Episode 121: EMF and internal resistance
Lesson for 16-19
- Activity time 130 minutes
- Level Advanced
The starting point for the theory can be either Kirchhoff’s second law or conservation of energy in the circuit (the same thing really) but a general discussion based on the circuit diagram below should use a variety of approaches.
- Discussion: Deriving an equation (15 minutes)
- Discussion: Practical effects of internal resistance (10 minutes)
- Student questions: Internal resistance of a power supply (20 minutes)
- Student experiment: Measuring internal resistance and EMF (45 minutes)
- Student questions: Practice questions (30 minutes)
- Discussion: More about the practical importance of internal resistance (10 minutes)
Discussion: Deriving an equation
There are three ways to arrive at the equation relating EMF, terminal PD, current and internal resistance. It is worth discussing all three, to show their equivalence. The order you take will depend on the approach used previously with the class:
Kirchhoff’s 2nd Law: As charge goes around the circuit the sum of EMFs must equal the sum of voltage drops leading to:
E = IR + Ir
The terminal voltage is equal to IR so this can be rearranged to give:
V = E − Ir
and interpreted as
terminal voltage = EMF −
Energy is conserved. Imagine a unit of charge, Q, moving around the circuit:
QE = QIR + QIr
This leads to the same equations as above.
Use Ohm’s law with E
driving current through the combined resistance (R + r):
I = ER + r
Multiplying throughout by (R + r) leads to the same equations and conclusions as in (1).
Discussion: Practical effects of internal resistance
At this point it might be worth pausing to illustrate the effects. Take a car as an example. The headlamps are connected in parallel across a twelve-volt battery. The starter motor is also in parallel controlled by the ignition switch. Since the starter motor has a low resistance 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. Ask them what happens when the engine is started (switch to starter motor closed for a short time). Look for an answer in general terms:
- sudden demand for more current
- large lost volts (around 0.05 Ω × 60 A = 3 V)
- terminal voltage drops to 12 V – 3 V = 9 V
- headlamps dim
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.
As an aside, a lot of students seem to think the engine is powered by the battery! Point out that its main purpose while the engine is running is to provide the sparks for ignition and that while the car is driving the alternator continually recharges the battery, the energy for both headlamps and driving comes ultimately from the fuel that is burnt (since the car has to work a little bit harder to turn the alternator).
Student questions: Internal resistance of a power supply
Some simple questions about the internal resistance of a power supply.
Episode 121-1: Internal resistance of power supplies (Word, 30 KB)
Student experiment: Measuring internal resistance and EMF
There are two experiments here, in which students determine the EMF E and internal resistance r of cells – one involving a potato cell (leading to a high internal resistance) and one involving a normal C cell (much lower internal resistance). You could get them to do both or ask some students to do one and some the other. Beware that, if you use an alkaline, high power C cell, it will run down quickly when there is a low load resistance, so you are advised to use cheap, low power cells which polarise quickly, they will depolarise over night. An alternative is to construct an artificial cell with a larger internal resistance by adding a higher series resistance (e.g. 100 Ω ) to a standard cell.
Episode 121-2: Internal resistance of a source of EMF (Word, 48 KB)
Episode 121-3: Internal resistance of a C cell (Word, 28 KB)
To determine E and r from the experimental results, there are various approaches. The simplest is to measure terminal voltage (V) and current (I) and to plot V against I. This gives an intercept at V = E on the y-axis and has a gradient of − r.
Student questions: Practice questions
Questions on EMF and internal resistance.
Episode 121-4: Questions on EMF and internal resistance (Word, 29 KB)
Discussion: More about the practical importance of internal resistance
Sometimes it is desirable to have a high internal resistance. Ask the class what happens if a 5 V cell is shorted – i.e. its terminals are connected together by a wire of zero resistance? Some might think I = VR with R = 0 should mean that an infinite current would flow (limited by other physical factors!)
Remind them of the internal resistance r. This limits the cell to a maximum (short-circuit) current of:
I = Er
We can use this to prevent EHT supplies giving the user an unpleasant shock. Take an EHT power supply off the shelf and show the connections for the series
internal resistance. It is usually 5 M Ω .
These supplies are designed to provide a high voltage to a high resistance load (e.g. cathode ray tube) but if the terminals or wires connected to them were accidentally touched this could provide a nasty shock (lower resistance in the load and higher current). One way to deal with this is to connect a large resistance in series with the output (positive) terminal. If the terminals are shorted (e.g. by contact through a person) the current drawn is limited to I = Er. A typical EHT supply (up to 5000 V) is protected by a 5 M Ω resistor so the maximum current if shorted is just 1 mA. That shouldn’t kill you! Be aware however that HT supplies (0-300 V) have a much lower internal resistance, and could kill you, so special shrouded leads should be used.
EHT supplies often have a further
safety resistor (e.g. 10 M Ω ) to reduce the maximum current still further. This resistor can be by-passed when necessary. No school EHT supply is allowed to provide more than 5 mA.