College Board AP 2023 (style)-style practice: Section I (multiple choice). Three 6\, resistors are connected in parallel. Their equivalent resistance is (A) 18\, (B) 6\, (C) 3\, (D) 2\,. Justify your reasoning.

A 1-point MCQ on parallel resistors. The answer is (D). 1R_eq = 16 + 16 + 16 = 36 = 12, so R_eq = 2\,. For n equal resistors in parallel, R_eq = R/n = 6/3 = 2\,. The trap is (A): adding them (the series rule) instead of combining reciprocals.

College Board AP 2024 (style)-style practice: Section II (FRQ, quantitative). A 12 V battery (negligible internal resistance) connects to a 4.0\, resistor in series with a parallel pair of 6.0\, and 3.0\, resistors. (a) Find the equivalent resistance. (b) Find the total current from the battery. (c) Find the current in the 6.0\, resistor.

A 5-point FRQ on reducing a compound network. (a) Equivalent resistance (2 points): parallel pair 1R_p = 16 + 13 = 12, so R_p = 2.0\,. In series with 4.0\,: R_eq = 4.0 + 2.0 = 6.0\,. (b) Total current (1 point): I = VR_eq = 126.0 = 2.0 A. (c) Current in the 6.0\, (2 points): voltage across the parallel pair = IR_p = (2.0)(2.0) = 4.0 V. So I_6 = 4.06.0 = 0.67 A. Markers reward the reduction, the total current, and using the parallel voltage to split the branch current.

United StatesPhysics C: Electricity and MagnetismSyllabus dot point

How do you reduce a network of series and parallel resistors to find currents and voltages?

Topic 11.5 Compound Direct Current Circuits: combine resistors in series and parallel to find equivalent resistance, currents and voltages in multi-resistor networks.

A calculus-based answer to AP Physics C E&M Topic 11.5, covering series and parallel resistor rules, equivalent resistance, reducing networks step by step, and voltage and current dividers.

Generated by Claude Opus 4.810 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: $R_{eq} = R_1 + R_2 + \cdots$ . Resistors in parallel share the same voltage and their reciprocals add: $\dfrac{1}{R_{eq}} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \cdots$ (for two, $R_{eq} = \dfrac{R_1 R_2}{R_1 + R_2}$ ). To analyze a network, reduce it step by step to one equivalent resistance, find the total current from $I = \dfrac{V}{R_{eq}}$ , then expand back, using each combination to find branch currents and voltages. A voltage divider splits a series voltage in proportion to resistance; a current divider splits a parallel current in inverse proportion to resistance. (Resistors combine oppositely to capacitors.)

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What this topic is asking
Series resistors
Parallel resistors
Reduce, then expand
Try this

What this topic is asking

The College Board (Topic 11.5) wants you to analyze compound DC circuits of several resistors by combining them in series and parallel to find an equivalent resistance, then work back to the current and voltage in each element. This is the standard toolkit before the more general Kirchhoff's rules.

Series resistors

Because the same current passes through each, the voltages add ( $V = IR_1 + IR_2 + \cdots$ ), giving $R_{eq} = \sum R_i$ . Series always increases the total resistance above the largest single resistor. The voltage splits in proportion to resistance, which is the voltage-divider rule: across two series resistors, the fraction of the total voltage on $R_1$ is $\dfrac{R_1}{R_1 + R_2}$ . A resistor that is twice as large drops twice the voltage, because the same current flows through both and $V = IR$ .

Parallel resistors

Because the same voltage drives each, the currents add ( $I = \dfrac{V}{R_1} + \dfrac{V}{R_2} + \cdots$ ), giving the reciprocal rule. Parallel always decreases the total resistance below the smallest single resistor. For two resistors, the handy form is $R_{eq} = \dfrac{R_1 R_2}{R_1 + R_2}$ . The current splits in inverse proportion to resistance, the current-divider rule: more current takes the path of lower resistance, so the smaller resistor in a parallel pair carries the larger share. For $n$ equal resistors in parallel, $R_{eq} = R/n$ , a fast result worth recognizing.

Reduce, then expand

The method for any series-parallel network:

Reduce. Replace each clearly series or parallel group with its equivalent, repeating until one resistance remains.
Total current. $I = \dfrac{V}{R_{eq}}$ from the source.
Expand back. Step outward, applying $V = IR$ to each combination: series elements share the current, parallel elements share the voltage.

Reducing a three-resistor network

A $9.0$ V battery (no internal resistance) drives a $2.0\,\Omega$ resistor in series with a parallel pair of $4.0\,\Omega$ and $4.0\,\Omega$ . Find the total current and the voltage across the parallel pair.

step 1 Combine the parallel pair

Two equal $4.0\,\Omega$ in parallel: $R_p = \dfrac{4.0}{2} = 2.0\,\Omega$ .

step 2 Add the series resistor

$R_{eq} = 2.0 + 2.0 = 4.0\,\Omega$ .

step 3 Find the total current

$I = \dfrac{V}{R_{eq}} = \dfrac{9.0}{4.0} = 2.25$ A from the battery.

step 4 Find the voltage across the parallel section

This current flows through the series $2.0\,\Omega$ and then the parallel block. Voltage across the parallel pair $= IR_p = (2.25)(2.0) = 4.5$ V, so each $4.0\,\Omega$ branch carries $\dfrac{4.5}{4.0} = 1.13$ A (summing to $2.25$ A, as a check).

Try this

Q1. Two $8.0\,\Omega$ resistors are in series. Find the equivalent resistance. [1 point]

Cue. $R_{eq} = 8.0 + 8.0 = 16\,\Omega$ .

Q2. A $12\,\Omega$ and a $6.0\,\Omega$ resistor are in parallel. Find the equivalent resistance. [2 points]

Cue. $R_{eq} = \dfrac{(12)(6.0)}{12 + 6.0} = \dfrac{72}{18} = 4.0\,\Omega$ .

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 2023 (style)1 marksSection I (multiple choice). Three

6\,\Omega

resistors are connected in parallel. Their equivalent resistance is (A)

18\,\Omega

(B)

6\,\Omega

(C)

3\,\Omega

(D)

2\,\Omega

. Justify your reasoning.

Show worked answer →

A 1-point MCQ on parallel resistors. The answer is (D).

$\dfrac{1}{R_{eq}} = \dfrac{1}{6} + \dfrac{1}{6} + \dfrac{1}{6} = \dfrac{3}{6} = \dfrac{1}{2}$ , so $R_{eq} = 2\,\Omega$ . For $n$ equal resistors in parallel, $R_{eq} = R/n = 6/3 = 2\,\Omega$ . The trap is (A): adding them (the series rule) instead of combining reciprocals.

AP 2024 (style)5 marksSection II (FRQ, quantitative). A

12

V battery (negligible internal resistance) connects to a

4.0\,\Omega

resistor in series with a parallel pair of

6.0\,\Omega

and

3.0\,\Omega

resistors. (a) Find the equivalent resistance. (b) Find the total current from the battery. (c) Find the current in the

6.0\,\Omega

resistor.

Show worked answer →

A 5-point FRQ on reducing a compound network.

(a) Equivalent resistance (2 points): parallel pair $\dfrac{1}{R_p} = \dfrac{1}{6} + \dfrac{1}{3} = \dfrac{1}{2}$ , so $R_p = 2.0\,\Omega$ . In series with $4.0\,\Omega$ : $R_{eq} = 4.0 + 2.0 = 6.0\,\Omega$ .
(b) Total current (1 point): $I = \dfrac{V}{R_{eq}} = \dfrac{12}{6.0} = 2.0$ A.
(c) Current in the $6.0\,\Omega$ (2 points): voltage across the parallel pair $= IR_p = (2.0)(2.0) = 4.0$ V. So $I_{6} = \dfrac{4.0}{6.0} = 0.67$ A.

Markers reward the reduction, the total current, and using the parallel voltage to split the branch current.

Related dot points

Sources & how we know this

AP Physics C: Electricity and Magnetism Course and Exam Description — College Board (2024)