts" due to internal resistance.
• e.m.f. is crucial for understanding the true capacity of a power source.

Common Mistakes to Avoid:
- Don't confuse e.m.f. with P.D.; they are different, though related, concepts.
- Remember that e.m.f. is only equal to terminal P.D. when no current is flowing (or if the internal resistance is zero, which is ideal).
- Don't treat "electromotive force" as a literal force like gravity; it's energy per charge.
- Forgetting to account for internal resistance in calculations involving real-world power sources.

5. Now Try It

You have a car battery with an e.m.f. of 12.0 V. When you start the car, the starter motor draws a current of 100 A, and the terminal potential difference across the battery drops to 11.5 V.

Your task is to calculate the internal resistance of the car battery. What does this internal resistance tell you about how efficiently the battery delivers power to the starter motor? (Hint: Consider the "lost volts".) Success looks like a correct value for internal resistance and a brief explanation of its implication.