How to Perform in the new G3 SEC Examinations | Learner’s Guide Vol 0120 | Science Physics: Parallel-Circuit Workshop — Trace the Nodes Before You Divide the Current

In a parallel circuit, components belong to separate paths between the same pair of nodes. A node is a set of points joined by ideal conducting wire without a component between them. The shape of the drawing can change while the electrical connections remain the same.

For G3 Science Physics revision, start by labelling the two ends of every resistor. Two resistors are in parallel only if both corresponding ends connect to the same two nodes. Two resistors are in series when the same current must pass through them along a path with no branching between them. “Side by side” and “one above another” are drawing descriptions, not electrical definitions.

This workshop uses explicit path descriptions, so you can redraw the examples on paper. Mark source positive as node A and source negative as node B. Keep A and B distinct: joining them directly with ideal wire would bypass the resistors and create a short circuit, not another useful parallel branch.

Contents · Next chapter

Current is the rate of flow of charge. Potential difference is energy transferred per unit charge between two points. In a steady parallel circuit, current divides at a junction and recombines later. Charge does not disappear inside a resistor; the sum of the branch currents equals the current entering the junction.

Each branch connected between the same nodes A and B has the same potential difference across its ends. This does not mean it has the same current. For an ohmic resistor, I = V/R. At the same potential difference, a lower resistance carries a larger current.

Avoid saying that the current is “used up” or that every parallel branch receives half the supply current. An equal split occurs for two equal branch resistances under this ideal model. With unequal resistances, calculate each branch current from its own resistance and the shared potential difference. Name whether you mean the source current, one branch current or a potential difference before using a number.

Contents · Previous chapter · Next chapter

Connect an ideal 12 V source across nodes A and B. One branch from A to B contains a 6 Ω resistor. A second branch from A to B contains a 3 Ω resistor. Each resistor therefore has 12 V across it.

The 6 Ω branch carries I₁ = 12/6 = 2 A. The 3 Ω branch carries I₂ = 12/3 = 4 A. The current in the wire from the source before the split is I = 2 + 4 = 6 A. The 3 Ω branch carries twice the current because it has half the resistance at the same potential difference.

Find the effective resistance from the whole circuit: R = V/I = 12/6 = 2 Ω. The reciprocal rule gives the same result: 1/R = 1/6 + 1/3 = 1/2. Do not write R = 1/2 Ω; 1/R, rather than R, equals 1/2 in the reciprocal calculation. With positive finite resistances in parallel, the effective resistance is smaller than either individual resistance.

Contents · Previous chapter · Next chapter

For the previous circuit, an ideal ammeter in the source lead before the junction reads 6 A. An ideal ammeter inserted in the 6 Ω branch reads 2 A. One inserted in the 3 Ω branch reads 4 A. “The ammeter reads 6 A” is incomplete unless its position is known.

An ideal voltmeter connected across A and B reads 12 V. Across either resistor it also reads 12 V because each resistor spans those nodes. Across two points on the same uninterrupted ideal wire, it reads 0 V. A voltmeter measures between two locations; an ammeter measures current through its own location in the path.

Represent an ammeter in series with the path whose current is measured and a voltmeter in parallel across the component. In the ideal model, the ammeter has zero resistance and the voltmeter infinite resistance. These are calculation assumptions. They explain why placing an ammeter directly across a source is a short circuit rather than a voltage measurement.

Contents · Previous chapter · Next chapter

Open a switch in the 3 Ω branch while leaving the 6 Ω branch connected across the ideal 12 V source. The open branch carries zero current. The remaining branch still has 12 V across it and carries 2 A. Total source current falls from 6 A to 2 A; effective resistance rises from 2 Ω to 6 Ω.

The current in the surviving branch does not rise to “take over” all 6 A. The ideal source maintains the voltage, while the connected resistance determines the current drawn. If a question includes internal resistance or a source voltage that changes under load, the predictions require that additional model.

A short across A and B is a different change. It supplies a near-zero-resistance path between the source terminals. An ideal voltage source with an ideal zero-resistance short does not give a finite current through I = V/R. Do not treat an open circuit and a short circuit as interchangeable examples of something being disconnected.

Contents · Previous chapter · Next chapter

Place a 2 Ω resistor in series with a parallel pair of 6 Ω and 3 Ω resistors, all powered by an ideal 12 V source. Describe the paths precisely: current first passes through the 2 Ω resistor, then splits through the 6 Ω and 3 Ω branches, then recombines on the way back to the source.

The parallel pair has effective resistance 2 Ω. Adding the series resistor gives a total of 4 Ω. Source current is therefore 12/4 = 3 A. The potential difference across the series resistor is 3 × 2 = 6 V, leaving 6 V across the parallel network.

The branch currents are 6/6 = 1 A and 6/3 = 2 A. Their sum is 3 A, matching the current through the series resistor. Applying the full 12 V separately to these two branches would overlook the voltage across the series component. Check both relationships: branch currents sum correctly, and potential differences along a complete path sum to the source voltage.

Contents · Previous chapter · Next chapter

A: An ideal 9 V source supplies 3 Ω and 6 Ω resistors in parallel. Find both branch currents, the total current and effective resistance. B: Open the 6 Ω branch. State the new total current and the current in the surviving branch. C: An ideal 12 V source supplies a 4 Ω series resistor followed by two 8 Ω resistors in parallel. Find the source current and each branch current.

Answers: A gives 3 A and 1.5 A, total 4.5 A, effective resistance 2 Ω. B gives total 3 A; the surviving 3 Ω branch remains at 3 A because it still spans the ideal 9 V source. C has a parallel equivalent of 4 Ω and total resistance 8 Ω. Source current is 1.5 A, the parallel network has 6 V across it, and each branch carries 0.75 A.

D: A student says “parallel resistors always carry equal current”. Give a numerical counterexample from this article and the condition that would make the currents equal. The 6 Ω and 3 Ω branches at 12 V carry 2 A and 4 A. Equal branch resistance would give equal current at the shared voltage.

Contents · Previous chapter · Next chapter

Redraw the mixed circuit with one branch curved above the other, then redraw it rotated. Label the same nodes and predict all readings again. If your answers change while the connections remain identical, the drawing’s appearance is driving your reasoning. Return to the node labels.

Use a brief checking routine: identify the network, state shared voltages or shared currents, calculate, then check the junction and complete path. In a positive-resistance parallel pair, an effective resistance larger than both branches signals an error. In a mixed circuit, a branch voltage larger than the ideal source voltage also needs investigation.

For a parent or tutor, ask where the current divides and which two nodes define the branch voltage. Those questions reveal more than asking the student to recite a formula. Continue through the EMS guide for related measurement and explanation work. Circuit success means that the calculation remains attached to the actual connections from the first node label to the final unit.

Contents · Previous chapter · Continue learning