Resistors in Series and Parallel

1. Series and parallel conditions

Resistors in series

Series components lie on the same unbranched path, so they carry the same current. Their potential differences add:

Rₑq = R₁ + R₂ + cdots

Resistors in parallel

Parallel branches connect between the same two nodes, so they have the same potential difference. Their branch currents add:

1/Rₑq = 1/R₁ + 1/R₂ + cdots

For two resistors only,

Rₑq = R₁R₂/R₁ + R₂.

2. Checks that catch errors

• A series equivalent resistance must be larger than every individual resistance.
• A parallel equivalent resistance must be smaller than the smallest branch resistance.
• The source current equals the sum of currents entering parallel branches.
• Potential difference is the same across every branch connected to the same two nodes.

3. Reducing a mixed network

1. Mark the nodes; components are parallel only if both ends share the same pair of nodes.
2. Combine the innermost clear series or parallel group.
3. Redraw the simpler circuit if the topology is not obvious.
4. Find the source current from I = mathcalE/Rₜₒₜₐₗ for an ideal source.
5. Work back through the network to find branch currents and p.d.s.

4. Worked examples

Example 1: Two resistors in seriesCore

A 4.0 Ω resistor and 8.0 Ω resistor are connected in series across a 12 V ideal source. Find the current and the p.d. across each resistor.

Show Answer

Rₑq = 4.0 + 8.0 = 12 Ω

I = 12/12 = 1.0 A

Therefore V₄Ω = IR = 4.0 V and V₈Ω = 8.0 V. The check is 4.0 + 8.0 = 12 V.

Example 2: Two resistors in parallelCore

A 6.0 Ω resistor and 3.0 Ω resistor are connected in parallel across 12 V. Find the equivalent resistance and branch currents.

Show Answer

Rₑq = (6.0)(3.0)/6.0 + 3.0 = 2.0 Ω

Each branch has 12 V across it, so I₆ = 12/6.0 = 2.0 A and I₃ = 12/3.0 = 4.0 A. The source current is 6.0 A, consistent with 12/2.0.

Example 3: A mixed series–parallel networkCore

A 4.0 Ω resistor is in series with a parallel pair of 6.0 Ω and 3.0 Ω. The network is connected to a 12 V ideal source. Find the source current.

Show Answer

The parallel pair has resistance 2.0 Ω. Hence

Rₜₒₜₐₗ = 4.0 + 2.0 = 6.0 Ω

Iₛₒᵤᵣcₑ = 12/6.0 = 2.0 A

5. Common mistakes

• Adding parallel resistances directly.
• Assuming series components have the same p.d.; they have the same current.
• Assuming parallel components have the same current; they have the same p.d.
• Using the two-resistor product-over-sum shortcut for three or more branches.
• Forgetting to include source internal resistance when the source is not ideal.

Next: Potential Divider Principle

8. Practice, Quiz and Next Step

Close your notes and use Resistors in Series and Parallel in the supplied context below. This requires a constructed explanation or working, not recognition of an option.

Fresh context: An unfamiliar data set or physical system requires you to apply Resistors in Series and Parallel while stating the model, regime and assumptions.

1. Retrieve: define resistors in series and parallel in your own words, including units, sign or conditions where relevant.
2. Represent: Choose and label an appropriate diagram, graph, table or symbolic model; derive or justify the relationship used.
3. Apply: Reach a conclusion, then evaluate it using units, uncertainty, a limiting case and one practical or modelling limitation.

Check the response before looking back

• The model, regime, coordinates and assumptions are explicit.
• The derivation or multi-step reasoning is visible rather than implied.
• The conclusion is tested against units, data quality and a limiting case.
• A practical control, uncertainty or model limitation is evaluated where applicable.

If one check fails, name that exact gap, revisit the matching explanation or worked example, and redo the task with different values or a different situation. Then use theA-Level Physics course hub orpractice browser for an independent re-test.