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Chapter 23: Resistance Please remember to photocopy 4 pages onto one sheet by going A3→A4 and using back to back on the photocopier. The resistance of a conductor is the ratio of the potential difference across it to the current flowing through it.
Mathematically: The unit of resistance is the Ohm Symbol is Ω
Ohm’s Law* states that the current flowing through a conductor is directly proportional to the potential difference across it, assuming constant temperature.
Resistors in series and in parallel
Resistors in Series*
Derivation: For resistors in series VTotal = V 1 + V 2
But V = IR (Ohm’s Law) ⇒ I RTotal = I R 1 + I R 2 (We can now cancel the I’s because the current is the same for resistors in series) ⇒ RTotal = R 1 + R 2
Resistors in Parallel*
Derivation : For resistors in parallel ITotal = I 1 + I 2
But I = V/R (Ohm’s Law) ⇒ V/RT = V/R 1 + V/R 2 (But we can cancel the V’s because the voltage is the same for resistors in parallel) ⇒ 1/RTotal = 1/R 1 + 1/R 2
RTotal = R 1 + R 2
Resistivity*
Resistivity is defined as the resistance of a cube of material of side one metre. or Resistivity is defined as the resistance of a material of unit length and unit cross sectional area. The symbol for resistivity is ρ (pronounced “rho”). We have come across ρ twice already in the course – can you remember where?) The unit of resistivity is the Ωm (can you see why by looking at the formula below?)
**Formula for resistivity: ***
You might need to revise how to convert from millimetres square to metres square.
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The Potential Divider Circuit
- In the diagram across if R 2 is four times greater than R 1 ; it means that the potential difference across R 2 will be four times greater than the potential difference across R 1 (4 times more work needed to bring the same amount of charge through both resistors).
- The total potential difference across both resistors would be the same as the potential difference between Vs and ground (the 0 Volt mark), because voltages in parallel are the same.
- In this scenario, if the supply voltage was 30 Volts, then the voltage across R 1 would be 6 Volts and the voltage across R 2 would be 24 Volts.
The Potentiometer A potentiometer is a variable potential divider. Potentiometers are variable voltage dividers with a shaft or slide control for setting the division ratio.
In the diagram for the potential divider circuit above, the voltage- out is the voltage across R 2 and will only change if the temperature of R 2 or R 1 was to change. A simple way to gradually vary the voltage out is to replace R 1 and R 2 by one variable resistor. The output voltage is then the voltage between the ground (the bottom line above) and the middle contact.
Mandatory Experiments:
- To verify Ohm’s Law
- To Investigation the relationship between the potential difference across a filament bulb and the current through it.
- To Investigation the relationship between the resistance of a metallic conductor and temperature.
- To measure the Resistivity of the Material of a Wire.
Solving Electric Circuit Problems
Step One: Find the Total Current flowing in the circuit (i) First establish the Total Voltage supplied, which causes the electrons to move. You are usually told this directly. (ii) Next we need to calculate the Total Resistance of the circuit. To find total resistance you must remember the rules for adding resistors in series and adding resistors in parallel. To test yourself try page 261, no. 11. (iii) Total Current can then be found using I = VTotal/RTotal
Step Two: Look for the isolated resistor. (i) There will be one resistor on its own, so it’s easiest to start with this. (ii) Remember the total current will flow through this resistor, so use V = IR to find V.
Step Three: Look at remaining resistors. (i) Subtract the voltage associated with the isolated resistor from the total voltage to find the voltage across the remaining resistors. (ii) Now use your knowledge of voltages in series and in parallel, and current in series and parallel to solve the question in hand.
To test yourself try page 250, no.12. Now look over Problems 4 and 5, page 260 and try No.s 1 – 14, page 261, concentrating on 13 and 14.
Example The diagram shows a number of resistors connected to a 12 V battery and a bulb whose resistance is 4 Ω. Calculate the current flowing through each resistor. Solution Step One: Find the Total Current flowing in the circuit Total voltage = 12 V Total resistance: First calculate the combined resistance of the 15 Ω and 3 0 Ω resistors in parallel. 1/R15,30 = 1/15 + 1/ R15,30 = 10 Ω Total resistance of the circuit = 10 + 10 + 4 = 24 Ω
Calculate the current flowing in the circuit I = V/R = ITotal = VTotal/RTotal = 12/24 = 0.5 A Step Two: Look for the isolated resistor(s). 10 Ω resistor: I = 0.5 A, R = 10 Ω. V = IR = (0.5)(10) = 5 Volts Lightbulb (4 Ω): I = 0.5 A, R = 4 Ω, V = IR = (0.5)(4) = 2 Volts Step Three: Look at remaining resistors. V = 12 – 5 – 2 = 5 Volts, so the potential difference across both the15 Ω and the 30 Ω resistors is 5 V. To find I use I = V/R Current through 15 Ω resistor = 5/15= 0.33 Amps Current through 30 Ω resistor = 5/30 = 0.17 Amps
Note that the current through the 15 Ω resistor plus the 30 Ω resistor adds up to 0.5 A, which was the current going in.
Note that you could have used ratios to split up the current: 30 is twice 15 so 2/3 of the total current goes through the 15 Ω resistor while 1/3 goes through the 30 Ω resistor. I am reluctant to encourage this approach because students often make the mistake of putting the 2/3 through the 30 Ω and 1/3 through the 15 Ω resistor. The ratio won’t always work out to be nice simple numbers either, but if you want to go down that route it is certainly allowed.
TO MEASURE THE RESISTIVITY OF THE MATERIAL OF A WIRE
APPARATUS:
Length of nichrome wire, micrometer or digital calilpers, ohmmeter, metre stick.
DIAGRAM:
PROCEDURE:
- Tie a length of nichrome between the bars of the two stands as shown above. Stretch the wire enough to remove any kinks or ‘slack’ in the wire.
- Note the resistance of the leads when the crocodile clips are connected together.
- Connect the crocodile clips to the wire. Read the resistance from the ohmmeter. Subtract the resistance of the leads to get the resistance R of the wire.
- Measure the length (L) of the wire between the crocodile clips with the metre stick.
- Find the diameter (d) of the wire at different points, taking the zero error into account. Find the average value of the diameter d. We use a digital callipers instead of the micrometer.
- Calculate the resistivity using the formula
- Repeat using different lengths of wire and calculate an average value for ρ.
RESULTS: Run no.
Diameter d (m)
RTotal (Ω)
RLeads (Ω)
Rnet (Ω)
L
(m) (Ω m) **1.
3.**
Average value for resistivity = ________________ Ω m.
CONCLUSION:
We found an average value for the resistivity of the wire as __________ Ω m. This is reasonably accurate because all our answers are close together and are reasonably close to the accepted value of 100 × 10-8^ Ω m (at 20 °C).
SOURCES OF ERROR / PRECAUTIONS:
1. Ensure that the wire is straight when measuring its length.
2. Measure only between the inside of the ohmmeter contacts.
- Make sure to zero the reading on the callipers before starting.
NOTE: When setting up lay the wire over a metre stick and clamp it to using G-clamps. The length of wire can now be measured directly.
TO INVESTIGATE THE VARIATION OF THE RESISTANCE OF A METALLIC CONDUCTOR WITH
TEMPERATURE
APPARATUS:
Coil of wire, glycerol, beaker, heat source, thermometer, ohmmeter, boiling tube
DIAGRAM:
PROCEDURE:
- Set up the apparatus as shown in the diagram.
- Use the thermometer to note the temperature of the glycerol, which we assume to be the same as the temperature of the coil.
- Record the resistance of the coil of wire using the ohmmeter.
- Heat the beaker and for each 10 °C rise in temperature record the resistance and temperature using the ohmmeter and the thermometer.
- Plot a graph of resistance against temperature.
RESULTS: R (Ω) θ (^0 C)
CONCLUSION:
- From the graph we can see that there is a linear relationship between the temperature of the wire and its resistance, and that as one increases so does the other.
- We believe our data to be reliable because it resulted in a straight line which the theory predicted.
PRECAUTIONS / SOURCES OF ERROR:
- Check for the resistance of the connecting leads and contacts on the ohmmeter. Subtract from later readings
- Heat very slowly to try to maintain thermal equilibrium between the water and glycerol and coil. When the bunsen is removed wait until the temperature is steady before taking the resistance readings.
- Use glycerol in the test tube as it is a better heat conductor than water.
TO INVESTIGATE THE VARIATION OF CURRENT (I) WITH P.D. (V) FOR A FILAMENT BULB AND
APPARATUS :
Low voltage power supply, rheostat, voltmeter, ammeter, filament bulb.
DIAGRAM:
PROCEDURE:
- Set up the circuit as shown (replace the letter X with A for ammeter above, and don’t write this into your report).
- Record the potential difference (V) and the current (I) using the voltmeter and ammeter respectively.
- Adjust the potential divider to obtain different values for V and I.
- Obtain at least six values for V and I.
- Plot a graph of I against V.
RESULTS:
V (V) I (A)
CONCLUSION :
- As the potential difference across the bulb increases so does the current, but not in a linear fashion therefore the two are not directly proportional.
- We believe our results to be reliable because they resulted in a smooth curve, as the theory predicted.
QUESTION:
- Why does the graph start to level off at the end?
- Why does this happen for the filament bulb but not for the coil of wire (metallic conductor) in the last experiment?
NOTES:
A varying voltage can be obtained from a fixed supply voltage by using a potential divider. It consists of a variable resistor or fixed resistors in series. Move the slider to change the output voltage. This results in the output voltage from the potential divider being a fraction of the input voltage.
*Extra Credit Ohm’s Law Remember from Junior Cert that a resistor was defined as something which resists the flow of current? Well in Leaving Cert Physics we like to quantify things (that means put numbers on them), hence the formula. If you think about it the formula does make sense, because if V is small (you use a small effort to push the electrons through) and you find that I (the current) is large, this suggests that there couldn’t have been much resistance in the circuit, i.e. if the ratio of Voltage to Current is small, so is the Resistance. Similarly, if you use a lot of energy to try and push the electrons through (V is large), and yet you still only get a small current (I is small), this suggests that there must be a large resistance in the circuit, i.e. if the ratio of Voltage to Current is large, so is the Resistance. “What power! To condense all the meaning in that long sentence into a simple four symbol equation. That is the art of our science.”
“at Constant Temperature” In practice most metals heat up when a current passes through them, which means that the resistance increases. This is because the atoms in the metal gain energy and ‘jiggle’ up and down, making it harder for the electrons to get by. Note that if you omit the phrase ‘at constant temperature’ you will lose 3 marks out of the total 6.
*Resistors in Series and Resistors in Parallel derivations The key here is to begin by putting the R on same side of the equation as the variable which is constant for that particular section. Because this variable is constant it can then be cancelled across the line.
*Resistivity Why do we have a concept called ‘resistivity’? Because you can’t just say that ‘the resistance of copper is 3 Ohms’; you would need to specify the length and the width of the material. It’s similar to the reason why we have the concept of Density. It wouldn’t make any sense to say that iron is heavier than paper, because you might have a tiny piece of iron and a very large piece of paper (say a Golden Pages directory). What we mean when we say that iron is heavier than paper is ‘ if we have the same volume of both, then the iron would have a greater mass’. Density is a shorthand way of saying this. In a similar way, saying that the resistivity of copper is greater than that the resistivity of silver is shorthand for saying; ‘ If the two materials are of equal length and equal cross-sectional area, then silver would have a greater resistance than copper’.
_The formula for resistivity is_*
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From the following; The resistance of a material is proportional to the length ⇒ R ∝ l The resistance of a material is inversely proportional to the Cross Sectional Area (C.S.A.) ⇒ R ∝ 1/A Putting this together ⇒ R = k l/A
The proportional constant is given the symbol ρ (rho– same as for density);
A
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R
Cross-multiplying to get ρ on its own;
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I have something called ‘Silly Putty’ which is used to demonstrate the relationship between Resistance, length and C.S.A.; Remind me to demonstrate it.
Resistivities of some common metals (You don’t need to know these) Silver : 1.6 × 10-8^ Ωm Copper: 1.7 × 10-8^ Ωm Nichrome: 1 × 10-6^ Ωm
