Electromotive Force and Internal Resistance

The electromotive force (EMF) of a cell is the amount of energy transferred per coulomb of charge. This is given in the equation:

\varepsilon = \dfrac{E}{Q}

• \varepsilon= the EMF of the cell in volts \text{(V)}
• E= the energy supplied in joules \text{(J)}
• Q= charge in coulombs \text{(C)}

Terminal pd is the potential difference measured across the terminals of the power source when the circuit is under load. It is the maximum potential difference the circuit actually receives. 

The EMF will always be greater than the terminal pd whilst connected to the circuit as the cell has to overcome its own internal resistance. The difference between EMF and terminal pd is known as the lost volts (v). 

As V=IR for the terminal pd, we can write a second equation for the lost volts caused by the internal resistance where:

v=Ir

• v= lost volts in volts \text{(V)}
• I= current in amps \text{(A)}
• r= internal resistance in ohms (\Omega) 

Combining these two equations gives us a third equation:

\varepsilon = IR+Ir = I(R+r)

• \varepsilon= EMF in volts \text{(V)}
• I= current in amps \text{(A)}
• R= total resistance in ohms (\Omega)
• r= internal resistance in ohms (\Omega)

The internal resistance (r) is the resistance caused by the materials inside of the battery. As with conventional resistance in wires and cell components, the internal resistance of the battery causes it to warm up when in use. The thermal energy that is dissipated comes from the lost volts (v) due to the internal resistance inside the battery. 

In most questions, you are told to consider internal resistance to be negligible and therefore, ignore its effects. However, when asked a question about EMF and internal resistance, it is important to ensure you consider their effects. 

Example: A circuit with a terminal pd of 1.5 \text{ V} is powered by a battery with internal resistance 1.3 \text{ k}\Omega. The current in the circuit is 0.6 \text{ mA}. Calculate the EMF.

[3 marks]

\bold{\varepsilon = IR+Ir}

As V=IR we can substitute V into the equation to make:

\begin{aligned} \bold{\varepsilon}&=\bold{V+Ir} \\ &= \textcolor{00bfa8}{1.5} + (\textcolor{00d865}{0.6 \times 10^{-3}} \times \textcolor{10a6f3}{1.3 \times 10^3}) \\&= \bold{2.28} \textbf{ V} \end{aligned}. 

Doing the experiment:

1. Set up the circuit as shown in the diagram above.
2. With one of the wires disconnected from the circuit, record the reading of V on the voltmeter.
3. Set the variable resistor to its maximum resistance.
4. Switch on the power supply and reconnect the wire and record a pair of values for V and I, then switch off the power supply. 
5. Repeat for 10 values of resistance over the whole range of the variable resistor and repeat 3 times for each resistance. This should give 10 pairs of readings for V and I.