What are the key learning points about calculating resistance?
The resistance of a metallic conductor at constant temperature depends on length.
The graph of resistance versus length is a straight line that passes through the origin, this shows that for a metal wire at constant temperature the resistance and length of wire are When one variable is zero so is the other. As one variable increases the other does at the same rate. When 𝒚 is plotted against 𝒙 this produces a straight-line graph through the origin..
The resistance of a metallic conductor at constant temperature also depends on the area of cross section and the material it is made from.
What is electrical resistance?
Electrical resistance is the opposition offered by a conductor to the flow of current around a circuit.
A good conductor has low resistance.
An insulator, or poor conductor, has high resistance.
What is the equation for calculating resistance?
Current, voltage and resistance are linked by an equation which comes from Ohm’s law:
voltage = current x resistance
V= IR
V = voltage in V
I = current in A
R = resistance in Ω
| V = IR | V = I x R |
| I = \(\frac{\text{V}}{\text{R}}\) | I = V ÷ R |
| R = \(\frac{\text{V}}{\text{I}}\) | R = V ÷ I |
Question
The current in a torch lamp is 0.3 A when the voltage across it is 3 V.
Calculate the resistance of the lamp.
Answer
V = 3 V
I = 0.3 A
R = \(\frac{\text{V}}{\text{I}}\)
R = \(\frac{\text{3 V}}{\text{0.3 A}}\)
R = 10 \(\Omega\)
The resistance of the lamp is 10 \(\Omega\)
Question
3 A flows through a 240 V lamp. What is the resistance of the lamp?
Answer
R = \(\frac{\text{V}}{\text{I}}\)
V = 240 V
I = 3 A
I = \(\frac{\text{240 V}}{\text{3 A}}\)
R = 80 \(\Omega\)
The resistance of the lamp is 80 \(\Omega\)
Question
What is the voltage across a 20 \(\Omega\) resistor when a current of 2A flows through it?
Answer
V = IR
I = 2 A
R = 20 Ω
V = 2 × 20
V = 40 V
The voltage across the resistor is 40 V.
Prescribed practical P2: The resistance of a metallic conductor
A guide to carrying out a practical to investigate resistance
What is the purpose of prescribed practical P2?
To investigate experimentally how the resistance of a metallic conductor at constant temperature depends on length and obtain sufficient values to plot a graph of resistance (y-axis) and length (x-axis).
What are the main variables in this practical?
The main variables in a science experiment are the independent variable, the dependent variable and the control variables.
The independent variable is what we change or control in the experiment.
The dependent variable is what we are testing and will be measured in the experiment.
The control variables are what we keep the same during the experiment to make sure it’s a fair test.
In this experiment the:
Independent variable is the length of wire.
Dependent variable is the resistance of the wire.
Control variables are the material, the cross section area and the temperature of the wire, and these are kept the same by not changing the wire during the experiment, by keeping the current small and by opening the switch in between readings.
Remember - these variables are controlled (or kept the same) because to make it a fair test, only one variable can be changed, which in this case is the length of wire.
What is the equation used for calculating resistance?
Resistance R = \(\frac{voltage~V}{current~I}\)
What is the prediction for this practical?
As the length of wire increases, the resistance will increase.
What is the justification for this prediction?
The greater the length of wire the greater the number of collisions between the Negatively charged sub-atomic particles that can move through the structure of a substance, usually a metal or graphite. A material with many free electrons is a good conductor. and metal Electrically charged particle, formed when an atom gains or loses electrons..
This will result in greater resistance.
How to carry out this experiment safely
| Hazard | Consequence | Control measures |
|---|---|---|
| Water | Electric shock | Do not set up the experiment near taps, sinks etc. |
| Wire gets hot | Minor burns | Do not handle the wire. Switch off between readings. |
What apparatus is needed for this practical?
1m length of constantan wire, a metre rule, a low voltage power pack, a voltmeter, an ammeter, connecting leads, a switch, 2 crocodile clips, Sellotape.
How to carry out prescribed practical P2?
- Set up the circuit, as shown above. Attach one of the crocodile clips at the 0 cm mark and the other at the 20 cm mark so that the length of wire the current flows through is 20 cm. Record this length in a suitable table
- Adjust the power pack until the current on the ammeter is 0.4 A. Record the current in the table
- Read the corresponding value of voltage across the wire on the voltmeter and record in the table
- Switch the switch off to prevent the temperature of the wire rising
- Switch on again and repeat the reading of voltage. Record in the table. Switch off and calculate the average voltage
- Calculate the resistance of this length of wire and record in the table
- Switch on again. Ensure that the current is still 0.4 A and repeat current and voltage reading for lengths of 40 cm, 50 cm, 60 cm, 80 cm and 100 cm
- Calculate the resistance for each length, remembering to switch off between each reading
Error
The temperature of the wire must be kept constant.
Whenever a current flows through a conductor there is a heating effect.
Electrical energy is converted to heat energy.
To ensure the temperature of the wire does not increase, switch off between readings and keep the current as low as possible.
Read the ammeter and voltmeter accurately by reading the scale from directly above the pointer or use digital instruments.
Results
| Length l / cm | Current I / A | Voltage V / V | Voltage V / V | Voltage V / V | Resistance R / Ω |
|---|---|---|---|---|---|
| Reading 1 | Reading 2 | Average voltage | |||
| 20 | |||||
| 40 | |||||
| 50 | |||||
| 60 | |||||
| 80 | |||||
| 100 |
Graph
Plot a graph of resistance, R, in Ω on the y-axis against length, l, in cm on the x-axis. Draw the line of best fit.
Conclusion
We can see from the graph that as the length of the wire, l, increases, the resistance, R, also increases.
This agrees with our prediction.
In fact, since the line of best fit is a straight line through the origin, we can be even more precise.
We can say that, for a metal wire at constant temperature, the resistance is When one variable is zero so is the other. As one variable increases the other does at the same rate. When 𝒚 is plotted against 𝒙 this produces a straight-line graph through the origin. to the length of the wire.
If you double the length of the wire you double its resistance.
Key points
- The resistance of a metallic conductor at constant temperature depends on the length of the conductor.
- Resistance is directly proportional to length.
Question
A 5 m length of wire is found to have a resistance of 40 Ω.
What is the resistance of each of the following identical wires of different length?
a) 10 m
b) 25 m
c) 2.5 m
Answer
Since each wire is identical resistance and length are directly proportional.
a) 10 m is twice the original length, so this wire will have twice the resistance = 2 x 40 Ω = 80 Ω
b) 25 m is five times the original length, so this wire will have five times the resistance = 5 x 40 Ω = 200 Ω
c) 2.5 m is half the original length, so this wire will have half the resistance = \(\frac{1}{2}\)x 40 Ω = 20 Ω
What is a variable resistor? (Higher tier only)
Variable resistors can be used to control the current that is flowing in a circuit.
They do this by changing the length of some resistance wire in a circuit.
Moving the position of the slider on this resistor, changes the length of wire in the circuit which changes the resistance.
A variable resistor is used in some dimmer switches and volume controls.
As the slider moves towards B the length of wire in the circuit increases.
This increases the resistance in the circuit and decreases the current.
It would dim a lamp.
As the slider moves towards A the length of wire in the circuit decreases.
This decreases the resistance in the circuit and increases the current.
It would make a lamp shine more brightly.
If connections are made between the two lower terminals it will act as a fixed resistor.
The electrical symbol for a variable resistor:
Resistance and area of cross-section (Higher tier only)
A second experiment can be carried out to investigate experimentally how the resistance of a metallic conductor at constant temperature depends on the area of cross section.
The above experiment is repeated but with six, equal lengths of constantan wire, of different thickness.
In this experiment the:
Independent variable is the cross section area of the wire.
Dependent variable is the resistance of the wire.
Control variables are the material, the length and the temperature of the wire. These are kept the same by not changing the wire during the experiment, by keeping the current small and opening the switching between readings.
Remember - these variables are controlled (or kept the same) because to make it a fair test, only one variable can be changed, which in this case is the cross section area of the wire.
Record voltage, current and The distance across the middle of a circle. of the wire, d (supplied by the manufacturer).
Calculate resistance and cross section area, A, in mm2 (A = \(\frac {{\pi }d^2}{4}\)).
Plot a graph of resistance, R, in Ω on the y-axis against cross section area, A, in mm2 on the x-axis.
Draw the line of best fit.
We can see from the graph that as the cross section area, A, increases, the resistance, R, decreases.
A thicker wire has a smaller resistance than a thin wire.
A more detailed investigation shows that resistance and cross section area are Two variables are said to be inversely proportional when one increases by a factor of two (doubles) the other decreases by a factor of two (halves)..
If you double the cross section area you halve the resistance of the wire.
A final experiment can be carried out to investigate experimentally how the resistance of a metallic conductor at constant temperature depends on the material of the conductor.
The experiment is repeated again but with six, equal lengths and thicknesses of wire of different materials.
In this experiment the:
Independent variable is the material of wire.
Dependent variable is the resistance of the wire.
Control variables are the length, the cross section area and the temperature of the wire. The temperature is kept the same by keeping the current small and opening the switching between readings
Remember - these variables are controlled (or kept the same) because to make it a fair test, only 1 variable can be changed, which in this case is the material of wire.
Record voltage, current and calculate resistance.
A comparison of results in the table shows that wires of different material have different resistance.
Key points
The resistance of a metallic conductor at constant temperature depends on:
The length l. Resistance is When one variable is zero so is the other. As one variable increases the other does at the same rate. When 𝒚 is plotted against 𝒙 this produces a straight-line graph through the origin. to length.
The cross section area A. Resistance is Two variables are said to be inversely proportional when one increases by a factor of two (doubles) the other decreases by a factor of two (halves). to cross section area.
The material of the conductor
Resistance increases as:
The length of the wire increases.
The thickness of the wire decreases.
An electric current flows when Negatively charged sub-atomic particles that can move through the structure of a substance, usually a metal or graphite. A material with many free electrons is a good conductor. move in one direction through a conductor, such as a metal wire.
The moving electrons can collide with the Electrically charged particle, formed when an atom gains or loses electrons. in the metal.
This makes it more difficult for the current to flow, and causes resistance.
The resistance of a long wire is greater than the resistance of a short wire because electrons collide with more ions as they pass through a longer wire.
Resistance and wire length are When one variable is zero so is the other. As one variable increases the other does at the same rate. When 𝒚 is plotted against 𝒙 this produces a straight-line graph through the origin..
The resistance of a thin wire is greater than the resistance of a thick wire because a thin wire has fewer gaps for the free electrons to pass through.
Resistance and the area of cross section of a wire are Two variables are said to be inversely proportional when one increases by a factor of two (doubles) the other decreases by a factor of two (halves).
Worked example (Higher tier only)
Question
A 2 m length of wire is found to have a resistance of 36 \( \Omega\)
What is the resistance of each of the following wires of equal length and made of the same material but having different cross section areas?
Double the area of cross section.
Four times the area of cross section.
Quarter the area of cross section.
Answer
Since each wire is identical, resistance and area of cross section will be Two variables are said to be inversely proportional when one increases by a factor of two (doubles) the other decreases by a factor of two (halves)..
Doubling the area of cross section halves the resistance, so this wire will have half the resistance = \(\frac {1}{2}\) x 36 \(\Omega\) = 18 \(\Omega\).
Four times the area of cross section will quarter the resistance, so this wire will have a quarter the resistance = \(\frac {1}{4}\) x 36 \(\Omega\) = 9 \(\Omega\).
Quartering the area of cross section will increase the resistance by a factor of four, so this wire will have 4 times the resistance = 4 x 36 \(\Omega\) = 144 \(\Omega\).
Question
A wire of length 50 cm is found to have a resistance of 15 \(\Omega\)
What is the resistance of a wire made from the same material but 150 cm long and half the area of cross section?
Answer
Resistance is directly proportional to length.
The new wire is three times the length of the original and so this will increase the resistance by a factor of three.
Resistance is inversely proportional to area of cross section.
The new wire has half the area of cross section of the original and so this will increase the resistance by a factor of two.
The combined effect is to increase the resistance by a factor of six = 6 x 15 = 90 \(\Omega\).
The resistance of the new wire is 90 \(\Omega\).
How much do you know about resistance?
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