Current, potential difference and resistance: three linked quantities
Definition
Current
Current, III, is the flow of electric charge through a component. It is measured in amperes, A\text{A}A.
Definition
Potential difference
Potential difference, VVV, is measured across a component in volts, V\text{V}V, and it drives charge through the component.
Definition
Resistance
Resistance, RRR, is a measure of how strongly a component opposes the current. It is measured in ohms, Ω\OmegaΩ.
- The current through a component depends on the potential difference across it and on its resistance.
- For a fixed resistance, a greater potential difference gives a greater current, because it pushes more charge through each second.
- For a fixed potential difference, a greater resistance gives a smaller current, because the component opposes the flow of charge more strongly.
Key Idea
For the same potential difference:
- a greater resistance gives a smaller current
- a smaller resistance gives a greater current.
Through, across and has: use the language precisely
- Current flows through a component.
- Potential difference is measured across a component.
- A component has resistance.
- The term potential difference is used throughout, and the word voltage means the same thing here and is credited when used correctly.
The equation that links them: V=IRV = IRV=IR
- Potential difference, current and resistance are linked by V=IRV = IRV=IR, which in words is potential difference = current ×\times× resistance, where:
- VVV is the potential difference in volts, V\text{V}V
- III is the current in amperes, A\text{A}A
- RRR is the resistance in ohms, Ω\OmegaΩ
- This is an equation you need to recall and select for yourself, so learn it.
- It rearranges to find the current, I=VRI = \dfrac{V}{R}I=RV, or the resistance, R=VIR = \dfrac{V}{I}R=IV.
Example
A resistor has a resistance of 8 Ω8\ \Omega8 Ω and a potential difference of 12 V12\ \text{V}12 V across it. Calculate the current.
State and rearrange:
V=IR⇒I=VR V = IR \quad\Rightarrow\quad I = \frac{V}{R} V=IR⇒I=RVSubstitute and calculate:
I=128=1.5 A I = \frac{12}{8} = 1.5\ \text{A} I=812=1.5 AThe current through the resistor is 1.5 A1.5\ \text{A}1.5 A.
Example
The current through a component is 0.40 A0.40\ \text{A}0.40 A when the potential difference across it is 6.0 V6.0\ \text{V}6.0 V. Calculate its resistance.
State and rearrange:
V=IR⇒R=VI V = IR \quad\Rightarrow\quad R = \frac{V}{I} V=IR⇒R=IVSubstitute and calculate:
R=6.00.40=15 Ω R = \frac{6.0}{0.40} = 15\ \Omega R=0.406.0=15 ΩThe resistance of the component is 15 Ω15\ \Omega15 Ω.
Common Mistake
- Do not say a greater resistance causes a greater current: for a fixed potential difference the opposite is true, and the current falls.
- Do not mix up the units: potential difference is in volts, current in amperes and resistance in ohms.
Exam technique
- Begin a calculation by writing V=IRV = IRV=IR, rearrange it before you substitute, show each step and give the unit.
- In a written answer, include the condition for a given potential difference, for example: "For a given potential difference, increasing the resistance decreases the current."
Self review
- State the equation linking potential difference, current and resistance.
- In what units are potential difference, current and resistance measured?
- For a fixed potential difference, how does increasing the resistance change the current?
- Rearrange V=IRV = IRV=IR to make resistance the subject.
- Which quantity flows through a component, and which is measured across it?
2.1.3b Investigating resistance (required practical)
Measuring resistance: ammeter in series, voltmeter in parallel
Definition
Resistance
Resistance is a measure of how difficult it is for current to flow through a component. It is measured in ohms, Ω\OmegaΩ.
- Resistance is found from R=VIR = \dfrac{V}{I}R=IV, where:
- RRR is the resistance in ohms, Ω\OmegaΩ
- VVV is the potential difference in volts, V\text{V}V
- III is the current in amperes, A\text{A}A
- To measure a component’s resistance, connect an ammeter in series with it to read the current through it.
- Connect a voltmeter in parallel across the component to read the potential difference across it.
- Then calculate the resistance with R=V/IR = V/IR=V/I, building the circuit from a diagram of standard symbols and checking it before switching on.
- You investigate two things that affect the resistance of a circuit: the length of a wire at constant temperature, and combinations of resistors in series and in parallel.
Practical
Investigation: how the resistance of a wire depends on its length
- Independent variable: the length of the wire.
- Dependent variable: the resistance of the wire.
- Control variables: the material, cross-sectional area and temperature of the wire, because each of these also affects resistance.
Apparatus
- a battery or low-voltage power supply
- an ammeter and a voltmeter
- a length of resistance wire, such as constantan or nichrome (about 22 swg), taped along a metre ruler
- two crocodile clips and connecting leads
- a fixed resistor in series is useful to limit the current
Method
- Tape the resistance wire tightly along the metre ruler so it lines up with the zero mark.
- Build a series circuit of the power supply, the ammeter and the wire, with one crocodile clip fixed at the zero end of the wire.
- Connect the voltmeter in parallel across the section of wire being tested, from the fixed clip to a movable clip.
- Use the free lead as a switch so you can disconnect the battery between readings.
- Set the first length, for example 10 cm10\ \text{cm}10 cm, by placing the movable clip at the 10 cm10\ \text{cm}10 cm mark.
- Close the circuit briefly, read the current and the potential difference, then disconnect.
- Calculate the resistance with R=V/IR = V/IR=V/I.
- Move the clip to the next length (20 cm20\ \text{cm}20 cm, 30 cm30\ \text{cm}30 cm, and so on up to about 100 cm100\ \text{cm}100 cm) and repeat.
- Repeat the whole set of readings and take a mean resistance at each length.
Working carefully and accurately
- Only close the circuit for the moment you take a reading: a thin wire heats up quickly, and a hotter wire has a higher resistance, which would make the test unfair.
- Use a low potential difference, especially for the short lengths where the current is largest.
- It is hard to fix the clip exactly at the zero of the wire, and the clips add a little contact resistance, so the line of best fit may not pass exactly through the origin.
Result
- Plot resistance on the vertical axis against length on the horizontal axis.
- The points lie on a straight line: at constant temperature the resistance is directly proportional to the length, because the charges pass through more wire.
Example
For a 0.40 m0.40\ \text{m}0.40 m length of wire a student measures a potential difference of 1.8 V1.8\ \text{V}1.8 V and a current of 0.30 A0.30\ \text{A}0.30 A.
Calculate the resistance:
R=VI=1.80.30=6.0 Ω R = \frac{V}{I} = \frac{1.8}{0.30} = 6.0\ \Omega R=IV=0.301.8=6.0 ΩDoubling the length to 0.80 m0.80\ \text{m}0.80 m gives a resistance of 12.0 Ω12.0\ \Omega12.0 Ω: doubling the length doubles the resistance, so resistance is directly proportional to length when temperature and the other variables are kept constant.
Practical
Investigation: the resistance of resistors in series and in parallel
- In a series arrangement the resistors are in one loop, so the total resistance is the sum of the separate resistances, because the current passes through every resistor.
- In a parallel arrangement each resistor is on its own branch, so adding a branch gives the current another path and the total resistance is less than the smallest single resistor.
Apparatus
- a battery or power supply and a switch
- an ammeter and a voltmeter
- two resistors of the same value, for example 10 Ω10\ \Omega10 Ω wire-wound resistors, which resist overheating
- connecting leads and crocodile clips
Method
- Build a series circuit with the two resistors, the ammeter in series and the voltmeter across the whole combination, working from a circuit diagram.
- Close the switch, read the current and the potential difference, then open the switch.
- Calculate the total resistance with R=V/IR = V/IR=V/I.
- Rebuild the circuit with the two resistors in parallel and take the readings again.
- Compare the total resistance of the series and parallel arrangements.
Sample readings
- In series, about 0.22 A0.22\ \text{A}0.22 A at 4.45 V4.45\ \text{V}4.45 V gives roughly 20 Ω20\ \Omega20 Ω, so the two 10 Ω10\ \Omega10 Ω resistors add.
- In parallel, about 0.79 A0.79\ \text{A}0.79 A at 3.94 V3.94\ \text{V}3.94 V gives roughly 5 Ω5\ \Omega5 Ω, which is less than a single resistor.
Keeping it fair
- Use equal, wire-wound resistors and take repeats so heating does not distort the comparison.
Common Mistake
- Do not connect an ammeter in parallel: its very low resistance could give a dangerously large current, and never put a voltmeter in series.
- Adding a resistor does not always increase the total resistance: it increases it in series but decreases it in parallel.
- A warmer wire has a higher resistance, so letting the wire heat up would make the length investigation unfair.
Exam technique
- For a method question, give the circuit, the independent, dependent and control variables, and the measurements taken.
- State that the ammeter is in series and the voltmeter is in parallel, explain how you keep the temperature constant, include repeats, and calculate resistance with R=V/IR = V/IR=V/I.
Self review
- How is a resistance found from measurements of potential difference and current?
- Where do the ammeter and voltmeter go when measuring the resistance of a component?
- What happens to the resistance of a wire as its length increases at constant temperature?
- Which variables must you control in the wire-length investigation?
- How does the total resistance change when resistors are added in series, and when they are added in parallel?
