Quick answer. Identify complete conducting paths before using formulas. In series, current is the same and potential differences add; in parallel, potential difference is the same and branch currents add. Connect an ammeter in series with the current being measured and a voltmeter across the two points being compared. For 2027, the G3 SEC Physics subject code is K323.
This is a learning chapter with original worked examples, not an official SEAB paper. It links the earlier PSLE circuit foundation to upper-secondary examination reasoning.
Start here: the three questions a circuit answer must settle
A direct-current circuit can look complicated because the drawing contains several devices, wires and numbers. The reasoning is simpler when the student asks three questions in order. First, where can charge move through a complete conducting path? Second, which components share the same current, and which share the same potential difference? Third, what is the measuring instrument actually connected across or through? These questions prevent more errors than memorising a page of isolated statements.
This chapter develops a method for G3 Physics D.C. circuits. It begins with the Primary-school idea of a complete circuit, builds the language of current, potential difference and resistance, then moves to series networks, parallel branches, meter connections, and unfamiliar mixed circuits. Worked examples show how to keep equations attached to the correct portion of the network. The final section turns knowledge into timed examination decisions.
The 2027 G3 SEC Physics syllabus is K323, replacing the earlier reference code 6091. Its D.C. circuits topic includes circuit diagrams, current and potential-difference rules, equivalent resistance, and applications involving potential dividers, thermistors and light-dependent resistors. These are not all equivalent tasks: recognising a series arrangement is the foundation; choosing an explanation under changing conditions is the advanced application. This guide uses original learning examples rather than representing them as official examination questions.
What a complete path means
An electric circuit is a connected arrangement that allows charge to move. In a basic classroom circuit, a source transfers energy to charges and the charges transfer energy in circuit components. A complete conducting path is necessary for sustained current. A gap in a required series path prevents current through that path. The important qualification is “that path.” If one branch of a parallel circuit is opened, another intact branch can still conduct current.
A student who says “the whole circuit is broken” whenever a switch opens has not yet considered topology. Topology means which points connect to which. A diagram can be redrawn into a different shape without changing its electrical connections. Two lines that merely cross are not automatically joined unless a connection is indicated. A junction is not the same thing as a bend. Before calculating anything, trace the conducting route from one terminal of the source through the connected components and back to the other terminal.
Try a mental test: a cell supplies two lamps on separate parallel branches. A switch opens only the branch containing Lamp A. Lamp A goes out. Lamp B can remain lit because its loop is still complete. If the switch instead lies in the common supply path before the branches, both lamps go out. The same number of symbols appears in the two drawings, but the position of one switch changes the outcome. This is why the student must read the connections, not the picture’s appearance.
Current: what flows and where it divides
Electric current is the rate of flow of electric charge. The relationship is I = Q/t, where I is current in amperes, Q is charge in coulombs and t is time in seconds. A reading of 0.40 A means that 0.40 C of charge passes a point each second. The current is not a store of energy and is not consumed by a lamp. Electrical energy is transferred, but conservation of charge still governs how current moves through a steady-state circuit.
In one unbranched series path, the current is the same at all points. If a source, resistor and lamp are connected in a single loop and an ammeter reads 0.25 A before the resistor, an ideal ammeter placed after the lamp will read 0.25 A. This statement remains true even though the potential differences across the resistor and lamp may differ. Confusing current with potential difference is a common source of wrong answers.
At a junction, current divides. If 1.8 A enters a junction and branches carry 0.7 A and an unknown current, the missing branch carries 1.1 A. The equation is 1.8 = 0.7 + 1.1. The branches need not carry equal currents; equal currents require appropriate electrical conditions, not simply two drawn branches. When branches rejoin, their currents add. On a well-drawn answer, mark a junction equation beside the actual junction. Avoid writing a sum without telling the reader what each term represents.
Potential difference: energy transferred per unit charge
Potential difference is measured in volts and expresses energy transferred per unit charge between two points. The relationship V = E/Q helps explain why potential difference belongs between points rather than at a single location. If 6 J of energy is transferred for each 2 C of charge across a component, its potential difference is 3 V. In a circuit calculation, the unit V is part of the meaning, not decoration.
Series circuits have a potential-difference sum rule. A 12 V source connected across two series resistors may produce a 4 V drop across one and an 8 V drop across the other. The drops add to the source potential difference: 4 + 8 = 12 V, assuming the other circuit elements do not contribute appreciable potential differences. It would be incorrect to write 12 V across each component merely because both are connected to the same battery in the overall loop.
Parallel branches are connected across the same two junctions. Therefore their potential differences are equal. If two complete branches connect directly across an ideal 9 V source, each branch has 9 V across its endpoints. Within a branch that contains two series components, those components may divide the 9 V between them. The student must distinguish “across the whole branch” from “across one resistor within the branch.” Those two phrases refer to different pairs of points. This one distinction unlocks many multi-stage network questions.
Resistance and Ohm’s law without shortcuts
Resistance R is the ratio of potential difference V to current I for a component at a particular condition: R = V/I. The unit is ohm, symbol Ω. For an ohmic resistor at constant temperature, V is proportional to I and R remains constant. For some other components, such as filament lamps and semiconductor diodes, the current–voltage relationship changes with operating condition. The Physics syllabus explicitly includes interpretation of such characteristics, so a universal “every V–I graph is straight” rule is unsafe.
Suppose a resistor has 6.0 V across it and carries 0.30 A. Its resistance is 6.0/0.30 = 20 Ω. If the same ohmic resistor is then connected across 3.0 V while temperature is held constant, I = V/R = 3.0/20 = 0.15 A. The reason is a constant resistance under the stated condition, not a blanket law that current always halves when voltage halves in every device.
Always name the object for which R is calculated. If 6.0 V is the potential difference across the entire arrangement but 0.30 A is the current through only one parallel branch, calculating 20 Ω may not describe any intended component. Pair measurements from the same part of the circuit. A useful writing habit is “For resistor R1, V1 = … and I1 = …, therefore R1 = ….” This looks slower, but it is faster than correcting a wrong equation after three lines of arithmetic.
Series resistors: one path, one current
Two resistors of 4 Ω and 8 Ω connected in series have an effective resistance of 12 Ω. The current flows through one and then the other, so the effective opposition in this simple model adds: Rtotal = R1 + R2. Across a 12 V ideal supply, I = 12/12 = 1.0 A. The potential differences are 4 V and 8 V. Check: 4 + 8 = 12. This example illustrates the full circuit relationship, not three disconnected formula substitutions.
To solve a series question, first identify that the components really form a single current path. Next add their resistances. Then use the source potential difference and total resistance to calculate the current. Finally calculate the potential difference across each resistor if needed. Check their sum against the supply. A student who calculates each resistor’s current using the full source potential difference has silently treated the two series resistors as separate parallel branches.
What happens if a third 6 Ω resistor is added in series to that arrangement and the source remains 12 V? The total resistance rises to 18 Ω. Current becomes 12/18 = 0.667 A to three significant figures. Each existing resistor then has a smaller potential difference than before because the current through all components has fallen. This is an exam-ready explanation: state the changed total resistance, the change in current, and the resulting change in potential difference. Do not rely on a vague claim that “more components use more electricity.”
Parallel resistors: shared endpoints, divided current
For two resistors connected in parallel, 1/Rtotal = 1/R1 + 1/R2. Consider 6 Ω and 3 Ω resistors connected directly across a 12 V ideal supply. The current through the 6 Ω branch is 2 A; the current through the 3 Ω branch is 4 A. The source supplies 6 A in total. The effective resistance is 12/6 = 2 Ω, which is less than either individual resistance. That is a useful reasonableness check.
Some students remember the reciprocal formula but forget why it works. Parallel connections provide additional paths for current. For a fixed potential difference across the network, total current is the sum of the branch currents. Since I = V/R, more total current at the same voltage means lower effective resistance. You can derive the reciprocal expression from this branch-sum idea instead of treating it as a mystery.
Now remove the 3 Ω branch without changing the source. The 6 Ω branch still has 12 V across it, so its current remains 2 A in an ideal model. Source current falls from 6 A to 2 A. Notice the distinction: source current changes, but current in the remaining unchanged branch does not. In some real circuits the source voltage may change under load. The examination question normally provides or implies the conditions, and the answer should follow those conditions rather than inventing complications.
Mixed networks: reduce the diagram in stages
Suppose an 8 Ω resistor is in series with a parallel pair of 12 Ω and 6 Ω resistors across a 24 V supply. Begin inside the parallel pair. Its effective resistance is 4 Ω, because 1/R = 1/12 + 1/6 = 3/12. The entire network now behaves like 8 Ω in series with 4 Ω. Total resistance is 12 Ω and source current is 24/12 = 2 A. The 8 Ω resistor therefore has a potential difference of 16 V; the parallel section has 8 V.
Within that parallel section, each branch has 8 V across its endpoints. The 12 Ω branch carries 8/12 = 0.667 A, and the 6 Ω branch carries 8/6 = 1.33 A. More precisely the currents are two-thirds of an ampere and four-thirds of an ampere, which sum to 2 A. Do not add rounded values first and then claim an exact equality. Keep sufficient digits during working and round only the stated answers.
This model exposes a reliable five-step routine: identify the network blocks; replace one simple block with its effective resistance; calculate the total source current; work backwards to recover voltages and branch currents; verify both sum rules. The strongest exam answers show the reduction, not just the final current. The method transfers to unfamiliar layouts because it works from connections, not picture patterns.
Ammeter and voltmeter placement
An ammeter measures current through a path, so it is placed in series with the path whose current is required. A voltmeter measures potential difference between two points, so it is connected in parallel across the component or branch of interest. In introductory circuit problems, ideal meters are often assumed: an ideal ammeter has negligible resistance and an ideal voltmeter very high resistance. The student should not treat these practical conventions as reasons to ignore the diagram.
To measure current through a lamp, put the ammeter in the lamp’s series path. To measure potential difference across the lamp, connect the two voltmeter terminals to the two ends of the lamp. A common mistake is placing a voltmeter in series and asking why the lamp behaves differently. Another is placing an ammeter directly across a cell; this can create a very low-resistance path and is unsafe in real experiments. Use only supervised, low-voltage educational circuits; never test these ideas on mains wiring.
A question may show an instrument already connected and ask what it reads. Read the connections literally. A voltmeter placed across two points joined by an ideal wire has approximately zero potential difference. An ammeter in the common path before a parallel junction measures total source current, while one inside a branch measures only that branch. Make a two-word annotation beside each meter: “total current,” “branch current,” “R2 voltage,” or “whole-network voltage.” A labelled quantity is harder to confuse than an isolated number.
Meters, ranges and measurement limits
A measured value depends on both the circuit and the instrument. In practical work, the choice of meter range affects readable precision. Record measurements with units and a consistent number of decimal places appropriate to the instrument. A digital display reading 0.26 A should not be transcribed as 26 A. A current value obtained by using a 200 mA range may need conversion: 150 mA = 0.150 A. Unit conversion is not a separate administrative step; it is part of the physics.
Think about the following log: supply 6.0 V, ammeter 0.30 A, lamp 2.4 V, resistor 3.6 V. The series potential-difference sum works: 2.4 + 3.6 = 6.0 V. The lamp and resistor share 0.30 A in the series path. The resistor’s resistance is 3.6/0.30 = 12 Ω. Do not calculate the lamp’s resistance by dividing the entire 6.0 V by 0.30 A. That would return the effective resistance of the series pair under these conditions, not the lamp’s individual resistance.
When a reading looks surprising, check the units, placement, range, and whether the circuit was changed between readings. Experimental uncertainty can justify small differences, but it is not a license to dismiss a large contradiction. A good learner can say precisely what remains unknown: “The source voltage was measured, but the potential difference across this branch was not.” That sentence is more scientific than fabricating a missing measurement.
How switches alter circuits
Switches are often used to test whether a student understands connected paths. For a single series circuit, opening any switch in the only conducting loop interrupts current throughout that loop. For two parallel branches with separate switches, each branch may be controlled independently. If a switch is placed in a shared supply path, it controls all downstream branches. The switch’s location matters more than its name.
Imagine two identical 6 Ω resistors in parallel across 12 V, with one switch in each branch. Both switches closed: each branch carries 2 A and total current is 4 A. Open one branch: the remaining branch still carries 2 A, but total current is 2 A. Open the common main switch instead: both branch currents become zero. This simple case should be practised until the learner can reason it out without writing the reciprocal formula every time.
The language of an explanation should follow the sequence of physical changes. “Opening S2 disconnects the second branch. The first branch remains connected across the same 12 V source, so its potential difference and current are unchanged. The source supplies less current because one branch no longer draws current.” Each sentence has one job. Avoid saying “the current chooses the shortest path,” a phrase that is both vague and misleading. Charge moves through available conducting paths according to the electrical conditions of the network.
Potential dividers: turning a ratio into a voltage
Two resistors in series across a fixed supply divide the supply potential difference in proportion to their resistances. If a 4 Ω resistor and an 8 Ω resistor are in series across 12 V, the current is 1 A, so the potential differences are 4 V and 8 V. The potential difference across the 8 Ω resistor may also be found using 12 × 8/(4 + 8) = 8 V. This is the basic potential-divider idea.
A potentiometer provides an adjustable output by moving a contact along a resistive track. Many examination tasks use a fixed resistor with a component whose resistance changes with a physical quantity. A light-dependent resistor usually has lower resistance in brighter light. An NTC thermistor has lower resistance at higher temperature. The output change depends on which component the voltmeter is connected across. It is unsafe to memorise “voltage increases when temperature increases” without saying which component receives the output voltage.
Example: an NTC thermistor in series with a 5 kΩ fixed resistor is connected across 10 V. The voltmeter is across the fixed resistor. At one temperature, the thermistor is 5 kΩ, so the output is 5 V. As temperature rises and thermistor resistance falls to 2 kΩ, current rises and the output across the fixed 5 kΩ resistor becomes 10 × 5/7 ≈ 7.14 V. If the voltmeter were instead across the thermistor, the measured voltage would fall. The component location decides the trend.
Avoiding the equation trap
The equation R = V/I can be rearranged as V = IR and I = V/R. Yet an equation is not enough until the variables refer to the same circuit element. In examination preparation, write a subscript whenever the diagram has more than one relevant component. Use V1, I1, R1 for one resistor, then Vtotal, Itotal and Rtotal for the entire network. Once the working is secure, the notation can become compact.
Consider a 9 V source with a 3 Ω resistor in series with a 6 Ω resistor. The total current is 1 A. If the question asks for the potential difference across the 6 Ω resistor, V2 = I × R2 = 1 × 6 = 6 V. Writing 9/6 = 1.5 A would apply the full source voltage to the wrong object. The arithmetic would be neat but the model would be false.
Another trap is dividing a total current equally at a junction merely because two branches are drawn. In parallel, the branch with lower resistance carries more current for the same potential difference. Two equal resistances produce equal branch currents under the same branch voltage, but otherwise no equal-split rule follows. Examiners test reasoning through changing configurations precisely because memorised patterns fail when the circuit is rearranged.
Reading circuit diagrams under time pressure
Start by locating the source terminals, then identify continuous nodes. A node is a region connected by ideal wire, with no component between the points being treated as one node. Mark nodes A and B across a parallel section. If two components have both terminals connected to nodes A and B, they are in parallel even if one is drawn above and one below. If the same charge path must pass through two components with no intervening branch, they are in series.
After identifying topology, predict something qualitative before calculating. For a fixed-voltage source, adding a parallel resistor generally reduces the effective network resistance. Adding a series resistor increases it. In a steady-state series path, a current reading does not change merely because the ammeter is moved to a different point in the same path. These predictions work as error checks. If the final numerical result violates them, revisit the drawing.
Keep separate the geometry of a page and the physics of a network. Two resistors may be side by side in the drawing yet actually be in series. A long wire may bend around the page without creating another branch. A voltmeter symbol may make the drawing look branched, but its purpose is to measure potential difference. In the usual ideal-meter model it draws negligible current. Understanding the model assumptions is part of reading the diagram correctly.
Worked challenge A: a missing branch reading
An ideal 15 V source feeds a 10 Ω resistor in parallel with an unknown resistor. The total source current is 4.0 A. Calculate the unknown resistance. Since both branches connect across the same source, each has 15 V. The 10 Ω branch carries 15/10 = 1.5 A. The other branch must carry 4.0 − 1.5 = 2.5 A. Its resistance is 15/2.5 = 6.0 Ω.
There is a tempting but wrong shortcut: divide 15 V by 4.0 A to obtain 3.75 Ω and call that the unknown resistor. The result 3.75 Ω is instead the total effective resistance of both branches together. It is useful as a check: for a parallel arrangement the total should indeed be less than 6 Ω and 10 Ω. But the question asked for one component, not the whole network.
An excellent solution says what each intermediate value means. “Current in 10 Ω branch = 1.5 A”; “current in unknown branch = 2.5 A”; “unknown resistance = 6.0 Ω.” With these labels, even an arithmetic error is easier to find and correct. Under time pressure, clarity of identity is a practical skill, not an ornament.
Worked challenge B: reading across the wrong points
A 20 V source is connected to a 2 Ω resistor and a parallel combination of 6 Ω and 3 Ω resistors. Calculate the potential difference across the 3 Ω resistor. First reduce the parallel pair: (1/6 + 1/3) = 1/2, so the pair has 2 Ω effective resistance. The whole circuit contains 2 Ω in series with 2 Ω, giving 4 Ω total. Source current is 20/4 = 5 A.
The separate 2 Ω series resistor takes 5 × 2 = 10 V. The parallel section therefore receives the remaining 10 V. Both branches share those two junctions, so the 3 Ω resistor has 10 V across it. Its branch current is 10/3 ≈ 3.33 A if required. Do not confuse the source current of 5 A with the 3 Ω branch current. The other branch carries 10/6 ≈ 1.67 A; the two branch currents add to 5 A.
This example contains three different “twos”: a physical 2 Ω resistor, an effective 2 Ω parallel block, and two branches inside that block. Good notes say “series 2 Ω” and “parallel block 2 Ω,” instead of putting two unlabeled values in a list. A one-word label can save an entire solution.
Worked challenge C: a sensor’s direction of change
A 12 V supply feeds an LDR and a fixed 3 kΩ resistor in series. A voltmeter measures the potential difference across the fixed resistor. In dim light the LDR is 9 kΩ; in bright light it falls to 3 kΩ. In dim light the output is 12 × 3/(9 + 3) = 3 V. In bright light the output is 12 × 3/(3 + 3) = 6 V.
The output across the fixed resistor rises as light increases. Explain why: light reduces the LDR’s resistance, which reduces total series resistance and increases current through the fixed resistor. Since its resistance remains constant, V = IR makes its potential difference rise. This answer has a clear causal chain and names the measured component.
Now reverse the voltmeter so it measures across the LDR instead. In dim light it reads 9 V; in bright light it reads 6 V. The trend reverses although the circuit components and lighting change are identical. The measured pair of points has changed. If the task asks a prediction rather than exact numbers, the same causal reasoning is enough. Practise saying the reason in complete sentences before relying on abbreviated formula working.
What the examination actually asks
The official K323 G3 Physics specification requires candidates to enter three papers: Paper 1 multiple-choice, Paper 2 structured and free-response, and Paper 3 practical. For 2027, the document lists Paper 1 as one hour and 40 marks with 30% weighting, Paper 2 as one hour 45 minutes and 80 marks with 50% weighting, and Paper 3 as one hour 50 minutes and 40 marks with 20% weighting. Circuit understanding can be tested through recall, interpretation, unfamiliar applications, and practical measurement; the same concept may appear in different response forms.
In a multiple-choice question, eliminate results that violate circuit rules before investing time in long arithmetic. In structured questions, show the network reduction, not just a final current. In practical questions, a student may need to set up, measure, record or evaluate a circuit, and the precise task instructions always control. Do not use an “examination trick” that requires ignoring a safety instruction or changing apparatus without permission.
The syllabus distinguishes knowledge from handling information and solving problems. This is the reason to practise unfamiliar drawings. The task is not to memorise a single ideal circuit diagram but to recognise the principles when the question changes the number, position or role of components. A learner who can explain what happens after one switch moves has more robust knowledge than a learner who can reproduce only an untouched textbook picture.
A method for six command words
When a question says “state,” give the requested fact directly. Example: “In a series circuit, current is the same at every point.” When it says “calculate,” identify the relevant quantities, substitute, show working and give the unit. When it says “explain,” supply a mechanism: adding a series resistor raises total resistance, causing current to fall for a fixed source voltage. When it says “predict,” say what will happen under the specified new condition. When it says “compare,” put both cases on the same basis. When it says “suggest,” apply known principles to a new arrangement rather than claiming an observation you were never given.
These are not rigid templates that replace thought. They remind the learner to answer the demand of the verb. A “why” answer that merely repeats a number is incomplete. A “calculate” answer that describes a trend without a numerical result is incomplete. A “compare” answer that explains only one case gives the reader no comparison. Precise answers tend to be concise because each sentence serves the question.
A productive exercise is to take one solved 12 V series example and write four questions about it: state the current rule, calculate a potential difference, explain what happens if another resistor is added, and suggest where to connect a voltmeter. The circuit is unchanged, but the thinking task is different. This develops flexible performance without requiring dozens of near-identical diagrams.
Mistake clinic: diagnose the first wrong move
If the answer wrongly gives the full source voltage to each series component, the misunderstanding is about shared endpoints. Return to the two-junction test. If the answer divides total current equally between unequal parallel resistors, the misunderstanding is about branch resistance. Return to I = V/R using the common parallel voltage. If the answer uses an ammeter in parallel with a lamp, the misunderstanding concerns what the instrument measures. Return to “through” for current and “across” for potential difference.
If the student knows the formulas but loses marks through substitutions, label every quantity by its physical location. If the student cannot draw a circuit but can solve a ready-made diagram, practise symbolic representation and connection tracing. If the student gives correct final numbers with missing working, rehearse a short chain of stated equations so reasoning remains inspectable. If the student is accurate without timing but erratic under timing, reduce the size of timed sets, establish a stable opening routine, and lengthen only when accuracy holds.
Correct one kind of failure at a time. More practice on a large mixed paper is not always the next sensible move. After correcting an error, retest the same principle with a new layout, changed source voltage or changed measurement point. The test of learning is whether the rule survives the change. A learner who merely remembers the original answer has not yet gained a transferable method.
Independent problem set: predict before calculating
Question 1. Two identical 5 Ω resistors are in series across 10 V. Find the current and potential difference across each. Answer: total 10 Ω; current 1 A; each has 5 V. Check that the two drops sum to the source. Question 2. The same two 5 Ω resistors are connected in parallel across 10 V. Find branch currents and source current. Answer: each branch has 10 V and carries 2 A; total current 4 A; effective resistance 2.5 Ω. Contrast these answers with Question 1 rather than memorising the final numbers.
Question 3. A 3 Ω resistor is in series with a pair of 6 Ω parallel resistors across 12 V. Find the voltage across the parallel pair. Answer: pair equivalent 3 Ω, total 6 Ω, current 2 A, parallel voltage 6 V. Question 4. In a parallel arrangement the source current is 3.0 A; one branch carries 0.8 A and the second 1.1 A. A third branch carries 1.1 A. Check: 0.8 + 1.1 + 1.1 = 3.0 A.
Question 5. An ohmic resistor carries 0.20 A at 4.0 V. Predict current at 10 V at constant temperature. Answer: R = 20 Ω; I = 0.50 A. Explain the constant-temperature assumption. Question 6. A voltmeter across a 7 Ω resistor reads 14 V. Find its current: 2 A. Then explain why 2 A cannot automatically be assigned to an adjacent parallel branch. The second part tests whether a solved number is being applied with care.
Independent problem set: control the explanation
Question 7. An ammeter is moved to a different point in the same unbranched series loop. Predict its ideal reading. It is unchanged because there is no junction at which current divides. Question 8. A lamp in one of two parallel branches fails open-circuit. Predict the ideal potential difference across the intact branch. It remains equal to the source potential difference if the ideal supply and connections remain as stated. Explain that total source current may nevertheless decrease.
Question 9. Two different resistors share both endpoints. Are they series or parallel? Parallel. The answer must refer to shared endpoints, not where they appear on the page. Question 10. A student reports that a 6 Ω and 12 Ω parallel pair has an effective resistance of 18 Ω. Why is that unreasonable? Adding a second path should make effective resistance lower than the smaller individual resistance; recalculation gives 4 Ω.
Question 11. A fixed resistor in a divider is monitored by a voltmeter. The resistance of the series NTC thermistor falls as the circuit warms. With a fixed supply, the fixed resistor’s potential difference rises because current rises. Question 12. The voltmeter is moved across the thermistor. Now the thermistor’s potential difference falls. The pair of questions is intentionally similar: the challenge is to locate the measuring points, not to repeat an unqualified phrase about temperature.
Independent problem set: extend the method
Question 13. A 24 V source is connected to a 4 Ω resistor in series with two 8 Ω resistors in parallel. Determine total current and current in each parallel branch. The parallel pair is 4 Ω. Total is 8 Ω. Source current is 3 A; series resistor drops 12 V; each parallel branch has 12 V and carries 1.5 A. The two currents sum to 3 A.
Question 14. A circuit has 5 V across a fixed 10 Ω resistor and 2 V across another resistor in series. Determine the source potential difference and current, then the second resistance. The source is 7 V; current is 0.5 A; the second resistance is 2/0.5 = 4 Ω. Explain why the answer needs the word “series.” Without it, the shared-current step is unjustified.
Question 15. A battery supplies 18 V across a 9 Ω resistor. Find current. Answer: 2 A. Add an identical 9 Ω resistor in parallel. The original branch still carries 2 A, the new branch carries 2 A, and source current is 4 A. Add it instead in series: total resistance becomes 18 Ω and source current becomes 1 A. One component, two connection choices, three different outcomes. This is a useful transfer test because it demands structural understanding.
Question 16. A parallel network contains resistances 12 Ω, 6 Ω and 4 Ω across 12 V. Branch currents are 1 A, 2 A and 3 A, totaling 6 A. The effective resistance is 2 Ω. Check using reciprocals. Next, open the 4 Ω branch: new source current is 3 A, and the remaining branch currents are still 1 A and 2 A under the stated ideal supply.
A two-session practice plan
Session one is for secure foundations. Spend ten minutes drawing and relabelling a simple series circuit. Spend ten minutes on two-junction parallel identification. Work six short calculations that alternate current, potential difference and resistance. For each, draw a box around the component whose value is requested. End with a short spoken explanation of why an ammeter goes in series and a voltmeter across a component. The target is not speed: it is accurate connection reading and unit use.
Session two introduces unfamiliarity. Start with a mixed network, then a switch change, then one potential-divider trend and one diagram with an unexpected meter position. Set a reasonable time limit, but stop if the learner begins guessing connections. Compare the first attempt with a corrected attempt and classify the first wrong move: reading, topology, equation, units or explanation. Write one sentence stating the repair that would prevent the same mistake tomorrow.
A successful week does not require completing hundreds of circuit sums. It requires being able to identify topology, apply the correct rule, label quantities, show inspectable working and explain the effect of a change. Move on only after these actions remain reliable when the diagram is redrawn. These habits also support Mathematics graphs and English evidence-based answers, because all three subjects reward a correct relationship between a statement and the information that supports it.
Where to go next and official sources
For the bridge from primary learning, review the Primary 6 electrical circuits learning guide and the PSLE English, Mathematics and Science route. For progression across this G3 SEC series, begin from Volume 0612 on evidence and model limits, then move to the next Mathematics guide on electrical measurements. The G3 SEC EMS learner-guide hub gives the broader learning route.
For examination format and syllabus limits, consult the SEAB SEC overview, the 2027 G3 syllabus index and the K323 G3 Physics syllabus. The 2027 code is K323; the earlier 6091 code is listed by SEAB as a reference to 2026 and earlier. School instructions and the official paper always take priority over any general practice example in this guide.
