Quick answer. Diagnose a D.C. circuit by tracing conducting paths and comparing measured current and potential difference with the predictions of specific open-branch and bypass models. A dark lamp alone cannot identify the fault. This chapter uses original idealised problems aligned to the 2027 G3 SEC Physics K323 syllabus.
The useful question is not “which component is wrong?”
A circuit may produce an unexpected meter reading without telling you which part has failed. The first task is therefore not to name a broken component. It is to explain what a correctly connected circuit would do, compare that prediction with the observations, and decide which possible fault could produce the difference. This order protects the answer from guesswork.
Start with the simplest distinction: an open path has a break; a bypass provides a lower-resistance path around a component. Those changes are not opposites in every observable detail. Either can make a lamp go dark, but the surrounding currents and potential differences may be very different. The lamp’s darkness alone is weak evidence. Its location, the other lamp’s behaviour, and the measuring points matter more.
A good examination answer begins with a circuit model and ends with a measured conclusion. Between them is a chain of reasoning: trace the path, locate the branching points, identify the quantity the meter measures, predict each plausible fault, compare predictions with readings, and state the remaining uncertainty. This is the same discipline that made Volume 0613 on current, potential difference and meter placement useful. Here we take the next step, from reading a working circuit to diagnosing what a set of imperfect observations supports.
The examples below use idealised low-voltage classroom circuits and original practice data. They are not instructions for investigating household wiring, which requires qualified personnel. Never deliberately short a power supply or imitate a suspected electrical fault on live equipment. A safe examiner-style question can explore an unsafe physical condition without asking anyone to create it.
What the official 2027 examination requires
The Singapore-Cambridge Secondary Education Certificate begins in 2027, replacing the former separate GCE N(T), N(A) and O-Level naming. Students sit subjects at G1, G2 or G3 according to their subject level. The 2027 G3 Physics code is K323, with 6091 shown by SEAB only as the reference code used in 2026 and earlier. Do not put the old reference code into a 2027 registration or revision label as though it were the new code.
The official K323 syllabus includes the physics of electric current, resistance, D.C. circuits, potential differences, effective resistance, practical electricity and experimental investigations. It expects students to reason about series and parallel paths, apply R = V/I, and interpret measurements. In the documented assessment scheme, Paper 1 is a one-hour multiple-choice paper worth 40 marks and 30%; Paper 2 is a one-hour-45-minute structured and free-response paper worth 80 marks and 50%; Paper 3 is a one-hour-50-minute practical assessment worth 40 marks and 20%. These are the published 2027 arrangements, not estimates based on an older textbook.
Paper 2 also includes data-handling demands. Paper 3 can require measurements, data presentation, evaluation and planning. A student who knows the resistor formula but cannot explain why one observation rules out a proposed fault is not yet ready for the transfer that such questions invite. Conversely, a student who can justify a limited conclusion from measurements is building the intended scientific skill even when the surface story is unfamiliar. Read the official syllabus directly using the link at the end, and follow the actual examination instructions on the day.
Begin with the topology, not the drawing
Topology means the way circuit components are connected. Two circuit diagrams can look different on the page while being electrically equivalent. A long looping wire might simply join two points with negligible resistance. Two resistors drawn side by side might be in series if there is no branching junction between them. What decides the relationship is the existence of separate current paths, not artistic position.
Trace the circuit from one terminal of the source through each component and back. At a junction, ask whether there are two or more ways forward. At a reconnection point, recognise that branch currents add. Mark nodes—points joined by ideal wires—because the voltage across a component is the potential difference between its two nodes. A voltmeter attached across the wrong nodes may report a completely valid number that answers the wrong question.
Before calculating, write three reference rules in words. In series, current is the same through every component. At a junction, current entering equals current leaving. Around a closed series path, the potential differences across the components add to the source potential difference, under the assumptions of the model. For parallel branches connected across the same two nodes, the potential difference is the same. Then attach the relationships for resistance: series resistances add; reciprocal resistances add for parallel components. The meaning of each rule is more important than its appearance as a memorised equation.
A useful self-test is to redraw a complicated network with only node labels and resistor labels. If you cannot redraw it without changing which nodes connect, postpone numerical calculation. It is easier to correct a connection misunderstanding at the beginning than to trace a page of arithmetic built on the wrong circuit.
Open circuits: the missing conducting path
An open circuit has a break along a path that would otherwise conduct. An open switch, a disconnected lead, or a failed filament can create an open in an idealised examination diagram. If a component is in a single series path, an open there prevents sustained current through that path. If the same break occurs in one branch of a parallel network, other branches may still have complete routes back to the source.
Suppose two 3 Ω resistors are connected in series to an ideal 6 V supply. With the switch closed and no fault, total resistance is 6 Ω, current is 1 A and the potential difference across each resistor is 3 V. Now imagine an open in the connection between them. Under this ideal model, current falls to zero. With no current, there is no resistive voltage drop across either 3 Ω resistor; the potential difference appears across the open gap. A voltmeter across that gap can read the source voltage, depending on the stated circuit and ideal meter assumptions.
This result often surprises students who have learned the inaccurate rule that voltage is always zero where current is zero. Current and potential difference are different quantities. The 6 V source can maintain a potential difference across a gap even though there is no closed path for a sustained current. The correct explanation is that the open prevents charge flow through the original circuit, not that the source has stopped providing potential difference.
In a real measurement, source internal resistance, leakage, meter loading and imperfect connections complicate the picture. For an examination problem with ideal meters, use its stated assumptions. If the question offers non-ideal readings, do not force exact textbook values onto the data. Describe the discrepancy as a clue rather than automatically treating it as student calculation error.
Bypasses and short circuits: the wrong path can dominate
A bypass connects two nodes through an alternative path. If that path has a much lower resistance than the component it bypasses, more current tends to flow through the low-resistance route. In a simple idealised parallel arrangement, a zero-resistance connection across a resistor makes the potential difference across that resistor zero. The resistor receives no current because its voltage difference is zero. The consequence can be a large total current limited only by whatever other resistance and source properties remain.
This is not a harmless trick to demonstrate physically. A direct short across a power supply can draw dangerous current, damage wiring or equipment, and cause injury. In classroom answers, examine the hypothetical diagram; in practical lessons, use approved current-limited apparatus under supervision. Do not propose intentionally connecting a wire across a supply to test a theory.
An examination trap is to say that a shorted lamp is “broken in the same way” as an open filament. The observable result, no light, can be similar, but the circuit evidence differs. An open filament blocks that branch. A bypass provides another route for charge. If the lamp is in a series arrangement, an open can stop the entire circuit, while a bypass around that lamp can allow current through the remaining load. If the lamp is a parallel branch, the consequences depend on the exact common nodes.
Use the wording “effectively bypassed” when a low-resistance path shunts a component. Use “open circuit” when a required conducting path is broken. Reserve a claim of a physical short circuit for evidence that actually supports such a connection. The more accurately you name the change to the network, the fewer incorrect predictions you make about the other components.
The three-resistor benchmark model
Work with an ideal 12 V supply. A 6 Ω resistor, R1, is in series with a pair of parallel resistors, R2 = 12 Ω and R3 = 12 Ω. This one network will become our benchmark for several fault hypotheses. It has enough structure to test whole-circuit calculations while remaining simple enough to check.
First calculate the effective resistance of the two parallel resistors. Since 1/Rparallel = 1/12 + 1/12, their equivalent resistance is 6 Ω. Add the series resistor R1: total resistance is 12 Ω. The source current is I = 12/12 = 1 A. That full ampere passes through R1, so V1 = IR1 = 6 V. The parallel network therefore has 12 − 6 = 6 V across it. Each 12 Ω branch carries 6/12 = 0.5 A. At the junction, the currents recombine: 0.5 + 0.5 = 1 A.
These results give several cross-checks. The potential differences around the main route add to 12 V. Both parallel resistors share the 6 V potential difference. The branch currents add to the series current. All calculated resistances and currents are positive. If your answer fails any one of these tests, find the first equation that was applied to the wrong portion of the network.
Do not skip this fault-free calculation because the problem is called “fault finding.” There is no defensible diagnosis until you know what the normal circuit should produce. The baseline is not extra working; it defines what counts as abnormal. In a longer examination problem, a small table headed normal, proposed fault A and proposed fault B can prevent you from comparing mismatched situations.
Fault hypothesis A: R3 becomes open
Now assume R3 develops an open connection but R1 and R2 remain unchanged. The circuit is no longer a pair of conducting 12 Ω parallel branches. Only R2 remains as an available branch of that section. The total resistance becomes 6 + 12 = 18 Ω. Source current becomes 12/18 = 2/3 A, approximately 0.667 A. Current through R1 is the same. Its potential difference is (2/3) × 6 = 4 V. The remaining branch R2 receives 8 V and carries 8/12 = 2/3 A. Current in the open R3 branch is zero.
Compare those values with the benchmark. The source current falls from 1 A to about 0.667 A. The potential difference across R1 falls from 6 V to 4 V. Yet current through R2 increases from 0.5 A to about 0.667 A because more of the supply voltage appears across the surviving parallel branch. Many students predict that every current must fall when one branch fails. That intuition overlooks the way the series resistor and the equivalent branch resistance share the source potential difference.
The important result is conditional. If the parallel branches were connected directly across an ideal fixed-voltage supply, a fault in one branch would not change the voltage across the other branch. Here the parallel pair is fed through R1, so the voltage across the pair changes when its effective resistance changes. Circuit questions often test this difference. The same verbal phrase, “one parallel branch is opened,” cannot be used without noting what else lies in series.
State the fault only after connecting it to several observations. A lower source current together with a higher potential difference across the surviving branch supports this particular open-branch model. Neither observation alone uniquely identifies the failed component.
Fault hypothesis B: R3 is bypassed
Imagine instead an ideal wire bypasses R3 across the same two junctions. In the ideal model, those junctions are connected by zero resistance. They therefore have zero potential difference between them. R2, which is also across those junctions, has zero potential difference and zero current. The branch section has an effective resistance of zero in the simplified network, leaving only R1 = 6 Ω to limit current. The source current becomes 12/6 = 2 A, with the full 12 V across R1.
These values tell a completely different story from the open-branch fault. The source current rises rather than falls. The parallel section’s potential difference collapses rather than increasing. A meter across R2 reads zero, even though the R2 component itself has not necessarily failed. Saying “R2 is broken because its current is zero” would be an unsupported conclusion. The failure may be elsewhere in the network.
In a real circuit, a power source and its conductors are not perfectly ideal; protective devices may disconnect the circuit, temperature may rise and a low-resistance path may still have some resistance. The theoretical 2 A is therefore a calculation for the specified ideal circuit, not a prediction of what every physical device will safely do. If the question says a protective fuse has operated, that changes the active network again. Read the condition and reassess the circuit rather than applying the original ideal model blindly.
A useful distinction is between “no current in R2” and “no current from the source.” The first is a local statement about one path; the second concerns the entire supply. Fault-finding questions reward students who track these separately. It is possible for a component to receive no current precisely because a different path carries too much.
Meter evidence and measurement loading
An ammeter measures current in the path in which it is connected. A voltmeter measures potential difference between its two terminals. In school-level ideal circuit analysis, the ammeter is modelled as having negligible resistance and the voltmeter as having very high resistance. Connect the ammeter in series with the particular path whose current is required. Connect the voltmeter in parallel across the intended component or pair of nodes. Never connect an ammeter directly across a source as a practical test; its low resistance makes that connection hazardous.
Suppose a diagram gives a reading of 0.67 A through R1 and 8.0 V across R2 in the benchmark network. Those numbers agree closely with fault hypothesis A, an open R3 branch. The agreement has explanatory force because two independent predicted quantities line up. By contrast, a 2.0 A reading through R1 and zero volts across R2 fits the idealised bypass model, subject to the question’s source and protection assumptions.
A measurement can still be misleading if it was made across the wrong nodes. Ask the student to point to both voltmeter terminals on the diagram before using any formula. If a voltmeter is placed across R1 and the answer treats its reading as the voltage across the whole supply, the subsequent calculations may be internally neat but physically wrong. A source current reading also cannot be treated automatically as the current in each parallel branch.
Real meters can alter a circuit slightly, particularly when the circuit contains very high resistances or sensitive devices. This effect is called loading. At this level, the main requirement is to respect any provided meter assumptions; discussing loading is useful as an extension rather than an excuse to disregard the simpler circuit model. Your exam answer should be no more complicated than the evidence requires, but no simpler than the connections allow.
The fault table that separates competing explanations
When a fault question offers two or three candidate causes, use a compact qualitative table before writing a paragraph. Across the top, place the candidate conditions: normal; one branch open; one branch bypassed; main series path open. Down the side, place source current, current in the unaffected branch, potential difference across the branch network, and whether the other lamp remains on. Fill each cell with increases, decreases, stays the same or becomes zero, and use calculated values when available.
The table is a reasoning instrument, not an alternative to reasoning. For example, in the benchmark circuit, an open R3 branch gives source current 0.667 A and R2 current 0.667 A, while a bypass around R3 gives source current 2 A and R2 current zero. A source-current reading alone distinguishes these ideal cases, but a voltmeter reading adds independent confirmation and helps catch incorrect meter placement.
A question may also withhold enough evidence to make a unique diagnosis. Suppose only that “Lamp R3 is dark” is reported. You cannot conclude from darkness alone that its filament is open; a bypass, insufficient potential difference, switch position, protective disconnection or another fault could produce the same observation. A high-quality answer says what additional measurement would distinguish the possibilities, and why that measurement has diagnostic value.
The phrase “consistent with” is often more scientifically accurate than “proves.” A set of observations can be consistent with a model without uniquely establishing it. If the question asks you to suggest a possible cause, one plausible explanation is sufficient with its reasoning. If it asks you to deduce the only possible fault from complete data, eliminate the alternatives systematically. Read the command word before deciding how strongly to state your conclusion.
Question clinic 1: a lamp fails but the others remain lit
An original practice question states: Three identical lamps A, B and C are connected so that A is in series with the parallel combination of B and C. C goes dark, while A and B remain lit. What can you deduce? First, the source path through A still carries current. A complete circuit exists through at least one of the other branches. The failure is therefore not an open in the only series path through A.
Could C’s own branch be open? Yes. That leaves a route through B. Could some other arrangement or fault have caused C to receive little potential difference? Depending on the given circuitry, yes. Without measurements, the lamp observation may not establish a unique cause. Resist the temptation to claim that B must become dimmer. The brightness of B depends on how the circuit is arranged and on the lamp’s non-constant resistance with temperature.
Now the question supplies a simplified resistor model with all three lamps treated as fixed resistances and reports that supply current decreased. That supports an open C branch rather than a low-resistance bypass across the parallel section. If it additionally reports that the potential difference across B increased, the observation matches the behaviour of a parallel branch supplied through series resistance.
The exam-worthy skill is the logic of elimination: retain the faults consistent with the observed continuing path, test them against another independent reading, and state what the evidence can support. Do not turn a school circuit into a story about invisible broken parts. It is a model from which predictions can be derived and tested.
Question clinic 2: two measurements disagree
Another practice problem gives an ideal 10 V source, a 2 Ω resistor in series with a 3 Ω resistor, and states that the current is 2 A. A student records 6 V across the 2 Ω resistor. Which value should be challenged? By R = V/I, the 2 Ω resistor should have 2 × 2 = 4 V across it if 2 A passes through it. The 3 Ω resistor should have 6 V. The source voltage check is 4 + 6 = 10 V. The reported 6 V therefore matches the other resistor rather than the named one.
A weak answer says “the voltmeter is wrong.” That conclusion is premature. The meter may have been placed across the 3 Ω resistor instead; the label may have been copied incorrectly; the assumed ideal model may have been violated; or one measurement may be erroneous. The arithmetic identifies an inconsistency, not its precise cause.
An effective response first states the expected readings and the evidence gap. It then proposes a safe, meaningful check: verify which two nodes were used for the voltage measurement, check the component labels and, in a supervised low-voltage setup, repeat the appropriate measurement with the correctly configured instrument. If the exam simply asks what value is inconsistent, give the value and the calculation. If it asks for an improvement to the experimental method, state the action that would reduce ambiguity.
The lesson applies beyond electricity. Numbers become evidence only when the measurement quantity, position, unit and assumption are correct. An accurate calculation of the wrong physical quantity is not an accurate answer.
Question clinic 3: do not confuse a circuit model with a safety diagnosis
Suppose a textbook diagram shows a metallic case, a switch, a fuse and a supply. A candidate says, “The lamp went out, therefore the fuse blew because the current was too high.” A lamp can go out for several reasons; the observation does not demonstrate that the fuse operated. A fuse is a protective device intended to interrupt a circuit under excessive current, but the actual diagnosis requires evidence of the state of the fuse and circuit.
In the K323 practical-electricity section, students are expected to understand hazards such as damaged insulation, overheating cables and damp conditions, and the roles of fuses, circuit breakers, earthing and double insulation. The value of this knowledge is not just naming a device. Students should connect the device to the risk it addresses, and they should recognise the limits of what a drawing or a single symptom proves.
A school exam may ask you to explain why a fuse belongs in the live wire. A correct answer should describe how the fuse disconnects the live supply in an overcurrent situation, rather than claiming that any fuse makes all contact with electricity safe. Earthing and double insulation are different protective approaches. Neither makes unsafe experiments acceptable. The circuit’s potential difference, source characteristics and protective arrangement all matter.
In everyday life, do not troubleshoot damaged household mains equipment by touching exposed wiring, bypassing a fuse or re-energising a suspect circuit. Stop use and seek appropriate professional help. In the examination, keep the reasoning within the supplied scenario; when identifying an unsafe condition, explain the mechanism rather than providing a practical procedure for creating it.
From numerical answer to scientific explanation
A full-mark-style explanation should identify the changed path, connect that change to the electrical rule and then predict the observable consequence. For example: “When R3 becomes open, no current can pass through its branch. The remaining parallel section is effectively R2 only, so total resistance rises from 12 Ω to 18 Ω. At a fixed 12 V source, total current falls from 1 A to about 0.667 A.” This is better than the isolated phrase “resistance increases.”
When a question asks why current in the surviving branch has increased, go further: the series resistor now has a smaller potential difference, leaving a larger potential difference across R2. By V = IR at fixed R2, current through R2 rises. You may not need to write every numerical step if the command word asks for an explanation, but you do need the causal chain that makes the observation intelligible.
Separate observations from explanations. “The voltmeter reads 8 V” is evidence. “The increased branch voltage is consistent with R3 being open in the specified model” is an inference. “R3 has definitely broken” is a stronger claim that may not be supported by one reading. Small differences in words can represent large differences in scientific reasoning.
Avoid unsupported statements about brightness unless lamp behaviour is part of the model and sufficient information is supplied. A filament lamp is not necessarily an ohmic resistor at different temperatures. Its I–V relationship may be nonlinear. In a purely resistive numerical question, calculate using stated constant resistances; in a lamp question, qualify your explanation if the assumption changes. Good answers are faithful to the model they were given.
A disciplined 90-second examination method
In a timed structured question, spend your first moments locating the supply, junctions and return path. Mark series components with an S and branches with a P only after checking their nodes. Next, write the relevant rules alongside the diagram. Compute effective resistances in the smallest parallel groups first, then move outward. With the whole-circuit resistance found, calculate source current. Work back inward to find the potential difference and current of each branch.
When the question introduces a fault, do not carry all old values across. Redraw or mentally revise the topology. Recalculate the equivalent resistance under the new condition, because a change in one branch may alter the source current and the voltage shared with other components. If the question supplies meter readings, compare them with each hypothesis. State what the readings confirm, what they contradict, and what remains uncertain.
Reserve a few seconds for a check. Branch currents must recombine correctly; potential differences across genuinely parallel components must match; a result’s unit must belong to the quantity named. Two quick circuit-law checks often catch a major structural error.
For a practical or investigative item, record the relevant measured quantity and unit, explain how the setup supports a fair comparison, and identify one specific improvement if asked. The aim is not to perform a ritual list of “repeat readings” regardless of context. An improvement should repair a real weakness: unstable connection, wrong meter placement, uncertain scale reading, unrecorded starting conditions or an uncontrolled variable.
The method becomes faster when practised on unfamiliar drawings. You are training yourself to recognise the network, not memorising the decorative position of a lamp on a page.
Six original practice questions with worked answers
Practice 1. Two 4 Ω resistors are connected in series to an ideal 8 V source. Find the current. The total resistance is 8 Ω, so I = 8/8 = 1 A. Each resistor has 4 V across it. If one becomes an open circuit, the steady current is zero. The exact voltage across the break depends on the stated connection, but in the ideal simple series arrangement it can equal the source voltage.
Practice 2. Two 6 Ω resistors are connected in parallel across an ideal 12 V source. Find their total current. Each branch receives 12 V and carries 2 A. Total current is 4 A. If one branch opens, the other still carries 2 A because the source directly maintains 12 V across it. Do not use the 12 V network with a preceding series resistor as though it were the same circuit.
Practice 3. A 4 Ω resistor is in series with a 12 Ω resistor across 8 V. Find the potential difference across the 12 Ω resistor. Total resistance is 16 Ω, current is 0.5 A and the relevant potential difference is 6 V. Check that the other resistor receives 2 V, summing to 8 V.
Practice 4. A voltmeter placed across a branch reads 5 V. May the current in that branch be inferred without another value? Not in general. For a resistor of known constant resistance, I = V/R can be used. Without resistance or an appropriate component characteristic, the current remains undetermined.
Practice 5. A component is dark while the supply ammeter reads a nonzero current. Must the source be disconnected? No. At least some current is flowing through the measured path. The dark component may be in an open branch, bypassed, or otherwise receiving insufficient electrical power. The rest of the diagram determines the possibilities.
Practice 6. In the benchmark 12 V network, a total current of about 0.667 A and 8 V across R2 are measured. Which ideal fault hypothesis fits best: open R3 or bypassed R3? Open R3. Show the new total resistance 18 Ω and the remaining branch voltage 8 V. State that the agreement supports the model, rather than claiming it excludes every conceivable physical fault.
Practice schedule: repair the first weak link
On the first day, draw five networks with a source, one series component and two parallel branches. Do not calculate yet. Label each pair of common nodes and mark where currents split and recombine. The criterion for success is whether a second person can follow your connections without relying on where the lines sit on the paper.
On the second day, take one network and work out its normal current and voltage distribution. Change exactly one feature at a time: open a branch in the hypothetical diagram, move a component to another branch, or replace a stated resistance. Recalculate the whole network from the new condition. Compare your table with the original predictions. This trains the habit of treating a fault as a change of topology.
On the third day, do three short tasks in which the readings are provided but the fault is not. For every candidate fault, predict at least two observations. Explain why the chosen measurement distinguishes the candidates. When two faults make the same prediction, identify a different measurement that would separate them.
On the fourth day, write explanations in complete sentences. Use “because” to connect the physical principle to the observation; use “therefore” to connect the principle to the conclusion. Read each sentence and ask whether it refers to the whole circuit, a specific branch or a pair of nodes. Many apparently scientific errors are really failures to name the right part of the system.
On the fifth day, practise under short time limits. After marking, log not just wrong answers but wrong moves: confusing total with branch current, treating an open as a bypass, attaching a voltmeter reading to the wrong component, or claiming that darkness proves one fault. Correct each move with one contrasting example, then retry without notes. The goal is reliable reasoning, not a higher question count for its own sake.
How this chapter fits the English–Mathematics–Science route
The circuits are scientific models, but solving their questions also demands mathematical and language skills. The Mathematics of fractions, ratios, equations and measurement checks explains why the same source current can produce different branch voltages. The English of a precise explanation distinguishes an observation, a plausible inference and a demonstrated conclusion. Together those skills allow students to answer an unfamiliar question without treating every unfamiliar diagram as a new topic.
For the mathematical bridge, revisit Volume 0614 on current–voltage data and gradients and the Secondary Mathematics algebra guide. For clearer causal statements, see the English evidence and inference guide. For subject support, read the Secondary Physics owner rather than treating this one fault chapter as the entire Physics course.
Students bridging from Primary Six can return to the PSLE Science explanations route and the post-PSLE transition plan. Those earlier skills are not discarded when a student enters G3 work. A complete path, accurate observation and explanation of cause remain the foundations; the new work makes them more exact.
Continue through the G3 SEC learner-guide hub. The next Mathematics chapter will study whether records of experimental faults are being compared fairly; later chapters turn that evidence into responsible written conclusions.
Sources and final check
The official reference is the SEAB SEC overview and the 2027 G3 school-candidate syllabus index (updated 26 October 2026). For paper descriptions and the scope of circuits and investigations, consult the official 2027 K323 G3 Physics syllabus. Some students take combined Science rather than pure Physics: their exact subject code and assessment format must be checked separately against the official K326, K327 or K328 syllabus, not copied from K323.
Remember the core diagnostic sequence: model the normal circuit, change one path, predict two measurable consequences, compare the readings and state only what follows. Clear reasoning is the skill that transfers from one unfamiliar circuit to another. You do not need to guess the fault quickly. You need to find which explanation survives contact with the evidence.
