United StatesPhysics 2Syllabus dot point

How do resistors combine in series and parallel, and how does that change the current and voltage?

Topic 11.5 Resistors in Series and Parallel: find the equivalent resistance of series and parallel combinations and the resulting currents and voltages.

A focused answer to AP Physics 2 Topic 11.5, covering the equivalent resistance of resistors in series and in parallel, how current and voltage divide in each arrangement, the reasoning behind the combination rules, and how to reduce a network step by step, with full worked examples.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-04

Reviewed by: AI editorial process; not yet individually human-reviewed

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Quick answer

Resistors in series carry the same current and their resistances add: $R_{series} = R_1 + R_2 + \dots$ . The voltage divides among them in proportion to their resistances, and the total resistance is larger than any one. Resistors in parallel have the same voltage across them, and their reciprocals add: $\dfrac{1}{R_{parallel}} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dots$ , so the equivalent resistance is smaller than the smallest branch. The current divides among parallel branches, with more current through the smaller resistance. To analyze a network, reduce it step by step: combine parallel groups, then series groups, until you reach a single equivalent resistance, find the total current from the source, then work back to find the current and voltage of each resistor. These two rules, plus Ohm's law, solve almost every circuit.

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What this topic is asking
Resistors in series
Resistors in parallel
Reducing a network
Try this

What this topic is asking

The College Board (Topic 11.5) wants you to find the equivalent resistance of resistors in series and in parallel, and to work out how the current and voltage divide in each, reducing a network step by step.

Resistors in series

In series there is a single path, so the current has nowhere else to go: it is the same through every resistor. Each resistor takes a share of the voltage given by $V = IR$ , so the largest resistor drops the most voltage, and the shares add up to the source voltage. Because every resistor adds to the obstacle, the total resistance is always larger than any single one. Adding a series resistor reduces the current everywhere.

Resistors in parallel

In parallel the resistors share two common nodes, so each feels the same voltage. Adding a parallel branch gives the current an extra route, which lowers the overall resistance, just as opening another checkout lane speeds the queue. The current splits between branches inversely to their resistances: the easy path (small $R$ ) carries more. The result is always less than the smallest branch, a useful sanity check.

Reducing a network

The strategy for any series-parallel network is a step-by-step reduction. Identify the innermost parallel or series groups, replace each with its equivalent resistance, and repeat until the whole network collapses to one resistor. Then find the total current from the source with $I = V/R_{total}$ , and work back outward: the current through any series section is the total current, while the voltage across any parallel block is found from $V = I R_{block}$ , and that voltage is shared by each parallel branch. Tracking which quantity is shared, current in series, voltage in parallel, is the single most important habit. The strategic role of this topic is that it turns a tangle of resistors into a solvable chain of Ohm's-law steps, and it sets up Kirchhoff's rules (Topics 11.6, 11.7) for networks that cannot be reduced by series and parallel alone.

A series-parallel network

A $24$ V source drives a $2.0$ -ohm resistor in series with two parallel resistors of $4.0$ ohms and $4.0$ ohms. Find the total resistance, the source current, and the current in each parallel branch.

step 1 Combine the parallel pair

$\dfrac{1}{R_p} = \dfrac{1}{4.0} + \dfrac{1}{4.0} = \dfrac{2}{4.0}$ , so $R_p = 2.0$ ohms.

step 2 Add the series resistor

$R_{total} = 2.0 + 2.0 = 4.0$ ohms.

step 3 Source current

$I = \dfrac{V}{R_{total}} = \dfrac{24}{4.0} = 6.0$ A. This flows through the $2.0$ -ohm series resistor.

step 4 Split between branches

Voltage across the parallel block: $V_p = I R_p = (6.0)(2.0) = 12$ V. Each $4.0$ -ohm branch carries $I = V_p/R = 12/4.0 = 3.0$ A, and the two add back to $6.0$ A.

Try this

Q1. Find the equivalent resistance of $3.0$ ohms and $6.0$ ohms in series. [1 point]

Cue. $R = 3.0 + 6.0 = 9.0$ ohms.

Q2. Find the equivalent resistance of $3.0$ ohms and $6.0$ ohms in parallel. [2 points]

Cue. $R = \dfrac{(3.0)(6.0)}{3.0 + 6.0} = \dfrac{18}{9.0} = 2.0$ ohms.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2024 (style)7 marksSection II (long FRQ). A

12

V battery (negligible internal resistance) is connected to a

4.0

-ohm resistor in series with a parallel combination of a

6.0

-ohm and a

3.0

-ohm resistor. (a) Calculate the equivalent resistance of the parallel pair. (b) Calculate the total resistance and the current from the battery. (c) Calculate the voltage across the parallel combination.

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A 7-point FRQ on a series-parallel network.

(a) Parallel pair (2 points): $\dfrac{1}{R_p} = \dfrac{1}{6.0} + \dfrac{1}{3.0} = \dfrac{1}{6.0} + \dfrac{2}{6.0} = \dfrac{3}{6.0}$ , so $R_p = 2.0$ ohms.
(b) Total and current (3 points): in series with $4.0$ ohms, $R_{total} = 4.0 + 2.0 = 6.0$ ohms. The battery current is $I = \dfrac{V}{R_{total}} = \dfrac{12}{6.0} = 2.0$ A.
(c) Parallel voltage (2 points): the $2.0$ A flows through the parallel block, so $V_p = I R_p = (2.0)(2.0) = 4.0$ V.

Markers reward the parallel combination, the total resistance and current, and the voltage across the parallel block.

AP 2023 (style)1 marksSection I (multiple choice). Two identical resistors are connected in parallel. How does their equivalent resistance compare with one resistor alone? (A) twice as large (B) half as large (C) the same (D) four times as large. Justify your reasoning.

Show worked answer →

A 1-point MCQ on parallel resistance. The answer is (B).

For two equal resistors $R$ in parallel, $\dfrac{1}{R_p} = \dfrac{1}{R} + \dfrac{1}{R} = \dfrac{2}{R}$ , so $R_p = R/2$ , half of one. Adding a parallel path gives the current more routes, lowering the resistance. The trap is (A): parallel resistance is always less than the smallest branch, not more.

Related dot points

Sources & how we know this

AP Physics 2: Algebra-Based Course and Exam Description — College Board (2024)