   EMF and Internal Resistance
Introduction

Recent decades have seen a huge expansion in consumer electronics. While the functions of new devices differ widely, their power still largely comes from either batteries or mains power supplies.

In the simple circuits met so far, we have pictured power supplies as being able to provide as much current as the components require. In reality, this is not the case.

EMF basics
The
energy
A system has energy when it has the capacity to do work. The scientific unit of energy is the joule.
energy
provided by any
power
The power of system is a measurement of the rate at which energy is transferred from one form to another. The scientific unit of power is the watt.
power
supply, such as a battery, to each
coulomb
The coulomb is the unit of charge. One coulomb is approximately equivalent to the charge carried by 6.25 × 1018 electrons.
coulomb
of charge leaving a battery or
power
The power of system is a measurement of the rate at which energy is transferred from one form to another. The scientific unit of power is the watt.
power
supply is called the
electromotive force
The electromotive force (often abbreviated to EMF) of a battery is a measurement of its capacity to provide energy to the charges.
electromotive force
or
EMF
EMF is the common abbreviation for the electromotive force – a measurement of a battery's capacity to provide energy to the charges.
EMF
. The electromotive force is measured in units of joules per
coulomb
The coulomb is the unit of charge. One coulomb is approximately equivalent to the charge carried by 6.25 × 1018 electrons.
coulomb
(JC−1) or volts (V).

Click on the figure below to interact with the model.  Figure 1.  A series circuit.

Turn on the circuit in Fig.1 by clicking on the switch. Then connect the voltmeters across the buzzer and lamp by dragging them into the spaces in the circuit. Note the values for the p.d.s across the two components.

Select the correct values to complete the sentences below.

• In the circuit in Fig.1, batteries with EMFs of 3 V and V are connected together and joined in series with a buzzer and a lamp. The batteries provide p.d.s across the lamp of V and across the buzzer of V.
• Which statement below correctly describes what you observe in Fig.1?
• You should have discovered one example of a general rule which states that the sum of the battery EMFs is always equal to the sum of the p.d.s across the components. This can be summarized in the equation below: Confirm that the sum of the battery EMFs is always equal to the sum of the p.d.s across the components by changing the values of the battery voltages in Fig.1. To do this, click on the battery voltages and alter their values in the range 0–10 V.

Internal resistance
In Fig.1, the buzzer and the lamp were attached to the terminals of the batteries. They could just as easily have been connected to the output terminals of a power supply. When any set of components is connected across the terminals of a battery or a power supply like this, they are collectively called the external circuit. This is to distinguish such components from the internal circuit within the power supply. The
resistance
The opposition to the flow of current provided by a circuit is called resistance. Resistance is measured in units called Ohms.
resistance
of the
external circuit
All the components connected across the terminals of a battery or power supply are collectively known as the external circuit.
external circuit
is often referred to as the load resistance.

Use this new terminology to complete the sentences below.

• In Fig.1, the p.d. across the entire circuit is equal to the sum of EMF of the batteries. The resistance of the lamp and buzzer combined is called the resistance.
• In Fig.1 the load resistance is the only resistance in the circuit. However, this is not the case in the circuit in Fig.2 below.

Click on the figure below to interact with the model.  Figure 2.  Effect of internal resistance.

Again, move the voltmeters into place to measure the p.d. across the two components.

What is the p.d. across the bulb?
•  V   (to 2 d.p.)
• What is the p.d. across the red
LED
LED is the common abbreviation for a light emitting diode. When electrons and holes in an LED recombine, the excess energy produced is radiated as photons of light of one particular colour.
LED
?
•  V   (to 2 d.p.)
• What is the total p.d. across the load?
•  V   (to 2 d.p.)
• Complete the following statements for the circuit in Fig.2.

• The battery has an EMF of  V which provides p.d.s across the load components. The total of these p.d.s is  V and so it would appear that the EMF of the battery is greater than the sum of the p.d.s in the circuit. At first it might seem that  V has been lost somewhere in the circuit.
• We can explain the 'lost volts' in Fig.2 by the fact that the LED and the lamp are not the only sources of resistance in the circuit. In fact, the battery also has a small internal resistance.

Any battery or power supply with an
internal resistance
All batteries or power supplies have internal resistance. This resistance has the effect of reducing the output p.d. as the current supplied increases.
internal resistance
can be pictured as shown in Fig.3. Here, the internal resistance of 2 Ω is shown as a resistor in series with the power source.

 Figure 3. A representation of a battery with internal resistance. Internal resistance is a feature of all practical power supplies. Different designs of battery will have different internal resistances, and the manufacturers will often take steps to minimize the internal resistance.

The 12 V output from a car battery will have the same output
voltage
The voltage across a component is the electrical energy transferred by 1 coulomb of charge passing through the component.
voltage
as a 12 V domestic battery. However, the car battery needs to have a much lower internal resistance and the design needed to achieve this results in the car battery being larger and heavier.

The voltmeter in the Fig.4 below shows the 'lost volts', even though in practice this is a quantity that could not be measured experimentally.

Click on the figure below to interact with the model.  Figure 4.  Measuring 'lost' volts.

Set the variable resistor in the circuit of Fig.4 to 10 Ω. Close the switch and note the values of the p.d.s across the internal resistance and the load resistor.

Complete the following statement:

• The p.d. across the 10 Ω load resistor is V and the p.d. across the 2 Ω internal resistance is V. The EMF of the battery is V.
• The p.d. actually available at the battery terminals is called the terminal p.d. If the EMF of the battery is E, the
terminal p.d.
The terminal p.d. is the voltage available at the terminals of a battery or power supply. The terminal p.d. varies depending on how much current is being drawn.
terminal p.d.
is Vtpd and the p.d. across the internal resistance is Vlost, we can say:

The above equation holds, no matter what the value of the load resistor or the
current
The rate of flow of charge past any specific point in a circuit. The base unit of current is the Ampere.
current
drawn from the supply. We can verify this using Fig.4. Set the variable resistor to a value of 60 Ω.

What is the value of the p.d. across the load resistor when the variable resistor has a value of 60 Ω?
• What is the value of the 'lost' volts when the variable resistor has a value of 60 Ω?
• Check this again for another value of the variable resistor.

Effects of internal resistance
Fig.5 below shows two identical circuits allowing different numbers of lamps to be connected to the batteries. Circuit B has internal resistance in the battery, whereas Circuit A shows the hypothetical case where there is no internal resistance.

Click on the figure below to interact with the model.  Figure 5.  The effect of internal resistance.

Complete the following table for Circuit A.

 Switches closed Current / mA Voltage / V A1 A2 A3 A4 A5

Complete the following sentences to summarize the behaviour of the circuit with no internal resistance.

• When switch A1 is closed, the p.d. across the lamp is  V and the current taken from the battery is  mA. When subsequent switches are closed the current taken from the battery increases in steps of  mA.
• What happens to the p.d. across the terminals of the battery in Circuit A as the current taken from it increases?
• Now complete the following table for Circuit B.

 Switches closed Current / mA Voltage / V B1 B2 B3 B4 B5

Complete the following sentences to summarize the behaviour of the circuit with internal resistance.

• When switch B1 is closed the p.d. across the bulb is 9 V and the current taken from the battery is 90 mA. When subsequent switches are closed the current taken from the battery increases in unequal steps of 90 mA. With the same number of bulbs turned on, the current flowing in Circuit A is always the current flowing in Circuit B.
• What happens to the p.d. across the terminals of the battery in Circuit B as the current taken from it increases?
• Fig.5 demonstrates clearly that, when a battery has internal resistance, increasing the current taken from the battery reduces the voltage available to the external circuit. This is why manufacturers of power supplies and batteries often try to reduce the internal resistance of their products.

Determining the maximum current
When a power supply or battery has internal resistance, increasing the current provided to an external circuit reduces the p.d. across this circuit. As the current taken from the supply increases, the current through the internal resistance increases and so more of the p.d. will be across the internal resistance. Consequently less p.d. will be available for the external circuit. We can see this idea mathematically by using the equation from above: where E is the EMF, Vtpd is the p.d. across the terminals of the battery, and Vlost is the 'lost' voltage. Since the internal resistance r and the load resistor R are in series the same current, I, passes through both resistors. Therefore,  As I increases, Ir increases and, since E is constant, Vtpd must decrease.
 So as the current I drawn from a battery increases, the p.d. available at its output (Vtpd) decreases.

If the current becomes sufficiently large, the output p.d. will fall to zero. The value of current for which this happens is called Imax.

Click on the figure below to interact with the model.  Figure 6.  Determining the maximum current.

The battery in Fig.6 above has an internal resistance (which is not shown on the circuit diagram). Use the variable resistor to alter the current drawn from the battery and observe how the current varies.

What is Imax for the circuit in Fig.6?
•  A   (to 2 d.p.)
• When is the current in the circuit equal to Imax?
• When the resistance across the battery is zero, we say that the battery is short-circuited. Maximum current flows when the battery is short-circuited.

Determining the EMF
We have seen what happens when the maximum current is obtained from a power supply. Alternatively, if no current is drawn from the supply, the output voltage Vtpd and the EMF E of the battery are equal.

We can again show this idea mathematically by reusing the equation from above:  If no current is drawn from the supply I = 0 so Ir = 0
 Therefore E = Vtpd.

The EMF of the supply is equal to the output p.d. when no current is being supplied.

Click on the figure below to interact with the model.  Figure 7.  Determining the EMF of the battery.

The battery in Fig.7 above has an internal resistance (which is not shown on the circuit diagram). Use the variable resistor to alter the current drawn from the battery and observe how the p.d. across the bulb varies.

What is the EMF of the battery for the circuit in Fig.7?
•  V   (to 2 d.p.)
• When is the EMF of the battery equal to the p.d. across the variable resistor?
• Click on the figure below to interact with the model.  Figure 8.  Drawing a current from a battery.

Close the switch to see how the reading on the voltmeter in Fig.8 changes.

Complete the following sentences to summarize the operation of the circuit above.

• When the switch is open the voltmeter measures a p.d. across the terminals of the supply of  V and the current being supplied is  mA.

Closing the switch enables the battery to supply a current of  mA. When the switch is closed the output p.d. from the battery falls to  V. Since the EMF of the battery is the same as the output p.d. for zero current, the EMF of this battery is  V.
• Determining the internal resistance
In Fig.9 a variable resistor is connected to a battery. We will use this circuit to determine the EMF, the maximum current, and the internal resistance of the battery.

Click on the figure below to interact with the model.  Figure 9.  Battery and resistor.

Adjust the variable resistor to alter the current passing through the circuit in Fig.9 above.

What happens to the p.d. across the variable resistor when the current supplied by the battery in Fig.9 increases?
• What is the p.d. across the variable resistor in Fig.9 when the current supplied by the battery to the variable resistor is 0.6 A?
• Set the variable resistor in Fig.9 to a range of different positions and enter values for the terminal p.d. Vtpd, and the current I into the table of Fig.10.

 Figure 10. Terminal p.d. variation with current. The line in the graph of Fig.10 is given by the equation E = Vtpd + Ir. We can use this equation to explain the shape of the line. As I increases, Ir increases. Since E is constant, Vtpd must decrease. When the current increases and becomes equal to Imax, then the output p.d. will have fallen to zero. The current reaches Imax when the resistance across the output of the battery is zero.
 As we saw previously, this is the current that flows through a short circuit.

On the graph, the value of Imax is given by the point at which the line cuts the x-axis. By extending the line in Fig.10 to where it cuts the x-axis, we can find the value for Imax.

What is Imax of the battery?
•  A   (to 1 d.p.)
• We have already seen that the EMF is the largest voltage available from the battery, and that this occurs when I = 0. On the graph, this is illustrated by the point where the line cuts the graph's y-axis.

What is the EMF of the battery?
•  V
• Now that we know the values of the EMF and the short-circuit current Imax we can calculate the internal resistance of the battery. The steps below lead through the analysis of how to do this. We know that  We also know that when the current is equal to Imax, the output p.d. is zero, Vtpd = 0   We can rearrange this to find an expression for the short-circuit current.   We can rearrange once more to find an expression for the internal resistance r. What is the value of the internal resistance r of the battery in Fig.9?
•  Ω   (to 1 d.p.)
• Alternatively we can use the data in a different way to find the internal resistance of the power source. The line in the graph of Fig.10 is given by the equation:

Click on the Vtpd term to rearrange the equation. It is now in the standard form 'y = mx + c' where Vtpd is the variable plotted on the y-axis and I the variable plotted on the x-axis. The intercept gives the EMF of the supply and the negative of the gradient indicates its internal resistance.

 Figure 11. Terminal p.d. variation with current. The graph above shows the graph of terminal p.d. plotted against current for a power supply.

Calculate the values indicated below for the power supply using the graph in Fig.11 above.

• The EMF of the power supply:
V   (to 1 d.p.)

The short-circuit current of the power supply:
mA   (to 1 d.p.)

The internal resistance of the power supply:
Ω   (to 1 d.p.)
• Power in circuits
The experimental data generated from Fig.10 shows the different currents passing through a resistor for different values of the p.d. across it. We can multiply these values to get the power dissipated in the resistor at different currents. Similarly we can use
Ohm's Law
Ohm's Law states that at constant temperature the current in a conductor is directly proportional to the d.p. across the conductor. The constant of proportionality is called the resistance of the sample.
Ohm's Law
to get the values of the resistance connected to the battery by simply dividing the p.d. readings by the current readings.

 Figure 12. The maximum power. A graph of power delivered versus load resistance shows that the power delivered to the external circuit is not constant. The power delivered rises to a maximum when the internal and external resistances are equal.

Summary

All the components connected across the terminals of a battery or power supply are collectively known as the external circuit. The total resistance of these components is known as the load resistance.

All batteries or power supplies have internal resistance. This resistance has the effect of reducing the output p.d. as the current supplied increases.

The output p.d. Vtpd for any current I can be found using the equation:

where E is the EMF of the supply and r is the internal resistance of the supply.

If the EMF of the power supply and the short-circuit current Imax are known, the internal resistance of the supply can be determined using the equation below: Exercises
1. What is the EMF of a power supply?
• Figure 13. 2. In the circuit shown in Fig.13 a 6 V battery with an internal resistance of 0.2 Ω is connected to a 0.4 Ω load resistor. Calculate the terminal p.d. across the load.
•  V   (to the nearest whole number)
• Figure 14. 3. When a very high resistance
digital
A digital signal is recorded in discrete units, often in binary. This is in contrast to an analogue signal which is composed of a continuous range of values.
digital
voltmeter is connected across the terminals of a car battery, as shown in Fig.14, it reads 12 V. When the battery is supplying a current to a circuit of resistance 0.1 Ω the reading falls to 10 V. Calculate the internal resistance of the car battery.
• Ω   (to 2 d.p.)
• 4. In the circuit shown in Fig.15 a battery with an EMF of 6 V and an internal resistance of 0.5 Ω is connected to a 4 Ω load resistor. Calculate the current in the circuit.
•  A   (to 2 d.p.)
• Figure 15. 5. Calculate the 'voltage lost' due to the internal resistance shown in Fig.15.
•  V   (to 2 d.p.)
• 6. Calculate the terminal p.d. for the battery in Fig.15.
•  V   (to 2 d.p.)
• 7. A very high resistance digital voltmeter connected across the terminals of a heavy duty lorry battery, as shown in Fig.16, shows a reading of 24 V. When the battery is tested by supplying current to a 0.5 Ω load resistor, the voltmeter reading falls to 22 V. What is the EMF of the battery?
•  V   (to nearest whole number)
• Figure 16. 8. Calculate the current in the 0.5 Ω load resistor in Fig.16 during the testing of the battery.
•  A   (to the nearest whole number)
• 9. Calculate the power dissipated in the load resistor in Fig.16 during the battery testing.
•  W   (to the nearest whole number)
• 10. Calculate the internal resistance of the battery in Fig.16.
•  Ω   (to 3 d.p.)
• Figure 17. 11. Three identical cells each of EMF 1.5 V and internal resistance 0.3 Ω shown in Fig.17 are connected in series to make a battery. Which of the choices below shows the total EMF of the battery and the initial current if the battery were accidentally short-circuited?
• Figure 18. 12. In an experiment to measure the internal resistance (r) of a supply of EMF (E), a pupil connects the circuit shown in Fig.18 using a very high quality voltmeter and ammeter. She records the ammeter readings (I) and the voltmeter readings (V). She then makes three statements about the arrangement. Decide whether each statement is true or false.
•  The EMF is the reading V when the switch is open. False True With the switch open, V divided by I gives the value for resistance R. False True With the switch closed, the expression gives the internal resistance r. False True
• Well done!
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