11.12 Electromotive Force (emf)

The electromotive force (emf), denoted by the symbol $\epsilon$, is the energy provided by a source (such as a battery or generator) per unit of charge that passes through it. Despite its name, emf is not a force; it is an energy-related quantity that represents the work done to create a potential difference, which drives an electric current.

$\epsilon =\frac{W}{q}$

where $W$ is the work done and $q$ is the charge. The SI unit of emf is the Volt (V).

1. EMF vs. Terminal Potential Difference

It is essential to distinguish between the emf of a source and the actual potential difference (voltage) it supplies to an external circuit.

• EMF ($\epsilon$): The ideal or total potential difference a source can provide. It is the voltage across the terminals when no current is being drawn ($I=0$).
• Terminal Potential Difference ($V_T$): The actual voltage measured across the terminals of the source when it is connected to a circuit and current ($I$) is flowing.
• Internal Resistance ($r$): All real-world sources of emf possess some internal resistance, which opposes the flow of current within the source itself.

When current ($I$) flows from a source, some potential is dropped across this internal resistance. The relationship is:

$V_T=\epsilon -Ir$

This equation shows that the terminal voltage is always less than the emf when the source is supplying current. When the source is being charged (current forced in reverse), $V_T=\epsilon +Ir$.

2. Current in a Simple Circuit

For a simple circuit with an emf source ($\epsilon$), an internal resistance ($r$), and an external load resistance ($R$), the total resistance is $R+r$. By Ohm's Law:

$I=\frac{\epsilon }{R+r}$

Worked Example: A battery of emf $\epsilon =9\text{ V}$ and internal resistance $r=1\Omega$ is connected to a $R=8\Omega$ resistor. Find the current and terminal voltage.

$I=\frac{9}{8+1}=1\text{ A}$ $V_T=\epsilon -Ir=9-(1)(1)=8\text{ V}$

3. Power and Efficiency

Power Dissipation

The total power supplied by the emf source is split between the external load and the internal resistance:

By conservation of energy: $\epsilon I=I^{2}R+I^{2}r$

Efficiency

The efficiency of an emf source is the ratio of useful power delivered to the load to the total power supplied:

$\text{Efficiency}=\frac{P_R}{P_{total}}=\frac{V_{T}I}{\epsilon I}=\frac{V_T}{\epsilon }=\frac{R}{R+r}$

Maximum Power Output

The power delivered to the external load $P_R=I^{2}R$ is maximised when the load resistance ($R$) equals the internal resistance ($r$):

$R=r\text{(condition for maximum power transfer)}$

At this condition, the current is $I=\frac{\epsilon }{2r}$ and the maximum power delivered to the load is:

$P_{max}=I^{2}R={\left(\frac{\epsilon }{2r}\right)}^{2}r=\frac{{\epsilon }^2}{4r}$

Note: At maximum power transfer, the efficiency is only 50%, since equal power is dissipated in the internal resistance.

Summary