5.1.6 Electromotive force and internal resistance

Internal Resistance

Definition: Batteries have an internal resistance $r$ , which is due to collisions between electrons and atoms inside the battery. As a result, some energy is lost as heat within the battery itself before the electrons leave, which can be represented as a small internal resistor.

Electromotive Force (EMF)

Definition: EMF ( $\varepsilon$ ) is the energy transferred per coulomb of charge by a cell. It represents the maximum energy that the battery can provide per unit charge, ideally without losses.
Formula:

\varepsilon = \frac{E}{Q}

where $E$ is the energy transferred and $Q$ is the charge.

Total Circuit Resistance

In a circuit, the total resistance $R_t$ is the sum of the internal resistance $r$ and the load resistance $R$ (resistance of external components). Therefore:

R_t = R + r

EMF and Potential Difference Relationships

The EMF in the circuit can be expressed as:

\varepsilon = IR + Ir

where $IR$ is the potential difference across the load resistance (external components) and $Ir$ is the potential difference across the internal resistance.

$IR$ (across $R$ ) is also called the terminal p.d. $( V )$ .
$Ir$ (across $r$ ) is known as lost volts ( $v$ ), which represents the energy lost inside the cell. Thus, the relationship between EMF and potential differences is:

\varepsilon = V + v

Measuring EMF

To measure the EMF of a cell, use a voltmeter across the cell in an open circuit (i.e., no current is flowing). In this state, there are no lost volts, so the voltmeter reading is equal to the EMF.

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Worked Examples

Example 1 A cell with an EMF of 5 V has 2 V lost volts, and the external resistance $R\ is\ 10\ \Omega$ . Calculate the current in the circuit.

Solution:

Given: $\varepsilon = 5 \text{ V}$ , $v = 2 \text{ V}$ , $R = 10 \Omega$
We know:

\varepsilon = IR + v

Substitute values to find $IR$ :

5 = IR + 2 \Rightarrow IR = 3 \text{ V}

Using $V = IR$ :

3 = I \times 10 \Rightarrow I = 0.3 \text{ A}

Example 2 A cell with an EMF of $10$ V has a current of 2 A flowing through the circuit. The load resistances are $R_1 = 3.5 \Omega$ and $R_2 = 0.5 \Omega$ . Find the terminal p.d., lost volts, and the internal resistance of the cell.