2027 G3 SEC Science Physics | Learner’s Guide Vol 0159. A simple series circuit offers three reliable starting ideas: there is one path for charge flow, the current is the same at every point in that path, and the potential differences across components add to the source potential difference in the idealised model. From those ideas, you can solve many unfamiliar circuit diagrams without memorising a separate formula for every picture.
This chapter uses original, low-voltage theoretical examples and fictional numerical readings. It is not a set of instructions for unsupervised experiments or for opening household electrical wiring. Real practical work must follow the school’s equipment and safety procedures under qualified supervision. Learn the relationship on paper first; use apparatus only in an approved environment.
1. Check which SEC Science combination contains Physics
The SEAB 2027 G3 directory, updated on 26 October 2026, lists Science (Physics, Chemistry) as K326, Science (Physics, Biology) as K327, and Science (Chemistry, Biology) as K328. The Physics content in this guide directly supports K326 and K327, not the Chemistry–Biology pairing K328.
The official 2027 K326/K327/K328 Science syllabus covers electric charge and current, potential difference, resistance, circuit diagrams, series and parallel circuits, and applying R = V/I. It explicitly requires candidates to state and apply same current in series, addition of potential differences across a series circuit and effective resistance calculations.
The shared Science scheme has Paper 1 MCQ (1 hour, 40 marks, 20 per cent), two relevant subject theory papers (each 1 hour 15 minutes, 65 marks, 32.5 per cent) and Paper 5 practical (1 hour 30 minutes, 30 marks, 15 per cent). Which subject theory papers a candidate takes depends on K326, K327 or K328. This is a subject chapter, not a promise of a particular live question.
2. Start with a complete circuit, not an isolated resistor
A circuit needs an energy source and a complete conducting route for current in its intended model. A schematic diagram shows connections; its lines are not usually meant to reproduce the physical position or size of every wire. Before calculating, trace the path from the source through every component and back to the source.
If there is only one continuous route through all the components, the components are connected in series. If the route divides into branches that later rejoin, at least part of the circuit is parallel. A complicated drawing may contain both series and parallel groups, so identify the connections rather than relying on whether symbols happen to be drawn side by side.
In a school diagram, ideal wires are normally treated as having negligible resistance unless the question states otherwise. Real wires, internal source resistance and measuring instruments can matter in more advanced models, but add these complexities only when the problem provides information that makes them relevant.
3. Current is not used up by a lamp
Conventional current is the rate of flow of electric charge: I = Q/t, where current is measured in amperes, charge in coulombs and time in seconds. In an ordinary steady series circuit, the same charge flow passes successive components, so the current is the same at all points in the single path.
Suppose the current entering a resistor is 0.40 A. Under the steady series assumption, the current leaving that resistor is also 0.40 A, and the current entering the next series lamp is 0.40 A. The lamp may transfer electrical energy to light and heat, but it does not consume a fraction of the charge passing through it.
“Current gets smaller after every lamp” is an attractive but incorrect explanation of why a series circuit may become dimmer when additional resistive lamps are included. In an ideal fixed-voltage model, adding resistance reduces the same current everywhere in the circuit. The decrease is a whole-circuit change, not current disappearing step by step.
4. Potential difference is energy transfer per unit charge
Potential difference, measured in volts, describes energy transferred per unit charge as charge passes through a component. The formula V = W/Q expresses this idea: a potential difference of 3 V corresponds to 3 joules of energy transferred per coulomb in the relevant idealised process.
In a simple circuit with a 9 V source and two resistors in series, the source potential difference equals the sum of the component potential differences: Vtotal = V1 + V2. If one resistor has 4 V across it, the other has 5 V in that model. Do not divide the total voltage equally unless the components and assumptions justify an equal division.
The source provides the energy, the components transfer energy into other forms, and charge continues around the path. It is useful to keep these roles separate. Calling voltage “the amount of current” mixes two different physical quantities and makes later calculations unreliable.
5. Resistance connects current and potential difference
For an ohmic resistor under the assumptions of the model, R = V/I, or V = IR. Resistance is measured in ohms. If 2 V is across a resistor and the current through it is 0.5 A, its resistance is 2/0.5 = 4 Ω.
The equation does not mean every electrical component has a constant resistance at every operating condition. A filament lamp can heat up and change its resistance; non-ohmic components require attention to their actual characteristics. But for questions supplying fixed resistances, Ohm’s law-style calculations are a powerful route.
Look for units every time you substitute. Volts divided by amperes gives ohms. A resistance expressed as 0.004 kΩ must be converted carefully to 4 Ω if other values use base units. Do not let the calculator hide a factor of 1000.
6. Resistances add in a series path
For fixed resistors connected in series, the effective resistance is Rtotal = R1 + R2 + …. This follows from the same current through each component and the sum of the component voltages. If Vtotal = IR1 + IR2, then Vtotal = I(R1 + R2), so the source sees the sum.
Take resistors of 4 Ω and 8 Ω in series across an ideal 12 V supply. The total resistance is 12 Ω, so the current is I = 12/12 = 1.0 A. Across the 4 Ω resistor the voltage is 4 V, and across the 8 Ω resistor it is 8 V. The voltage drops add to 12 V.
Check the direction of change. Adding another positive series resistance to the same fixed-voltage supply increases total resistance and reduces current. That statement depends on the fixed source and simple resistance model. It should not be applied carelessly to an active electronic circuit with changing sources or components.
7. Voltage sharing depends on resistance
In a series circuit, the same current flows through each ideal fixed resistor. Therefore V1 = IR1 and V2 = IR2, so V1/V2 = R1/R2. Larger resistance gets a larger share of the total potential difference when the current is common.
If the resistances are 3 Ω and 9 Ω across 12 V, the total is 12 Ω and the current is 1 A. Their voltage drops are 3 V and 9 V respectively. The 9 Ω resistor does not receive more charge per second; it has a larger potential difference because its resistance is larger at the same current.
A frequent exam error is to confuse shared current with shared voltage. Current stays common in series; voltage is divided according to component properties. In a parallel section of a circuit, the relationship is different: potential difference across branches is common. Always identify topology before borrowing a rule.
8. Learn the voltage-divider relationship from the model
For two fixed series resistors, V1 = Vsource × R1/(R1 + R2), and V2 = Vsource × R2/(R1 + R2). This is not a separate physical law. It comes from I = Vsource/(R1 + R2) and multiplying that current by each resistance.
For a source of 18 V and resistors 2 Ω and 4 Ω, total resistance is 6 Ω, current is 3 A, and voltage drops are 6 V and 12 V. The divider expression gives the same numbers: 18 × 2/6 = 6 and 18 × 4/6 = 12. Two routes to the same answer strengthen confidence.
Do not use a voltage-divider formula when the components are not in the same series path or when another branch loads the midpoint in a way not included in the model. In a compound circuit, reduce the actual connections first or solve the whole network systematically.
9. Why component order usually does not change simple series results
Imagine swapping a 4 Ω and an 8 Ω resistor while keeping both connected in the same ideal series loop and the source fixed at 12 V. The total resistance remains 12 Ω, and the current remains 1 A. The potential difference across each resistor remains determined by its own resistance.
This means the 4 Ω resistor still has a 4 V drop whether it is drawn first or second. The 8 Ω resistor still has 8 V. The physical connection order of ideal fixed resistors in a single series loop does not change their effective sum.
However, moving a component from a series route into a parallel branch would change the topology, and inserting a connection to another part of the network can change the result. Distinguish rearranging the drawing from changing how the terminals are actually connected.
10. Read ammeters and voltmeters by what they measure
An ammeter measures current and is connected in series with the part of the circuit whose current is to be measured. In a simple steady series circuit, moving an ideal ammeter to another point along the same route gives the same current reading.
A voltmeter measures potential difference across two points and is connected in parallel across the component or section whose voltage is wanted. It does not need to be inserted into the single current path like an ammeter. In schematic questions, draw its two connections to the correct component terminals.
An incorrect meter placement changes what is measured and may alter a real circuit. For written exam work, identify what the instrument must compare: an ammeter tracks charge flow through a branch, while a voltmeter compares electrical potential between two nodes. Practical use belongs in teacher-supervised laboratory work.
11. The difference between an open switch and a failed resistor
When a switch opens the only route in a simple series circuit, no steady current flows through the interrupted path. A lamp goes out because the circuit is incomplete. Do not claim that the current becomes smaller only after the switch: in this model, current in the whole open loop is zero.
A failed component can interrupt the path or develop another fault, depending on the problem. Never infer the exact failure mode from “the lamp stopped glowing” alone. A lamp could be faulty, a connection loose or a source depleted. The right conclusion is tied to the evidence supplied.
A useful scientific explanation distinguishes direct observation from a possible cause. That distinction is part of the SEC Science assessment habit. A correct circuit equation cannot by itself determine which real-world fault happened when several explanations fit the symptom.
12. Parallel branches obey a different pair of rules
In a parallel circuit, the current from the source divides among branches and the potential difference across each branch is the same under the ordinary idealised assumptions. In series, by contrast, current is common and potential differences add. Students often exchange these rules because both involve addition somewhere.
Suppose two resistors are connected in parallel across 6 V. Each has 6 V across it, while their branch currents depend on their resistances. If those same resistors are reconnected in series across 6 V, the source current is common and the resistors share the 6 V according to resistance.
This chapter focuses on series so it does not reproduce Volume 0120 on parallel-circuit nodes. Use that earlier owner for branch splitting, then return here to compare the two topologies.
13. A mixed circuit: reduce the obvious group first
Consider a 12 V ideal supply feeding a 2 Ω resistor in series with a parallel group of two 6 Ω resistors. The parallel group has effective resistance 3 Ω, since 1/R = 1/6 + 1/6 = 1/3. The entire network then has 2 + 3 = 5 Ω.
The source current is 12/5 = 2.4 A. That current flows through the series 2 Ω resistor, giving a voltage drop of 4.8 V. The remaining 12 − 4.8 = 7.2 V is across the parallel group. Since each branch has 6 Ω, each branch current is 7.2/6 = 1.2 A.
Check conservation of current at the branch junction: 1.2 + 1.2 = 2.4 A. Check the potential differences around the series arrangement: 4.8 + 7.2 = 12 V. These independent checks help detect where a mixed-circuit calculation went wrong.
14. Work from a given current rather than a given voltage
Some questions supply the series current and the resistances, asking for the source voltage. Suppose a series loop contains a 5 Ω resistor and a 7 Ω resistor, with current 0.25 A. The total resistance is 12 Ω, so the required ideal source potential difference is V = IR = 0.25 × 12 = 3.0 V.
The individual drops are 0.25 × 5 = 1.25 V and 0.25 × 7 = 1.75 V. Their sum is 3.0 V. Show the route rather than treating 3.0 V as a guessed answer from a diagram.
Reversing a formula is part of reasoning. If a known current is the most reliable starting point, calculate each drop and add them. If the total source voltage is given, sum resistance first. Either route makes sense because the underlying circuit relationships are consistent.
15. A missing resistance can be found from a measured drop
Imagine two fixed series resistors across 15 V. One resistor has 5 Ω and a measured drop of 5 V. The series current is 5/5 = 1 A. The remaining resistor has 10 V across it, so its resistance is 10/1 = 10 Ω.
An independent check adds the resistances: 5 + 10 = 15 Ω. At 1 A that requires 15 V, matching the source. Both voltage and resistance checks confirm the result. Do not subtract 5 Ω from 15 V; different units cannot be directly subtracted.
Good numerical work makes the unit relationship explicit. A voltage difference can be subtracted from a voltage, and resistance from resistance, but not a resistance from a voltage. Dimensional consistency is an early warning system against incorrect shortcuts.
16. Current, charge and time form another useful chain
The equation I = Q/t gives charge transferred in a given time. If a steady series current is 0.40 A for 30 seconds, Q = It = 0.40 × 30 = 12 C. Under the steady model, that amount of charge passes every point in the series loop during the same interval.
Do not multiply by the number of lamps and claim more charge flows through the circuit as a whole. Every charge that passes one series component also passes the next during steady operation. Counting it once per component would double-count the same charge flow.
Potential difference connects charge to energy transfer: W = VQ. If a resistor has 3 V across it while 12 C passes, the energy transferred at that resistor is 36 J. This is different from the amount of charge. Keep coulombs, joules, amperes and volts labelled.
17. Power is a rate of energy transfer
For a component in the suitable DC model, electrical power is P = VI. If 3 V appears across a resistor with current 0.4 A, its power is 1.2 W. If those conditions hold for 30 seconds, energy transferred is E = Pt = 36 J, matching the independent W = VQ route when Q = 12 C.
In a series circuit, each fixed resistor may dissipate a different amount of power even though current is the same, because the voltage drops differ. Using P = I²R, the resistor with larger resistance dissipates more power at the common current under the stated fixed-resistance model.
Real component ratings and heating limits matter; do not attempt to reproduce these power examples with unapproved equipment. The purpose is quantitative reasoning. More detailed household electrical safety and cost questions belong to the relevant practical-electricity syllabus section.
18. The lamp-brightness trap
A student says, “The first lamp will be brighter because it receives the current first.” In a simple series circuit that explanation is wrong: the same current flows through both lamps. Their brightness depends on power transfer and actual lamp characteristics, not physical order along the wire.
If two identical lamps operate under the same conditions in series, they are expected to have equal power and comparable brightness in the simple model. If the lamps differ, the power and brightness may differ. For non-ohmic filament lamps, the operating resistance can depend on temperature, so avoid pretending a cold resistance alone always settles the outcome.
State the assumption that supports the prediction. A school physics answer should not use “all series lamps are equally bright” as a universal law. It should describe the components and the idealised conditions that make the conclusion valid.
19. Add one resistor: predict before calculating
Take an ideal 12 V source and one 6 Ω resistor. The current is 2 A. Add a second 6 Ω resistor in series while keeping the source at 12 V. The total becomes 12 Ω and current drops to 1 A everywhere in the new circuit. Each resistor now has 6 V across it.
The voltage drop across the original resistor changed from 12 V to 6 V because the complete circuit changed. The resistor did not “use up half the current”; the whole loop carries a smaller current. The source voltage is shared between two equal resistors.
This before-and-after reasoning is useful for unfamiliar diagram questions. First identify what was held constant, then what component or topology changed, and finally which dependent quantities respond. A memorised statement without these conditions can lead to the wrong comparison.
20. An error clinic: the five most common wrong paths
Error one: dividing source current equally among series components. Correct it: series current is the same throughout. Error two: assigning source voltage across every series component. Correct it: series drops add to the source total, with their shares determined by component behaviour.
Error three: treating two branches drawn side by side as series solely because the diagram looks narrow. Trace the connections instead. Error four: using voltage as a unit of charge or resistance. Restore the physical meanings and units before calculating.
Error five: declaring a component fault from one symptom without considering alternatives. A dim or extinguished lamp is an observation; the cause requires additional evidence. The strongest answer distinguishes what the model proves from what a real test would still need to establish.
21. Twenty-eight original series-circuit problems
The questions move from basic topology through whole-circuit calculations and simple mixed networks. They use idealised sources and fixed resistors unless stated otherwise; model answers are for practice, not an official SEAB mark scheme.
Worked question 1
Task. Two resistors, 4 Ω and 8 Ω, are in series. Total resistance? Worked answer. 12 Ω.
Reason. Series resistances add. Rebuild the circuit from the actual connections, check whether current or voltage is common, and carry each unit through the calculation. A correct number with the wrong circuit model is not a reliable answer.
Worked question 2
Task. The pair is connected across 12 V. Current? Worked answer. 1 A.
Reason. 12 V divided by 12 Ω. Rebuild the circuit from the actual connections, check whether current or voltage is common, and carry each unit through the calculation. A correct number with the wrong circuit model is not a reliable answer.
Worked question 3
Task. Voltage across the 4 Ω resistor in that circuit? Worked answer. 4 V.
Reason. I × R = 1 × 4. Rebuild the circuit from the actual connections, check whether current or voltage is common, and carry each unit through the calculation. A correct number with the wrong circuit model is not a reliable answer.
Worked question 4
Task. Voltage across the 8 Ω resistor? Worked answer. 8 V.
Reason. I × R = 1 × 8. Rebuild the circuit from the actual connections, check whether current or voltage is common, and carry each unit through the calculation. A correct number with the wrong circuit model is not a reliable answer.
Worked question 5
Task. Add the two resistor voltages. What total? Worked answer. 12 V.
Reason. Series drops add to source. Rebuild the circuit from the actual connections, check whether current or voltage is common, and carry each unit through the calculation. A correct number with the wrong circuit model is not a reliable answer.
Worked question 6
Task. Does current fall between the 4 Ω and 8 Ω resistors? Worked answer. No.
Reason. A steady series path has the same current. Rebuild the circuit from the actual connections, check whether current or voltage is common, and carry each unit through the calculation. A correct number with the wrong circuit model is not a reliable answer.
Worked question 7
Task. Resistors 3 Ω and 9 Ω, source 12 V: current? Worked answer. 1 A.
Reason. Their total resistance is 12 Ω. Rebuild the circuit from the actual connections, check whether current or voltage is common, and carry each unit through the calculation. A correct number with the wrong circuit model is not a reliable answer.
Worked question 8
Task. For that circuit, voltage across 9 Ω? Worked answer. 9 V.
Reason. The same 1 A passes the resistor. Rebuild the circuit from the actual connections, check whether current or voltage is common, and carry each unit through the calculation. A correct number with the wrong circuit model is not a reliable answer.
Worked question 9
Task. Resistors 2 Ω and 4 Ω, source 18 V: current? Worked answer. 3 A.
Reason. Total resistance 6 Ω. Rebuild the circuit from the actual connections, check whether current or voltage is common, and carry each unit through the calculation. A correct number with the wrong circuit model is not a reliable answer.
Worked question 10
Task. For that circuit, voltage across 2 Ω? Worked answer. 6 V.
Reason. 3 A × 2 Ω. Rebuild the circuit from the actual connections, check whether current or voltage is common, and carry each unit through the calculation. A correct number with the wrong circuit model is not a reliable answer.
Worked question 11
Task. For that circuit, voltage across 4 Ω? Worked answer. 12 V.
Reason. 3 A × 4 Ω. Rebuild the circuit from the actual connections, check whether current or voltage is common, and carry each unit through the calculation. A correct number with the wrong circuit model is not a reliable answer.
Worked question 12
Task. Current 0.25 A through 5 Ω: potential difference? Worked answer. 1.25 V.
Reason. V = IR. Rebuild the circuit from the actual connections, check whether current or voltage is common, and carry each unit through the calculation. A correct number with the wrong circuit model is not a reliable answer.
Worked question 13
Task. Current 0.25 A through 7 Ω: potential difference? Worked answer. 1.75 V.
Reason. V = IR. Rebuild the circuit from the actual connections, check whether current or voltage is common, and carry each unit through the calculation. A correct number with the wrong circuit model is not a reliable answer.
Worked question 14
Task. Those 5 Ω and 7 Ω resistors in series: supply voltage? Worked answer. 3 V.
Reason. Add the drops. Rebuild the circuit from the actual connections, check whether current or voltage is common, and carry each unit through the calculation. A correct number with the wrong circuit model is not a reliable answer.
Worked question 15
Task. Two series resistors share 15 V; 5 Ω has 5 V across it. Current? Worked answer. 1 A.
Reason. I = 5/5. Rebuild the circuit from the actual connections, check whether current or voltage is common, and carry each unit through the calculation. A correct number with the wrong circuit model is not a reliable answer.
Worked question 16
Task. In that circuit, voltage across the other resistor? Worked answer. 10 V.
Reason. 15 − 5. Rebuild the circuit from the actual connections, check whether current or voltage is common, and carry each unit through the calculation. A correct number with the wrong circuit model is not a reliable answer.
Worked question 17
Task. Resistance of that other resistor? Worked answer. 10 Ω.
Reason. R = 10/1. Rebuild the circuit from the actual connections, check whether current or voltage is common, and carry each unit through the calculation. A correct number with the wrong circuit model is not a reliable answer.
Worked question 18
Task. A 0.4 A current flows for 30 s. Charge transferred? Worked answer. 12 C.
Reason. Q = It. Rebuild the circuit from the actual connections, check whether current or voltage is common, and carry each unit through the calculation. A correct number with the wrong circuit model is not a reliable answer.
Worked question 19
Task. The same current through a series lamp for 30 s: charge passing? Worked answer. 12 C.
Reason. It is not added again for each component. Rebuild the circuit from the actual connections, check whether current or voltage is common, and carry each unit through the calculation. A correct number with the wrong circuit model is not a reliable answer.
Worked question 20
Task. A resistor has 3 V across it and 0.4 A through it: power? Worked answer. 1.2 W.
Reason. P = VI. Rebuild the circuit from the actual connections, check whether current or voltage is common, and carry each unit through the calculation. A correct number with the wrong circuit model is not a reliable answer.
Worked question 21
Task. That power for 30 seconds: energy? Worked answer. 36 J.
Reason. E = Pt. Rebuild the circuit from the actual connections, check whether current or voltage is common, and carry each unit through the calculation. A correct number with the wrong circuit model is not a reliable answer.
Worked question 22
Task. A 6 Ω resistor alone across 12 V: current? Worked answer. 2 A.
Reason. I = 12/6. Rebuild the circuit from the actual connections, check whether current or voltage is common, and carry each unit through the calculation. A correct number with the wrong circuit model is not a reliable answer.
Worked question 23
Task. Add another 6 Ω resistor in series across the same 12 V: current? Worked answer. 1 A.
Reason. Total is now 12 Ω. Rebuild the circuit from the actual connections, check whether current or voltage is common, and carry each unit through the calculation. A correct number with the wrong circuit model is not a reliable answer.
Worked question 24
Task. In the new circuit, voltage across each identical 6 Ω? Worked answer. 6 V.
Reason. Same 1 A through each. Rebuild the circuit from the actual connections, check whether current or voltage is common, and carry each unit through the calculation. A correct number with the wrong circuit model is not a reliable answer.
Worked question 25
Task. An open switch interrupts the only route. Steady current? Worked answer. 0 A.
Reason. No complete path. Rebuild the circuit from the actual connections, check whether current or voltage is common, and carry each unit through the calculation. A correct number with the wrong circuit model is not a reliable answer.
Worked question 26
Task. A 2 Ω resistor is in series with two parallel 6 Ω resistors. Total resistance? Worked answer. 5 Ω.
Reason. Parallel pair is 3 Ω; add 2 Ω. Rebuild the circuit from the actual connections, check whether current or voltage is common, and carry each unit through the calculation. A correct number with the wrong circuit model is not a reliable answer.
Worked question 27
Task. That network is across 12 V. Source current? Worked answer. 2.4 A.
Reason. 12/5. Rebuild the circuit from the actual connections, check whether current or voltage is common, and carry each unit through the calculation. A correct number with the wrong circuit model is not a reliable answer.
Worked question 28
Task. Voltage across the parallel group in that network? Worked answer. 7.2 V.
Reason. The 2 Ω series resistor drops 4.8 V. Rebuild the circuit from the actual connections, check whether current or voltage is common, and carry each unit through the calculation. A correct number with the wrong circuit model is not a reliable answer.
22. A safe, useful seven-day practice progression
Day one: identify a single path and a branching path from original schematic drawings. Day two: trace steady series current in five diagrams and explain why it is the same throughout. Day three: calculate series effective resistance and total current from a fixed source.
Day four: calculate voltage drops and check that they sum to the source. Day five: reconstruct V = IR from different known quantities, including an unknown resistor. Day six: solve mixed network examples on paper, reducing branches before combining the series section. Day seven: review errors and complete a timed mixed set.
Do not build household circuits to practise the calculations. All practical investigation should follow school supervision and approved low-voltage apparatus. Textbook calculations, diagrams and measured data supplied by a teacher are sufficient to develop the reasoning skills explained here.
23. Links to the right learning owners
Visit the Science Hub for electricity and scientific method, and the Mathematics Hub for equation substitution and units. For long-term examination preparation, use the G3 SEC Learner Guide, the Examination Craft route and the PSLE Learning Guide for earlier foundations.
Read Volume 0120 on parallel-circuit nodes for the complementary rule, and Volume 0149 on numerical accuracy to reduce arithmetic loss. The Secondary English Hub helps students explain evidence and conclusions in clear prose.
The official 2027 K326/K327/K328 Science syllabus remains the governing reference for paper details and content. This chapter is an original reasoning manual about a Physics strand, not a replacement for actual syllabus coverage.
24. The explanation you should be able to give unaided
With two fixed resistors in one ideal series path, current is common because charge has no branch at which to divide. Potential differences add because the source’s energy transfer per unit charge is accounted for by the components in that loop. Resistances add because Vtotal = I(R1 + R2) when both share the current.
That explanation is more valuable than a memorised trio of formulae. It lets a learner reconstruct the circuit method when a diagram is rotated, the values change or the unknown becomes a voltage rather than a current.
Accurate examination performance begins with the circuit path and ends with a defensible answer. Name the topology, choose the physical relationship, substitute carefully, check units and explain the conclusion without claiming evidence that was never supplied.
