2027 G3 SEC Science Physics | Learner’s Guide Vol 0163. In a series circuit, the same current passes through each component. In a parallel circuit, branches share the same potential difference while current divides among them. That one contrast explains most of the mistakes students make when reading a parallel-circuit diagram.
This guide builds from the meaning of a branch and a junction to combined resistance, current conservation, electrical power and careful circuit explanations. All numerical circuits are fictional, idealised low-voltage models for written learning. Do not attempt electrical wiring or mains experiments; practical work belongs in a suitably equipped, supervised school laboratory.
1. Check the correct 2027 G3 Science combination
The current SEAB 2027 G3 directory lists Science (Physics, Chemistry) as K326, Science (Physics, Biology) as K327, and Science (Chemistry, Biology) as K328. This chapter covers the Physics portion of K326 and K327; it is not a Physics-paper guide for K328, which does not include Physics.
The official 2027 G3 Science syllabus identifies D.C. Circuits as Physics topic 13. It explicitly expects candidates to explain that the source current equals the sum of currents in parallel branches, that the potential difference across separate parallel branches is the same, and to use effective-resistance formulae for series and parallel resistors.
For the paired Science subjects, Paper 1 is a 1-hour, 40-mark multiple-choice paper worth 20 per cent. The relevant Physics structured/free-response paper is Paper 2, lasting 1 hour 15 minutes and carrying 65 marks and 32.5 per cent. Paper 5 practical lasts 1 hour 30 minutes, carries 30 marks and contributes 15 per cent. The subject combination determines which theory papers candidates take. These facts come from the official syllabus rather than from a generic claim that all Science candidates take a Physics paper.
2. Start with the physical meaning of current
Current measures the rate at which electric charge passes a point: I = Q/t, where I is in amperes, Q is in coulombs and t is in seconds. A 2 A current means 2 coulombs of charge pass a specified cross-section each second. This is a rate, not the amount of charge permanently stored inside one resistor.
In a complete steady circuit, charge does not normally accumulate indefinitely at a junction. If 6 A arrives at a split, the currents leaving along the available branches total 6 A. The individual branch values depend on the branch paths and electrical conditions.
This observation is not a special memory trick for two resistors. It is conservation of charge applied to a network. A circuit with three or four branches obeys the same balance at every junction.
3. What counts as a branch
A branch is a distinct conducting path between junctions. If two resistors are connected between the same pair of circuit nodes, they are in parallel. Charge can pass through either path as it moves through the circuit. The word parallel concerns connectivity, not whether two symbols happen to be drawn beside each other.
A textbook diagram may be bent, rotated or drawn as a rectangle. The decisive question is whether the endpoints of the components connect to the same two nodes. If they do, they share the same potential difference. If there is only one path for the charge around the components, the elements are in series.
Mark the junctions mentally with letters A and B. Trace paths from A to B without crossing through the other component. If there are separate paths between the same two nodes, a parallel connection exists. This is safer than inferring the connection from a picture’s visual shape alone.
4. The central rule: branch potential differences are equal
In a simple ideal circuit with a 12 V source connected across two parallel resistor branches, each branch has a 12 V potential difference. You do not split the 12 V into two 6 V quantities merely because there are two branches. Both paths begin and end at the same node potentials.
Potential difference means energy transferred per unit charge across two points. Charge traversing one branch and charge traversing the other each experience the source’s full node-to-node potential difference, under the ideal model. Equal potential difference does not imply equal current.
Students familiar with series circuits may have learned that component voltages add to the source voltage around a loop. That rule remains true for each complete loop. In a parallel circuit, each branch has its own loop spanning the same source, which explains why both can have the full source potential difference.
5. The second rule: current splits and recombines
Suppose a source supplies a total of 5 A and a two-branch network carries 2 A through one path. The other path carries 3 A, assuming these are the only two branches and no charge is accumulating at the junction. When the paths rejoin, the combined steady current is again 5 A.
Do not write that each branch receives 5 A. That would imply 10 A leaving a 5 A junction, violating the conservation rule. Equally, do not assume that branch currents are always equal. Equality of branch currents requires additional conditions, such as identical resistances under the same voltage in the ideal model.
Treat a junction like a balance equation. Current entering equals current leaving, with directions and signs handled consistently. Once that relationship is clear, the numerical questions become straightforward.
6. Ohm’s law connects voltage, current and resistance
For an ohmic resistor under suitable conditions, V = IR, so I = V/R and R = V/I. These relationships allow you to calculate branch currents from the common potential difference and each branch’s resistance.
Imagine two resistors of 6 Ω and 3 Ω connected in parallel across an ideal 12 V source. Through the 6 Ω branch, I = 12/6 = 2 A. Through the 3 Ω branch, I = 12/3 = 4 A. The lower resistance takes more current at the same voltage.
Do not reverse that relation. High resistance does not “attract” more current because it offers a larger number in ohms. With the same potential difference, doubling resistance halves the ideal ohmic current.
7. The complete two-branch example
Continuing the 12 V example, the total source current is 2 + 4 = 6 A. The effective resistance of the parallel network is R = V/I = 12/6 = 2 Ω. This is less than either individual resistance, 3 Ω or 6 Ω.
That result should not be a surprise. Adding another conducting path allows greater total current for the same voltage under the ideal source assumption. From the source’s viewpoint, the two-path network behaves like a single 2 Ω resistor.
Use three independent checks: both branches have 12 V, their currents sum to 6 A, and 12/6 = 2 Ω. If a calculated equivalent resistance is 9 Ω, you may have accidentally applied the series formula.
8. The reciprocal rule gives the same equivalent resistance
For two parallel resistors, 1/R = 1/R₁ + 1/R₂. With 6 Ω and 3 Ω, 1/R = 1/6 + 1/3 = 1/2, giving R = 2 Ω. That agrees with the total-current method.
An often-used two-resistor shortcut is R = R₁R₂/(R₁ + R₂). Substituting gives 6 × 3/(6 + 3) = 18/9 = 2 Ω. This shortcut works for two parallel resistors. For three or more branches, use the general reciprocal relation or reduce a pair at a time.
Writing 1/R = 1/2 is not the same as R = 1/2. You must take the reciprocal. Confusing the reciprocal of resistance with resistance is a common last-line error.
9. More paths mean lower equivalent resistance in the ideal model
Two 4 Ω resistors in parallel give 1/R = 1/4 + 1/4 = 1/2, so R = 2 Ω. Three identical 4 Ω resistors in parallel give 1/R = 3/4, so R = 4/3 Ω, about 1.33 Ω.
The effective resistance decreases as another positive-resistance parallel path is added. The total source current rises at a fixed supply voltage. This conclusion assumes the source can sustain that voltage and the elements behave as intended. Real sources and wires have limitations, which is another reason the examples are theoretical.
Do not generalise “adding a resistor always raises resistance”. That statement is true for a simple series addition of a positive resistance, not for an added parallel branch.
10. Series and parallel are opposite in a useful way
Take the same two resistors, 6 Ω and 3 Ω, with an ideal 12 V source. In series, the effective resistance is 6 + 3 = 9 Ω. The current is 12/9 = 4/3 A, about 1.33 A, through both resistors. Their potential differences are 8 V across 6 Ω and 4 V across 3 Ω.
In parallel, each resistor has 12 V. Their branch currents are 2 A and 4 A, which total 6 A. The effective resistance is 2 Ω. The component values and source voltage have not changed, but the topology has changed the quantities.
That comparison explains why a candidate must inspect the connections before calculating. Identifying a circuit as series or parallel from the component count alone is a fundamental modelling error.
11. Mixed circuits demand a staged method
Suppose a 4 Ω resistor is in series with a parallel pair of 6 Ω and 3 Ω. The pair has equivalent resistance 2 Ω, so the whole network has 4 + 2 = 6 Ω. With a 12 V ideal source, total current is 12/6 = 2 A.
The series 4 Ω resistor carries 2 A, so its voltage drop is 8 V. The parallel pair therefore has 4 V across it. Its 6 Ω branch current is 4/6 = 2/3 A, and its 3 Ω branch current is 4/3 A. Their sum is 2 A, matching the source current.
This is a strong multi-step check. Reduce the parallel section, calculate total current, allocate the series voltage and then expand the branches. Jumping straight to “every resistor has 12 V” would ignore the series component.
12. Identify which components share two endpoints
Imagine three resistors on a page with the middle resistor drawn vertically and the others horizontally. The drawing alone cannot identify the true network. Label every wire-connected region with the same node name, ignoring wire bends. Components sharing both endpoints are parallel; components on a common unbranched path are series.
If two resistors share one endpoint but the other ends go to different nodes, they are not automatically parallel. If a wire directly bypasses an ideal resistor, the model may include a short path that changes the circuit entirely. Do not ignore wires because they look visually uninteresting.
This node-based method is transferable to unfamiliar diagrams, including those rotated or redrawn in examination questions. The logic follows connections rather than the appearance of the picture.
13. What an ammeter measures in a parallel network
An ammeter measures current through the path where it is correctly placed. An ideal ammeter just before a junction measures the total current arriving at the split. A meter within one branch measures that branch’s current. A meter after all paths rejoin measures the recombined total.
Thus in the 12 V example, a source-path ammeter would read 6 A, while branch ammeters would read 2 A and 4 A. A student who labels every meter 6 A has confused the location of measurement with the existence of a common source.
This is a theoretical interpretation, not a procedure for connecting physical meters. Actual measurement must follow school laboratory safety rules, the instrument’s ratings and qualified supervision.
14. What a voltmeter measures
A voltmeter measures potential difference between two points. Across the two nodes of a parallel network under the ideal 12 V example, a voltmeter reads 12 V. Across either branch between the same two nodes, it also reads 12 V.
Do not infer that a voltage reading is a flow of current. Volts and amperes measure different things. A branch can have 12 V across it while carrying 2 A, 4 A or another current depending on its electrical behaviour.
In diagrams, focus on the endpoints connected to the voltmeter rather than the position of its symbol on the page. Correct measurement interpretation starts by naming the two points whose difference is being recorded.
15. Charge conservation creates a useful time calculation
A 3 A branch current for 20 s corresponds to Q = It = 3 × 20 = 60 C of charge passing a point in that branch. If two branches carry 3 A and 2 A over the same steady 20 s interval, the total junction current is 5 A and the total charge passing the source path is 100 C.
The 60 C and 40 C branch amounts add to 100 C. This is the charge-accounting version of the current-splitting rule. It can be a helpful explanation when current seems abstract.
Do not say that the branch “contains” 60 C as if the charge were stored there. The calculation describes total charge that passes a specified cross-section during the interval.
16. Electrical power and energy follow the circuit topology
Electrical power is P = VI. In the 12 V, 6 Ω and 3 Ω parallel example, the branch powers are 12 × 2 = 24 W and 12 × 4 = 48 W. Total power is 72 W, matching source power 12 × 6 = 72 W under the ideal assumptions.
If the network operates at that power for 5 minutes, the energy transfer is E = Pt = 72 × 300 = 21,600 J. Minutes must be converted to seconds for joules when using watts.
Power is a rate of energy transfer. Do not report 72 W as energy or 21,600 J as power. The distinction is central to later questions about appliances, electrical costs and time intervals.
17. Connecting power to a fictional cost calculation
Under the ideal 72 W load, one hour uses 0.072 kWh. At a fictional educational tariff of $0.30 per kWh, the modelled energy charge for that hour is $0.0216. This illustrative price is not a current Singapore electricity tariff and does not represent a recommended circuit setup.
The point is to follow a complete unit chain: watts to kilowatts, hours to kilowatt-hours and energy units multiplied by a charge per kWh. A student who multiplies watts directly by dollars per kWh without converting may introduce a factor-of-1000 error.
For more on carrying units through multi-stage calculations, use Volume 0149 on precision.
18. Why two identical branches carry equal currents
With two equal 5 Ω resistors in parallel across 10 V, each branch current is 10/5 = 2 A. The source current is 4 A and equivalent resistance is 10/4 = 2.5 Ω. Equality of current here follows from equal resistance under equal potential difference.
If one branch’s resistance changes to 10 Ω while the supply remains 10 V, that branch current drops to 1 A. The original 5 Ω branch still carries 2 A under the ideal source, and the new total is 3 A.
Distinguish a property of the parallel network (equal branch voltages) from a special feature of identical branches (equal currents). Questions often change one component precisely to see whether students understand that distinction.
19. Open branches and changed topology
In an ideal parallel circuit, if one branch becomes open, current ceases in that branch. Other intact branches may still carry current if a complete path remains and the source continues to provide the same potential difference. This differs from a simple series circuit, where a break in the sole conducting path stops current through all series components.
Do not make an unrestricted claim about every appliance or wiring installation. Real devices may contain controllers, protective systems or interconnected elements that change their behaviour. The classroom principle concerns the ideal network described in the question.
When explaining a branch change, identify what stays constant and what changes: the connected path, the voltage across remaining branches, their individual currents and the resulting source current. Writing “everything stops” imports the series rule into the wrong topology.
20. Do not confuse a parallel branch with a short circuit
A very low-resistance unintended conducting path between the two source nodes can produce excessive current in a real system. This is not a helpful way to add an ordinary load. It is a safety hazard. Classroom questions may use ideal wires as an abstraction, but students should not treat those diagrams as instructions to make physical connections.
Keep theoretical calculations separate from laboratory practice. Real circuits have limits, heating and protective devices. Never experiment with mains wiring. Follow only approved school procedures and instructor directions for practical activities.
In written reasoning, the important distinction is that a wire bypass can change which components are effectively part of the active path. Identify the node connections before applying resistor formulae.
21. A fully worked alternative source
Take a 9 V ideal source with 3 Ω and 6 Ω in parallel. Each branch has 9 V. The 3 Ω branch carries 3 A and the 6 Ω branch carries 1.5 A. The total is 4.5 A; the equivalent resistance is 9/4.5 = 2 Ω.
Check with the reciprocal formula: 1/R = 1/3 + 1/6 = 1/2, so R = 2 Ω. The source power is 9 × 4.5 = 40.5 W. The branch powers are 27 W and 13.5 W, which add to 40.5 W.
Notice how the same logic works with new numbers. Students should resist memorising that a 6 Ω branch “always has 2 A”; its current depends on the potential difference across it. The stable rule is I = V/R with the correct branch V.
22. Check a proposed answer before accepting it
Suppose a learner says the 6 Ω and 3 Ω parallel network across 12 V has 9 Ω resistance because 6 + 3 = 9. That result is the series sum. It also predicts source current only 12/9 ≈ 1.33 A, smaller than the 4 A flowing through the 3 Ω branch alone, which is impossible for the stated two-branch ideal circuit.
A useful physical bound is that the equivalent resistance of positive resistors in parallel is less than the smallest individual branch resistance. This quick check immediately rejects 9 Ω and 3 Ω for the 6 Ω–3 Ω pair.
Do not use the bound as the whole solution when the question requires calculation and working. Use it to catch a wrong result and then show the reciprocal or current-sum method.
23. Thirty-two original Physics parallel-circuit questions
These exercises are original theoretical tasks. All networks have ideal wires and an ideal steady source unless a question states otherwise. Quantities are suitable for written calculation; no real-world wiring is required.
Practice 1
Question. What is equal across simple parallel branches?
Worked answer. Potential difference. Explanation. Both branches connect to the same two nodes. Identify the node-to-node voltage and track conservation at each junction before interpreting the result.
Practice 2
Question. What is conserved at a junction in steady state?
Worked answer. Electric current in the charge-balance sense. Explanation. The sum entering equals the sum leaving. Identify the node-to-node voltage and track conservation at each junction before interpreting the result.
Practice 3
Question. A source provides 5 A and one branch takes 2 A. Find current in the second and only other branch.
Worked answer. 3 A. Explanation. Subtract 2 A from the total 5 A. Identify the node-to-node voltage and track conservation at each junction before interpreting the result.
Practice 4
Question. Two branches each carry 1.5 A. Find source current.
Worked answer. 3.0 A. Explanation. Add the currents. Identify the node-to-node voltage and track conservation at each junction before interpreting the result.
Practice 5
Question. A 12 V source feeds a 6 Ω parallel branch. Find branch current.
Worked answer. 2 A. Explanation. I = V/R = 12/6. Identify the node-to-node voltage and track conservation at each junction before interpreting the result.
Practice 6
Question. The same source feeds a 3 Ω parallel branch. Find current.
Worked answer. 4 A. Explanation. I = 12/3. Identify the node-to-node voltage and track conservation at each junction before interpreting the result.
Practice 7
Question. Find total current for 6 Ω and 3 Ω in parallel across 12 V.
Worked answer. 6 A. Explanation. Add 2 A and 4 A. Identify the node-to-node voltage and track conservation at each junction before interpreting the result.
Practice 8
Question. Find equivalent resistance for that pair.
Worked answer. 2 Ω. Explanation. R = V/I = 12/6. Identify the node-to-node voltage and track conservation at each junction before interpreting the result.
Practice 9
Question. Check using the reciprocal formula for 6 Ω and 3 Ω.
Worked answer. 2 Ω. Explanation. 1/R = 1/6 + 1/3 = 1/2. Identify the node-to-node voltage and track conservation at each junction before interpreting the result.
Practice 10
Question. Two 4 Ω resistors in parallel have what equivalent resistance?
Worked answer. 2 Ω. Explanation. 1/R = 1/4 + 1/4. Identify the node-to-node voltage and track conservation at each junction before interpreting the result.
Practice 11
Question. Three 4 Ω resistors in parallel have what effective resistance?
Worked answer. 4/3 Ω (about 1.33 Ω). Explanation. Three reciprocal contributions give 3/4. Identify the node-to-node voltage and track conservation at each junction before interpreting the result.
Practice 12
Question. A 10 V source feeds a 5 Ω branch. Find current.
Worked answer. 2 A. Explanation. I = V/R. Identify the node-to-node voltage and track conservation at each junction before interpreting the result.
Practice 13
Question. Two such 5 Ω branches are in parallel. Find total current.
Worked answer. 4 A. Explanation. Each branch takes 2 A. Identify the node-to-node voltage and track conservation at each junction before interpreting the result.
Practice 14
Question. Find equivalent resistance of the pair at 10 V.
Worked answer. 2.5 Ω. Explanation. R = 10/4. Identify the node-to-node voltage and track conservation at each junction before interpreting the result.
Practice 15
Question. Replace one 5 Ω with 10 Ω, keeping 10 V ideal. New total current?
Worked answer. 3 A. Explanation. The 5 Ω branch carries 2 A; the 10 Ω branch 1 A. Identify the node-to-node voltage and track conservation at each junction before interpreting the result.
Practice 16
Question. In an ideal 9 V circuit, currents through 3 Ω and 6 Ω parallel branches?
Worked answer. 3 A and 1.5 A. Explanation. Each branch receives 9 V. Identify the node-to-node voltage and track conservation at each junction before interpreting the result.
Practice 17
Question. Find total current and effective resistance for the previous circuit.
Worked answer. 4.5 A; 2 Ω. Explanation. Add currents and divide 9 V by 4.5 A. Identify the node-to-node voltage and track conservation at each junction before interpreting the result.
Practice 18
Question. A 4 Ω series resistor precedes a 6 Ω–3 Ω parallel pair. Find total resistance.
Worked answer. 6 Ω. Explanation. The pair is 2 Ω, then add the series 4 Ω. Identify the node-to-node voltage and track conservation at each junction before interpreting the result.
Practice 19
Question. With 12 V across the mixed circuit, find total current.
Worked answer. 2 A. Explanation. I = 12/6. Identify the node-to-node voltage and track conservation at each junction before interpreting the result.
Practice 20
Question. Find voltage across the 4 Ω series resistor in that circuit.
Worked answer. 8 V. Explanation. V = IR = 2 × 4. Identify the node-to-node voltage and track conservation at each junction before interpreting the result.
Practice 21
Question. Find voltage across the parallel pair in that circuit.
Worked answer. 4 V. Explanation. 12 V source minus 8 V series drop. Identify the node-to-node voltage and track conservation at each junction before interpreting the result.
Practice 22
Question. Find currents in its 6 Ω and 3 Ω branches.
Worked answer. 2/3 A and 4/3 A. Explanation. Each branch has 4 V; sum is 2 A. Identify the node-to-node voltage and track conservation at each junction before interpreting the result.
Practice 23
Question. What charge passes a point carrying 3 A for 20 seconds?
Worked answer. 60 C. Explanation. Q = It. Identify the node-to-node voltage and track conservation at each junction before interpreting the result.
Practice 24
Question. Two branches carry 3 A and 2 A steadily for 20 s. Charge through the source path?
Worked answer. 100 C. Explanation. I = 5 A; Q = 5 × 20. Identify the node-to-node voltage and track conservation at each junction before interpreting the result.
Practice 25
Question. Find power through a 12 V branch carrying 2 A.
Worked answer. 24 W. Explanation. P = VI. Identify the node-to-node voltage and track conservation at each junction before interpreting the result.
Practice 26
Question. Find power through another 12 V branch carrying 4 A.
Worked answer. 48 W. Explanation. P = 12 × 4. Identify the node-to-node voltage and track conservation at each junction before interpreting the result.
Practice 27
Question. Find total power of those branches.
Worked answer. 72 W. Explanation. 24 + 48 = 72, also 12 × 6. Identify the node-to-node voltage and track conservation at each junction before interpreting the result.
Practice 28
Question. Find energy transferred at 72 W for five minutes.
Worked answer. 21,600 J. Explanation. Five minutes is 300 seconds; E = Pt. Identify the node-to-node voltage and track conservation at each junction before interpreting the result.
Practice 29
Question. Convert 72 W over one hour to kWh.
Worked answer. 0.072 kWh. Explanation. 72 W is 0.072 kW; multiply by one hour. Identify the node-to-node voltage and track conservation at each junction before interpreting the result.
Practice 30
Question. Is current automatically identical in all parallel branches?
Worked answer. No. Explanation. Only their potential difference is automatically equal in the ideal arrangement. Identify the node-to-node voltage and track conservation at each junction before interpreting the result.
Practice 31
Question. Can 9 Ω be the effective resistance of 6 Ω and 3 Ω in parallel?
Worked answer. No. Explanation. The actual value is 2 Ω, less than the smaller resistor. Identify the node-to-node voltage and track conservation at each junction before interpreting the result.
Practice 32
Question. If one ideal parallel branch opens but another path remains, must all branch currents become zero?
Worked answer. No. Explanation. The intact branch can still conduct under the stated ideal source conditions. Identify the node-to-node voltage and track conservation at each junction before interpreting the result.
24. Five common mistakes and their repairs
Error one: assuming branch voltage splits equally. Repair by naming the shared two nodes. Error two: saying branch current equals the source current. Repair with a junction equation. Error three: adding parallel resistances as if they were series. Repair by calculating branch currents or using reciprocals.
Error four: misreading a voltmeter as a current meter. Repair by identifying what the instrument measures and where it is connected in the diagram. Error five: obtaining a valid number with the wrong unit, such as labelling joules as watts. Repair by writing the relationship with its physical quantities before substituting.
Most errors can be diagnosed from the first wrong statement. Fix that statement rather than redoing ten mixed problems without an explanation. A clear model is stronger than repeated memorisation.
25. A seven-day practice progression
On day one, trace ideal node connections and label branches. On day two, solve current-splitting problems with two and three paths. On day three, calculate effective resistance by both reciprocal and total-current methods. On day four, compare the same resistors in series and parallel.
On day five, solve mixed circuits by reducing sections step by step. On day six, add power, charge and energy calculations, with explicit unit conversions. On day seven, take a new diagram, predict qualitative effects and then check numerically. Use teacher-supervised practical lessons, where appropriate, for actual equipment.
This is optional study structure, not official examination timing. Focus on whether you can explain why a branch takes a particular current, not just whether a calculator produces a familiar answer.
26. How this chapter connects to the series
Read Volume 0159 on series circuits first if the single-path rule is unfamiliar. The two volumes deliberately form a connected pair: same current and divided voltage in series; divided current and same voltage in parallel.
For related numerical work use Volume 0149 on significant figures and Volume 0162 on fractional relationships. Use the Science Hub, Mathematics Hub and Secondary English Hub for subject owners.
The G3 SEC learner guide provides the examination route, while the PSLE Learning Guide supports earlier science and quantitative foundations. Verify the actual Physics content and scheme in the official 2027 G3 Science syllabus.
27. The reliable circuit habit
Before calculating, find the nodes. Before dividing a current, check whether it is a source current or a branch current. Before adding a resistance, identify whether the components are connected in series or parallel. Then use V = IR, junction conservation and the appropriate equivalent-resistance relationship.
A strong answer can be checked three ways: electrical quantities agree with the topology, units match their meanings, and the summed branch values recover the total where they should. Those checks make novel diagrams less intimidating.
The final skill is not a memorised drawing of two resistors. It is the ability to reconstruct how a network behaves from its connections. Once that reasoning is stable, the algebra supports the Physics rather than replacing it.
