Circuits, current and potential difference
By the end of this chapter you will be able to
- Name and draw the common circuit symbols
- Calculate charge and current, and explain what current means
- Calculate the energy transferred by a charge moving through a potential difference
- State how current and potential difference behave in series and parallel circuits, and connect meters correctly
You should already know
- Energy stores and transfers (Chapter 1)
- Electrical transfers by a current (Chapter 1)
Where this leads
- Resistance and I-V characteristics (Chapter 6)
- Series and parallel resistance and power (Chapter 7)
GCSE Physics. Content follows the GCSE (9-1) subject content for England (Edexcel and AQA). Content references: Draw and recognise standard circuit symbols, Explain current as the flow of charge, measured in amperes; use Q = It, Explain potential difference (volts) as energy transferred per unit charge; use E = QV, State the rules for current and potential difference in series and parallel circuits, Connect an ammeter in series and a voltmeter in parallel.
Exercise P Do you remember?
Method. Multiply the two numbers.
- Multiply. Half of 60 is 30.0.5 x 60 = 30
- Write the answer. 30
Check. 30 x 2 = 60. ✓
Method. Divide the two numbers.
- Divide. Sixty goes into 120 twice.120 / 60 = 2
- Write the answer. 2
Check. 2 x 60 = 120. ✓
Method. Recall the unit.
- Recall. Energy is measured in joules, J.joule (J)
- Write the answer. joule
Check. ✓
Method. Subtract the two numbers.
- Subtract. 9 - 3 = 6, then take away 0.5 to get 5.5.9 - 3.5 = 5.5
- Write the answer. 5.5
Check. 5.5 + 3.5 = 9. ✓
Method. Recall the convention for current.
- Recall. Conventional current flows out of the positive terminal and back into the negative terminal.positive
- Write the answer. positive
Check. ✓
Circuit symbols and drawing circuits
A circuit diagram shows the parts of a circuit using standard symbols, joined by straight wires with a ruler. A complete circuit is a closed loop, so the current can flow all the way round. An open switch breaks the loop and stops the current.
Key ideas
- A cell has a longer thin line and a shorter thick line; the longer line is the positive terminal.
- A lamp is a circle with a cross inside it.
- A closed switch makes a complete loop; an open switch does not.
- Meters: ammeter (A), voltmeter (V). Both are drawn as circles.
Method. Recall the shape of the lamp symbol.
- Look. A lamp is a circle with a cross in it.circle with cross
- Name. That symbol means a lamp.lamp
- Write the answer. lamp
Check. A resistor is a rectangle, and a thermistor is a resistor with a slant line. ✓
Method. Think about whether the loop is closed.
- Look. An open switch breaks the loop.loop broken
- Decide. So the current cannot flow round the circuit.no current
- Write the answer. no
Check. Closing the switch completes the loop. ✓
Method. Recall the letter shown on each meter.
- Look. An ammeter has the letter A inside the circle.A
- Name. Ammeters measure current in amperes.ammeter
- Write the answer. A
Check. The voltmeter has the letter V inside the circle. ✓
Watch out
Watch out: a battery is more than one cell. In a diagram, one cell is one pair of lines; a battery has several pairs.
Exam tip
Exam tip. Draw circuits with a ruler, and use the same symbols as the diagrams in your textbook. Do not draw the wires crossing without a dot.
Exercise 5A Symbols and drawing
Method. Recall the resistor symbol.
- Recall. A resistor is drawn as a rectangle.rectangle
- Write the answer. rectangle
Check. ✓
Method. Recall the letter in the meter circle.
- Recall. An ammeter is a circle with the letter A.A
- Write the answer. A
Check. ✓
Method. Recall the letter in the meter circle.
- Recall. A voltmeter is a circle with the letter V.V
- Write the answer. V
Check. ✓
Method. Recall the component.
- Recall. A diode allows current to flow one way only.diode
- Write the answer. diode
Check. ✓
Method. Recall the protective component.
- Recall. A fuse melts if the current is too large, which breaks the circuit.fuse
- Write the answer. fuse
Check. ✓
Method. Recall the light-sensitive component.
- Recall. A light-dependent resistor (LDR) has a lower resistance in brighter light.LDR
- Write the answer. LDR
Check. ✓
Method. Recall the temperature-sensitive component.
- Recall. A thermistor changes resistance with temperature.thermistor
- Write the answer. thermistor
Check. ✓
Method. Recall the convention for crossing wires.
- Recall. A dot shows that the wires are joined; without a dot they do not join.dot = joined
- Write the answer. yes
Check. ✓
Method. Think about how many closed paths there are in a series circuit.
- Count. A series circuit has one loop for the current to follow.one loop
- Write the answer. 1
Check. ✓
Method. Look at the diagram and list the parts that are not the cell.
- List. The series circuit has an ammeter and a lamp, besides the cell.ammeter, lamp
- Write the answer. ammeter, lamp
Check. The circuit has two parts other than the cell, so the answer is two names. ✓
Method. An open switch breaks the loop.
- Check. The open switch breaks the loop, so no current flows.no current
- Write the answer. no
Check. ✓
Method. Recall which meter is across a component.
- Look. The voltmeter measures pd across the lamp.pd
- Symbol. A voltmeter is drawn with the letter V.V
- Write the answer. V
Check. ✓
Method. Think about what a diagram gives you.
- Parts. It shows which parts are needed.parts
- Connections. It shows how they are connected.connections
- Write the answer. Accept any sentence that says it shows the parts and how they are connected, so that others can build the circuit.
Check. ✓
Method. Think about the symbol.
- Look. The longer line marks the positive terminal of the cell.positive
- Link. The cell has a positive and negative end, so it can drive the current in one direction.one direction
- Write the answer. Accept any sentence that says the long line is the positive terminal.
Check. ✓
Current and charge
Current is the rate of flow of charge. It is measured in amperes (A), where 1 A is 1 coulomb of charge passing a point each second. The charge carriers in a metal wire are electrons, which move from the negative to the positive terminal. By convention, the current is shown flowing from the positive terminal to the negative. The total charge is found using Q = I x t, where Q is in coulombs (C), I is in amperes and t is in seconds.
Key ideas
- Current I is the charge passing a point each second: I = Q / t.
- Charge Q = I x t, in coulombs (C).
- 1 A = 1 C/s.
- Ammeters measure the current in amperes.
Method. Use Q = I x t.
- Take the numbers. I = 0.5 A and t = 60 s.I = 0.5, t = 60
- Multiply. Q = 0.5 x 60 = 30 C.0.5 x 60 = 30
- Write the answer. 30 C
Check. 30 / 60 = 0.5 A. ✓
Method. Use I = Q / t.
- Take the numbers. Q = 120 C and t = 60 s.Q = 120, t = 60
- Divide. I = 120 / 60 = 2 A.120 / 60 = 2
- Write the answer. 2 A
Check. 2 x 60 = 120 C. ✓
Method. Convert the time to seconds, then use Q = I x t.
- Convert. 5 minutes = 300 s.300 s
- Multiply. Q = 3 x 300 = 900 C.3 x 300 = 900
- Write the answer. 900 C
Check. 900 / 300 = 3 A. ✓
Watch out
Watch out: convert minutes into seconds before you use Q = I t. 5 minutes is 300 s, not 5 s.
Exam tip
Exam tip. The unit of current is the ampere (A). Do not write “amp” for the coulomb, which is the unit of charge.
Exercise 5B Calculating charge and current
Method. Use Q = I x t.
- Multiply. Q = 2 x 10 = 20 C.2 x 10 = 20
- Write the answer. 20 C
Check. 20 / 10 = 2 A. ✓
Method. Use Q = I x t.
- Multiply. Q = 0.3 x 20 = 6 C.0.3 x 20 = 6
- Write the answer. 6 C
Check. 6 / 20 = 0.3 A. ✓
Method. Use I = Q / t.
- Divide. I = 60 / 30 = 2 A.60 / 30 = 2
- Write the answer. 2 A
Check. 2 x 30 = 60. ✓
Method. Convert the time to seconds first.
- Convert. 2 minutes = 120 s.120 s
- Multiply. Q = 4 x 120 = 480 C.4 x 120 = 480
- Write the answer. 480 C
Check. 480 / 120 = 4 A. ✓
Method. Use Q = I x t.
- Multiply. Q = 1.5 x 4 = 6 C.1.5 x 4 = 6
- Write the answer. 6 C
Check. 6 / 4 = 1.5 A. ✓
Method. Convert milliamperes to amperes first.
- Convert. 250 mA = 0.25 A.0.25 A
- Multiply. Q = 0.25 x 10 = 2.5 C.0.25 x 10 = 2.5
- Write the answer. 2.5 C
Check. 2.5 / 10 = 0.25 A. ✓
Method. Use I = Q / t.
- Divide. I = 45 / 15 = 3 A.45 / 15 = 3
- Write the answer. 3 A
Check. 3 x 15 = 45. ✓
Method. Use Q = I x t.
- Multiply. Q = 0.8 x 25 = 20 C.0.8 x 25 = 20
- Write the answer. 20 C
Check. 20 / 25 = 0.8 A. ✓
Method. Convert minutes to seconds, then multiply.
- Convert. 3 minutes = 180 s.180 s
- Multiply. Q = 0.5 x 180 = 90 C.0.5 x 180 = 90
- Write the answer. 90 C
Check. 90 / 180 = 0.5 A. ✓
Method. Use the definition of the ampere.
- Define. One ampere is one coulomb passing a point each second.1 C/s
- Write the answer. Accept any sentence that says 1 coulomb of charge passes a point each second.
Check. ✓
Method. Think about the flow of charge.
- Charge. Charge is not used up, it flows round the loop.not used up
- Link. In a series loop, the current is the same at every point.same current
- Write the answer. Accept any answer that says the charge is conserved, so the current is the same at all points in a series loop.
Check. ✓
Method. Convert minutes to seconds first.
- Convert. 5 minutes = 300 s.300 s
- Multiply. Q = 2 x 300 = 600 C.2 x 300 = 600
- Write the answer. 600 C
Check. 600 / 300 = 2 A. ✓
Method. Convert the hour to seconds, then divide.
- Convert. 1 hour = 3600 s.3600 s
- Divide. I = 3600 / 3600 = 1 A.3600 / 3600 = 1
- Write the answer. 1 A
Check. 1 x 3600 = 3600 C. ✓
Method. Think about the energy transfer, not the electrons.
- Speed. Electrons drift slowly, but the energy is passed along the wire quickly.energy quick
- Link. So the lamp lights almost at once.almost instant
- Write the answer. Accept any sentence that says the electrons pass on the energy to each other, so the effect is almost instant.
Check. ✓
Potential difference and energy transferred
The potential difference (pd) between two points is the energy transferred for each coulomb of charge that passes between them. It is measured in volts (V), where 1 V = 1 J/C. The equation is E = Q x V, so V = E / Q. A component with a larger pd transfers more energy to each coulomb of charge.
Key ideas
- Potential difference V = energy transferred per coulomb: V = E / Q.
- Energy transferred: E = Q x V, in joules.
- 1 V = 1 J/C.
- Voltmeters measure pd in volts, connected across the component.
Method. Use the definition: 1 V = 1 J/C.
- Take the numbers. V = 12 V, so each coulomb gains 12 J.V = 12
- Write. E per coulomb = 12 J.12 J/C
- Write the answer. 12 J
Check. 12 J/C is the same as 12 V. ✓
Method. Use E = Q x V.
- Take the numbers. Q = 30 C and V = 12 V.Q = 30, V = 12
- Multiply. E = 30 x 12 = 360 J.30 x 12 = 360
- Write the answer. 360 J
Check. 360 / 30 = 12 V. ✓
Method. Use V = E / Q.
- Take the numbers. E = 240 J and Q = 20 C.E = 240, Q = 20
- Divide. V = 240 / 20 = 12 V.240 / 20 = 12
- Write the answer. 12 V
Check. 12 x 20 = 240 J. ✓
Watch out
Watch out: pd is measured in volts, not amperes. Energy is in joules, and charge is in coulombs.
Exam tip
Exam tip. Use the units to check: E = QV gives J = C x V, which is right because 1 V = 1 J/C.
Exercise 5C Potential difference and energy
Method. Use E = Q x V.
- Multiply. E = 10 x 6 = 60 J.10 x 6 = 60
- Write the answer. 60 J
Check. 60 / 10 = 6 V. ✓
Method. Use V = E / Q.
- Divide. V = 50 / 5 = 10 V.50 / 5 = 10
- Write the answer. 10 V
Check. 10 x 5 = 50 J. ✓
Method. Use E = Q x V.
- Multiply. E = 4 x 9 = 36 J.4 x 9 = 36
- Write the answer. 36 J
Check. 36 / 9 = 4 C. ✓
Method. Use V = E / Q.
- Divide. V = 100 / 25 = 4 V.100 / 25 = 4
- Write the answer. 4 V
Check. 4 x 25 = 100. ✓
Method. One coulomb through 1.5 V gains 1.5 J.
- Multiply. E = 1 x 1.5 = 1.5 J.1 x 1.5 = 1.5
- Write the answer. 1.5 J
Check. 1.5 / 1 = 1.5 V. ✓
Method. Use E = Q x V.
- Multiply. E = 600 x 12 = 7200 J.600 x 12 = 7200
- Write the answer. 7200 J
Check. 7200 / 600 = 12 V. ✓
Method. Use E = Q x V.
- Multiply. E = 2 x 230 = 460 J.2 x 230 = 460
- Write the answer. 460 J
Check. 460 / 2 = 230 V. ✓
Method. Use V = E / Q.
- Divide. V = 180 / 12 = 15 V.180 / 12 = 15
- Write the answer. 15 V
Check. 15 x 12 = 180. ✓
Method. Use E = Q x V.
- Multiply. E = 2 x 3 = 6 J.2 x 3 = 6
- Write the answer. 6 J
Check. 6 / 3 = 2 C. ✓
Method. Use the definition of the volt.
- Define. 1 V means 1 J of energy is transferred per 1 C of charge.1 J per C
- Write the answer. Accept any sentence that says 1 J of energy is transferred for each coulomb of charge.
Check. ✓
Method. Check the definition of the volt.
- Volt. Volts measure energy per coulomb.energy per coulomb
- Current. The rate of flow of charge is the current, in amperes.amperes
- Write the answer. Accept any answer that says volts are energy per unit charge, not a rate of charge flow (which is current in amperes).
Check. ✓
Method. Use V = E / Q.
- Divide. V = 360 / 12 = 30 V.360 / 12 = 30
- Write the answer. 30 V
Check. 30 x 12 = 360. ✓
Method. Use E = Q x V with the charge and pd.
- Multiply. E = 300 x 230 = 69000 J.300 x 230 = 69000
- Write the answer. 69000 J
Check. 69000 / 300 = 230 V. ✓
Series and parallel circuits: current and pd
In a series circuit there is one path for the current. The current is the same at every point. The pd of the cell is shared between the components, so the pd across all the components adds up to the cell pd.
In a parallel circuit there are two or more branches. The current splits at each junction, and the total current is the sum of the branch currents. The pd is the same across each branch, equal to the pd of the cell.
| Meter | What it measures | Unit | How it is connected |
|---|---|---|---|
| ammeter | current | amperes (A) | in series, in the loop, so all the current flows through it |
| voltmeter | potential difference (pd) | volts (V) | in parallel, across the component being measured |
| Rule | Series | Parallel |
|---|---|---|
| current | the same at every point | splits; the total is the sum of the branch currents |
| potential difference | shared between the components; the total is the sum | the same across each branch as the source |
Key ideas
- Series: current is the same everywhere. The pd of the cell is shared.
- Parallel: the current splits at junctions. The pd across each branch is the same as the cell.
- An ammeter is connected in series. A voltmeter is connected in parallel.
- The total pd of a series loop is the sum of the pd across each component.
Method. In series, the pd is shared, so the parts add to the total.
- Look. The pd is shared, so the two lamp pds add up to the cell pd of 9 V.shared
- Subtract. The other lamp has 9 - 3.5 = 5.5 V across it.9 - 3.5 = 5.5
- Write the answer. 5.5 V
Check. 3.5 + 5.5 = 9 V. ✓
Method. The current splits, so the branch currents add up to the total.
- Look. The branch currents add up to the total current from the cell.add to total
- Subtract. The other branch has 1.5 - 0.5 = 1.0 A.1.5 - 0.5 = 1.0
- Write the answer. 1.0 A
Check. 0.5 + 1.0 = 1.5 A. ✓
Method. The pd is shared equally between identical lamps in series.
- Look. The lamps are identical, so they share the pd equally.equal share
- Share. The 12 V is shared between two identical lamps, so 12 / 2 = 6 V each.12 / 2 = 6
- Write the answer. 6 V
Check. 6 + 6 = 12 V. ✓
Watch out
Watch out: a voltmeter never goes in series, and an ammeter never goes in parallel. Connecting them the wrong way can break the circuit or give a wrong reading.
Exam tip
Exam tip. In parallel, say “the current splits” and “the pd is the same across each branch”. In series, say “the current is the same” and “the pd is shared”.
Exercise 5D Series and parallel current and pd
Method. The current is the same at all points in series.
- Recall. In a series loop the current is the same everywhere.0.4 A
- Write the answer. 0.4 A
Check. ✓
Method. Add the branch currents.
- Add. 0.2 + 0.4 = 0.6 A.0.2 + 0.4 = 0.6
- Write the answer. 0.6 A
Check. 0.6 - 0.4 = 0.2 A. ✓
Method. The pd across each branch equals the pd of the cell.
- Recall. In parallel, each branch has the full pd of the cell.6 V
- Write the answer. 6 V
Check. ✓
Method. Share the pd equally.
- Divide. 9 / 3 = 3 V each.9 / 3 = 3
- Write the answer. 3 V
Check. 3 + 3 + 3 = 9 V. ✓
Method. Recall how ammeters are connected.
- Recall. An ammeter is always connected in series.series
- Write the answer. series
Check. ✓
Method. Recall how voltmeters are connected.
- Recall. A voltmeter is connected in parallel, across the component.parallel
- Write the answer. parallel
Check. ✓
Method. Subtract the known pd from the cell pd.
- Subtract. 12 - 4 = 8 V.12 - 4 = 8
- Write the answer. 8 V
Check. 4 + 8 = 12 V. ✓
Method. Add the three currents.
- Add. 0.1 + 0.25 + 0.15 = 0.5 A.0.1 + 0.25 + 0.15 = 0.5
- Write the answer. 0.5 A
Check. 0.5 - 0.25 = 0.25 A. ✓
Method. Parallel branches each have the cell pd.
- Recall. The pd across each branch equals the cell pd, 12 V.12 V
- Write the answer. 12 V
Check. ✓
Method. The current is the same in series.
- Recall. The current is the same at every point, so 0.5 A.0.5 A
- Write the answer. 0.5 A
Check. ✓
Method. Think about the branches.
- Branch. There is one less branch carrying current.one less branch
- Total. So the total current from the cell falls.total falls
- Write the answer. Accept any sentence that says the total current falls, because there is one less branch to carry current.
Check. ✓
Method. Recall the rule for parallel.
- Check. In parallel, each branch has the full cell pd, which is not shared.not shared
- Write the answer. no
Check. Sharing of pd happens in series. ✓
Method. Think about the pd across each branch.
- pd. Each branch has the full pd of the cell.full pd
- Effect. Removing the other branch does not change this pd.unchanged
- Write the answer. Accept any sentence that says each lamp has the full cell pd across it, so it is unaffected by removing the other.
Check. ✓
Method. Subtract the two known pds from the total.
- Add. The known pds add to 4 + 5 = 9 V.9 V
- Subtract. The third lamp has 12 - 9 = 3 V.12 - 9 = 3
- Write the answer. 3 V
Check. 4 + 5 + 3 = 12 V. ✓
Method. The branch currents add to the total.
- Subtract. 1.2 - 0.7 = 0.5 A.1.2 - 0.7 = 0.5
- Write the answer. 0.5 A
Check. 0.7 + 0.5 = 1.2 A. ✓
Chapter summary
| Skill | Method |
|---|---|
| Draw circuits | Use standard symbols, with a ruler; a dot shows a junction |
| Current and charge | I = Q / t; Q = I x t; 1 A = 1 C/s |
| Potential difference | V = E / Q; E = Q x V; 1 V = 1 J/C |
| Series rules | Current is the same; pd is shared and adds to the cell pd |
| Parallel rules | Current splits and adds to the total; pd is the same across each branch |
Can you…?
- I can draw and recognise the common circuit symbols
- I can use Q = It and I = Q/t
- I can use E = QV and V = E/Q
- I can state the current and pd rules in series and parallel
- I can connect an ammeter in series and a voltmeter in parallel
Key terms
- current
- the flow of charge; measured in amperes (A)
- charge
- a quantity measured in coulombs (C)
- potential difference (pd)
- energy transferred per unit charge, measured in volts (V)
- series circuit
- a circuit with one path for the current
- parallel circuit
- a circuit with two or more branches, each with the full pd of the cell
- ammeter
- a meter that measures current, connected in series
- voltmeter
- a meter that measures pd, connected in parallel
Exercise 5Z Check-up
Method. Use Q = I x t.
- Multiply. Q = 3 x 10 = 30 C.3 x 10 = 30
- Write the answer. 30 C
Check. ✓
Method. Use I = Q / t.
- Divide. I = 50 / 10 = 5 A.50 / 10 = 5
- Write the answer. 5 A
Check. ✓
Method. Use E = Q x V.
- Multiply. E = 8 x 5 = 40 J.8 x 5 = 40
- Write the answer. 40 J
Check. 40 / 8 = 5 V. ✓
Method. Use V = E / Q.
- Divide. V = 60 / 4 = 15 V.60 / 4 = 15
- Write the answer. 15 V
Check. 15 x 4 = 60. ✓
Method. Share the pd equally.
- Divide. 6 / 2 = 3 V each.6 / 2 = 3
- Write the answer. 3 V
Check. ✓
Method. Recall the connection.
- Recall. An ammeter is connected in series.ammeter in series
- Write the answer. ammeter
Check. ✓
Method. Add the branch currents.
- Add. 0.3 + 0.9 = 1.2 A.0.3 + 0.9 = 1.2
- Write the answer. 1.2 A
Check. ✓
Method. Each coulomb gains V joules.
- Recall. E per coulomb = V = 1.5 J.1.5 J
- Write the answer. 1.5 J
Check. ✓
Method. Compare the definitions.
- Compare. Current is the flow of charge, in amperes; pd is energy per coulomb, in volts.different
- Write the answer. no
Check. ✓
Method. Find the charge, then the energy.
- Charge. Q = 2 x 120 = 240 C.240 C
- Energy. E = 240 x 6 = 1440 J.240 x 6 = 1440
- Write the answer. 1440 J
Check. 1440 / 240 = 6 V. ✓
Total 25 marks. Allow about 25 minutes.
Exercise 5X Exam-style questions
Method. Recall the meter and its connection.
- Name. The meter is an ammeter.ammeter
- Connection. It is connected in series.in series
- Write the answer. ammeter, in series
Check. ✓
Method. Use Q = I x t.
- Multiply. Q = 0.25 x 40 = 10 C.0.25 x 40 = 10
- Write the answer. 10 C
Check. ✓
Method. Use E = Q x V.
- Multiply. E = 5 x 12 = 60 J.5 x 12 = 60
- Write the answer. 60 J
Check. ✓
Method. Share the pd equally between identical lamps in series.
- Divide. 12 / 3 = 4 V each.12 / 3 = 4
- Write the answer. 4 V
Check. ✓
Method. Subtract the known branch from the total.
- Subtract. 2.4 - 1.1 = 1.3 A.2.4 - 1.1 = 1.3
- Write the answer. 1.3 A
Check. 1.1 + 1.3 = 2.4. ✓
Method. Use V = E / Q.
- Divide. V = 1200 / 40 = 30 V.1200 / 40 = 30
- Write the answer. 30 V
Check. 40 x 30 = 1200. ✓
Method. Find the charge, then the energy.
- Time. Time = 2 minutes = 120 s.120 s
- Charge. Q = 3 x 120 = 360 C.360 C
- Energy. E = 360 x 6 = 2160 J.360 x 6 = 2160
- Write the answer. 2160 J
Check. 2160 / 360 = 6 V. ✓
Method. The pds in series add to the cell pd.
- Add. The two known pds add to 2.5 + 4 = 6.5 V.6.5 V
- Subtract. The third component has 9 - 6.5 = 2.5 V.9 - 6.5 = 2.5
- Write the answer. 2.5 V
Check. 2.5 + 4 + 2.5 = 9 V. ✓
Method. Place each meter correctly.
- Ammeter. Put the ammeter in series in the loop, so all the current flows through it.series
- Voltmeter. Put the voltmeter in parallel across the lamp, to read its pd.parallel
- Check. Make sure the switch is closed so the current can flow.closed switch
- Write the answer. Accept any answer that puts the ammeter in series and the voltmeter in parallel across the lamp.
Check. ✓
Method. Separate charge from energy.
- Current. A current of 2 A means 2 coulombs of charge pass each second.2 C per second
- Energy. The energy transferred per second depends on the pd as well: E = Q x V.E = Q x V
- Conclusion. So 2 A alone does not give the energy in joules.not just current
- Write the answer. Accept any answer that says the current is charge per second, and the energy depends on the pd as well.
Check. ✓