← Electricity and electromagnetism

Diagnosing a circuit fault or change

Every year on record, Question Two introduces a fault or a deliberate change partway through — a lamp fails, a rheostat is adjusted, a switch closes. This is the single most repeated question shape in the whole paper.

A four-step method for any fault question

  • 1. IDENTIFY the fault's location: is it in series or parallel with the component you are asked about?
  • 2. Recompute the TOTAL resistance of the circuit with the fault applied (a break removes a resistor from a sum or a parallel combination; a short-circuit sets that branch's resistance to ~0).
  • 3. Recompute the TOTAL current from the supply, using the new total resistance.
  • 4. Only THEN work out the effect on the specific branch or component asked about — its own current, voltage or power may be unchanged even though the total current changed.
Worked example A lamp fails in a mixed series-parallel circuit A 12 V supply drives a 4 Ω resistor in series with two 8 Ω lamps in parallel with each other. One lamp fails open. Describe the effect on the other lamp's brightness. BEFORE the fault: the two 8 Ω lamps in parallel give 4 Ω, in series with the 4 Ω resistor, total 8 Ω. Total current = 12/8 = 1.5 A. This splits evenly, so each lamp carries 0.75 A. AFTER the fault: with one lamp open, the parallel combination is now just the single working 8 Ω lamp, in series with the 4 Ω resistor — total 12 Ω. Total current = 12/12 = 1.0 A, and since it is a series circuit now, all 1.0 A flows through the remaining lamp. COMPARE: the remaining lamp's current rose from 0.75 A to 1.0 A, so it gets BRIGHTER, not unchanged and not dimmer. ✓ The remaining lamp gets brighter, because removing a parallel branch increases the total resistance seen by the whole circuit only when that branch was in parallel with the surviving one — the surviving lamp now carries the full current alone. ✕

Confusing "unaffected" (separate branch) with "affected" (same branch)

The two fault outcomes look similar on the page but are OPPOSITE: a fault in a branch that is genuinely SEPARATE (parallel to, but not sharing a node with, the component asked about) leaves that component's own voltage and current unchanged. A fault in a branch that SHARES a series or parallel relationship with the component asked about changes the total resistance the whole circuit sees, which changes everyone's current. Draw the circuit and trace which category applies before calculating — guessing from the question's wording alone is unreliable.

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