# Electromotive Force and Internal Resistance

## Electromotive Force and Internal Resistance Revision

**Electromotive Force and Internal Resistance**

It is the role of the power source to supply the circuit and its components with **energy**. If the **power** source is a **battery** or a **cell**, **chemical energy** is converted to **electrical energy**. However, this process is not always 100 percent **efficient**. In this section we look into the **Electromotive Force (****EMF)** and **internal resistance** of batteries and cells.

**Electromotive Force**

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)}

**EMF** can be directly measured in a circuit by placing a **voltmeter** in **parallel** across the terminals of the battery whilst not connected to a circuit.

**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)

**Internal Resistance**

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}.

**Required Practical 6**

**Investigating the relationship between internal resistance and emf.**

**Doing the experiment:**

- Set up the circuit as shown in the diagram above.
- With one of the wires disconnected from the circuit, record the reading of V on the
**voltmeter**. - Set the
**variable resistor**to its maximum resistance. - 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.
- 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.

**Analysis of Results:**

To best analyse the results, a graph of V against I can be plotted which should look similar to the one on the right hand side.

In the graph, the **y-intercept** represents the **emf** whilst the **gradient** represents the negative of the **internal resistance**.

## Electromotive Force and Internal Resistance Example Questions

**Question 1:** In the circuit below, the voltmeter shows a reading of 12 \text{ V}. However, when the switch is closed, the reading changes to 11 \text{ V}. Explain why.

**[3 marks]**

When the switch is open, the **voltmeter is reading the EMF of the cell to be **\bold{12} \textbf{ V}. This is the maximum voltage the cell is capable of producing when unloaded. However, when the circuit is completed by **closing the switch, the cell becomes under load.** The cell has some **internal resistance so some voltage is lost, known as lost volts**. This is due to the cell overcoming the internal resistance of the material inside.

**Question 2:** A student measures the voltage of a battery in a circuit by placing a voltmeter in parallel across the terminals of a battery. They find the voltage to be 22 \text{ V} when the switch in the circuit is open and 20.5 \text{ V} when the switch is closed. Calculate the lost volts and suggest where this energy has gone.

**[2 marks]**

The energy from the lost volts will heat up the battery and **dissipate into the surroundings. **

**Question 3:** A circuit with a terminal pd of 12.5 \text{ V} is powered by a battery with internal resistance 2 \text{ k}\Omega. The current in the circuit is 2.5 \text{ mA}. Calculate the EMF.

**[2 marks]**

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

\begin{aligned}\bold{\varepsilon} &= \bold{V+Ir} \\ &= 12.5 + (2.4 \times 10^{-3} \times 2 \times 10^3) \\ &= \bold{17.5} \textbf{ V} \end{aligned}