Objectives:

• To analyse circuits with more than one source of e.m.f.
• Identify and understand that a source of e.m.f. has an internal resistance
• To understand the term terminal p.d. and how it involves ‘lost volts’
• Be able to select and use the equations  $\epsilon = I (R + r)$ , and  $\epsilon = V + Ir$
• To identify techniques and procedures used to determine the internal resistance of a chemical cell or other source of e.m.f.

To fully understand and appreciate the concept of internal resistance it is important to get your head around e.m.f, p.d., Ohm’s law, charge, Kirchhoff’s laws and power.

What is internal resistance?

Current flows through all parts of the circuit, including the power sources/ cells. Since the cells themselves are made of material with resistance there will inevitably be some energy lost in the cell. This will increase with current, so there is more loss at higher currents. Consider the diagram below:

Here the ‘real’ cell is represented by two  components in the diagram:

• An ideal cell with no internal resistance
• A small resistor, connected to the ideal cell inside the lash-lined box, representing the internal resistance

Energy is conserved, so the energy transferred from chemical to electrical by the e.m.f. of the cell must ‘pay’ for the energy transferred to other forms in the circuit – this is the sum of the energy transferred in the cell itself and the energy transferred by the external load.

Clearly the energy transferred to heat in the cell is not doing useful work and so is in a sense ‘lost’ – it is lost due to the internal resistance inside the cell.

Where confusion can lie

When first dealing with internal resistance, many people find the concept confusing and difficult. The internal resistor drawn on circuits with an internal resistance should be treated as an independent resistor, that is all. Most resistors will have the letter  $R$  written next to them, the internal resistor is simply labelled with a lower case   $r$ .

Deriving the equation for internal resistance

There are three methods to derive the equation and the following diagram will help with this;

First method

Using Kirchhoff’s 2nd law (KVL): The sum of the e.m.f.’s around a loop in a circuit must be equal to the sum of the p.d.’s around that same loop, therefore;

$\epsilon = V_{r} + V_{R}$

Since  $V_{R} = IR$  and  $V_{r} = Ir$ ;

$\epsilon = Ir + IR$
$\epsilon = I(r + R)$

The terminal voltage,  $V$  is the total e.m.f. that the ‘real’ cell will output (in otherwise it is the e.m.f. that would be shown if a voltmeter were to be placed across the ‘real’ cell) and can be written as;

$V = \epsilon - V_{r}$
$V = \epsilon - Ir$

This last equation tells us that terminal p.d. is equal to the e.m.f. minus the lost volts due to the internal resistance.

Second method

Since energy is conserved, we can imagine a unit of charge,  $Q$ , moving around the circuit.

$latex Energy supplied = energy lost by ‘R’ \ + \ energy lost by ‘r’$

$W_{\epsilon} = W_{R} + W_{r}$

Using the definition of p.d.  $V = \frac{W}{Q} \ \ \rightarrow \ \ W = VQ$ ;

$V_{\epsilon}Q = V_{R}Q + V_{r}Q$

$V_{\epsilon} = V_{R} + V_{r}$

Since  $V_{R} = IR$  and  $V_{r} = Ir$ ;

$\epsilon = Ir + IR$
$\epsilon = I(r + R)$

Thus we have arrived at the same equation as in the first method.

Third method

Use Ohm’s law with the e.m.f. ‘driving’ current through the combined resistance (R + r):

Using  $V = IR \ \ \rightarrow \ \ I = \frac{V}{R}$ ;

$I = \frac{V_{\epsilon}}{R_{T}}$

where  $R_{T}$  is the total resistance of the circuit and since both the normal resistor and the internal resistor are in series;  $R_{T} = R + r$

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

Therefore. by multiplying through by $(R+r)$ we get;

$I (R+r) = \epsilon$

Thus we have arrived at the same equation as in the first method.

Practical effects of internal resistance

Lets use a car as an example. The headlamps are connected in parallel across a 12 V battery. The starter motor is also in parallel controlled by the ignition switch. The starter motor has a low resistance as it demands a very high current (say 60 A). The battery itself has a low internal resistance (say 0.01 Ω). The headlamps themselves draw a much lower current.

What happens when the engine is started? (the switch to starter motor is closed for a short time whilst the engine starts up)

• There is a sudden demand for more current (60 A to the starter motor)
• Large lost volts (around 0.01 Ω x 60 A = 6 V)
• Therefore the terminal voltage drops to 6 V (12 V – 6 V)
• The headlamps dim because they are connected to the battery via a parallel connection (Kirchhoff’s voltage law)

When the engine fires, the starter motor switch is opened and the current drops. The terminal voltage rises and the headlamps return to normal. It’s better to turn the headlamps off when starting the car.

Current, voltage and resistance

A cell has an internal resistance, and just like any resistor, the resistance value remains constant (providing the temperature does not vary).

If more current was required for the same component the voltage would decrease as a result,  $R = \frac{V}{I}$  because the internal resistance would remain constant.

Kirchhoff’s voltage law tells us that if the p.d. across the component decreases then the terminal voltage from the cell decrease and therefore the current would increase.

The following I-V graph can be drawn to show this;

This graph tells us a lot if we can identify key characteristics.

Using the equation for terminal p.d. that was derived earlier;

$V = \epsilon - Ir$

$V = -Ir + \epsilon$

This is in the form  $y = mx + c$ where;
$y \ \rightarrow \ V$
$x \ \rightarrow \ I$
$-r \ \rightarrow \ m (gradient)$
$c \ \rightarrow \ \epsilon$

This information tells us that the gradient should be negative, which it is, and also tells us that the y-intercept should show us what the maximum e.m.f. output of the cell should be.

The graph shows us that the terminal p.d. is maximum with minimum current, this should make sense since when there is minimal current, the voltage lost due to the internal resistance would be minimum due to  $R = \frac{V}{I}$ .