Physics · Electricity

Resistors in series and parallel

Ideal ohmic resistors in a lumped circuit model. Series requires common current without a branch; parallel requires the same two nodes. The strict parallel bound assumes at least two positive finite resistances.

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Formula

Calculate

R_{g es} = R_{1} + R_{2} \frac{1}{R _{g es}} = \frac{1}{R _{1}} + \frac{1}{R _{2}}

R_ges in ohms [Ω] · R₁ in ohms [Ω] · R₂ in ohms [Ω]

Variables & units

Variables & units
Symbol	Meaning	Unit
$R_{\mathrm{ges}}$	Total resistance (equivalent resistance)	Ω
R₁, R₂	Individual resistances	Ω

LaTeX

R_{ges} = R_1 + R_2 \qquad \frac{1}{R_{ges}} = \frac{1}{R_1} + \frac{1}{R_2}

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A 3 kilohm upper resistor in series with parallel 6 and 10 kilohm branches; two bars show 8 V unloaded and 6.67 V loaded on the same scale. — A 12 V divider with 3 kilohms above and 6 kilohms below is loaded by 10 kilohms. Output falls from 8 V to 20/3 V.

Equivalent resistance calculator

Combine resistors in series or in parallel: enter as many values as you need and the equivalent resistance appears straight away.

Calculation model

Resistance 1

Resistance 2

Resistance 3

Filled with example values. Type your own and the result updates straight away.

Working · rearrange for R, substitute, evaluate

1. Rearrange for R
$R_{\mathrm{eq}}=\sum_iR_i$
2. Substitute values (in SI units)
$R_{\mathrm{eq}}=\left(100\right)+\left(220\right)+\left(330\right)=650\,\Omega$
3. Result
R = 650 Ω

Result: Equivalent resistance R

R = 650 Ω

With 100 Ω, 220 Ω and 330 Ω, R = 650 Ω.

Quick tips

In parallel the equivalent resistance is always smaller than the smallest resistor, in series always larger than the largest.
Two equal resistors in parallel give half: 2 × 100 Ω in parallel is 50 Ω.

Assumptions and limits

At least two finite positive ohmic resistances carrying the same current, with no additional current-carrying branch at the connecting node. This does not solve an arbitrary mixed network.

Derivation & background

Mark electrically identical nodes; drawing proximity alone does not determine series or parallel. The same current flows through every resistor in an unbranched series connection. The loop rule adds voltage drops: $U=\sum_k U_k=I\sum_k R_k$ . Compare with $U=IR_{\mathrm{eq}}$ : in series, $R_{\mathrm{eq}}=\sum_kR_k$ . Parallel branches connect the same two nodes. With common voltage, the node rule gives $I=U\sum_k\frac1{R_k}$ . Hence $\frac1{R_{\mathrm{eq}}}=\sum_k\frac1{R_k}$ . Reduce only genuine series or parallel groups, then recover individual voltages and currents. Two parallel resistors also give $R_{\mathrm{eq}}=\frac{R_1R_2}{R_1+R_2}$ .

Exam blueprint

Validity range

Derivation steps

Common current adds voltage drops; common voltage adds branch currents.

1Mark electrically identical nodes; drawing proximity alone does not determine series or parallel.
2The same current flows through every resistor in an unbranched series connection.
3The loop rule adds voltage drops: $U=\sum_k U_k=I\sum_k R_k$ .
4Compare with $U=IR_{\mathrm{eq}}$ : in series, $R_{\mathrm{eq}}=\sum_kR_k$ .
5Parallel branches connect the same two nodes. With common voltage, the node rule gives $I=U\sum_k\frac1{R_k}$ .
6Hence $\frac1{R_{\mathrm{eq}}}=\sum_k\frac1{R_k}$ . Reduce only genuine series or parallel groups, then recover individual voltages and currents.

Rearrangements

Parallel resistance in product form

R_{ges} = \frac{R_1 \cdot R_2}{R_1 + R_2}

Holds for exactly two resistors and is quicker than the reciprocal form.

Missing resistor in a series circuit

R_2 = R_{ges} - R_1

This is how you size a series resistor.

Task variant

Two 60 Ω resistors are connected in parallel. Find R_total.

$R_{\mathrm{eq}}=\frac{(60\,\Omega)^2}{120\,\Omega}=30\,\Omega$ .
For n identical parallel resistors, $R_{\mathrm{eq}}=\frac Rn$ .

$R_{1} = 150 \Omega{}$ and $R_{2} = 50 \Omega{}$ in series at $U = 12 V$ . Find R_total and I.

$R_{\mathrm{eq}}=150\,\Omega+50\,\Omega=200\,\Omega$ .
Hence $I=\frac{12\,\mathrm V}{200\,\Omega}=0.06\,\mathrm A=60\,\mathrm{mA}$ .

Common mistakes

Leaving the reciprocal sum as the final answer for a parallel circuit.

Take the reciprocal at the end: $R_{\mathrm{eq}}=\frac1{\frac1{R_1}+\frac1{R_2}}$ .

Expecting the parallel resistance to lie between R₁ and R₂.

With at least two positive finite resistances, the parallel equivalent is smaller than every individual resistance. Open branches and ideal shorts are separate limiting cases.

Swapping the series and parallel rules.

Mnemonic: series = one path, resistances add; parallel = several paths, current splits, resistance drops.

Exam context

Mixed circuits are standard: first reduce subgroups, then find currents and partial voltages with $U = R\cdot{}I$ .

Formula cluster

Circuit analysis

The equivalent resistance is the key to every network problem.

U = R·I

Ohm's Law

P = U·I

Electric Power

C = Q/U

Capacitance of a Capacitor

Worked example

$R_{1} = 100 \Omega{}$ and $R_{2} = 300 \Omega{}$ : in series $R_{\mathrm{total}} = 400 \Omega{}$ . In parallel: $\frac{1}{R_{\mathrm{total}}} =\frac{ 1}{100} +\frac{ 1}{300} =\frac{ 4}{300}$ , so $R_{\mathrm{total}} = 75 \Omega{}$ .

Applications

Voltage dividers, household wiring (sockets in parallel), extending measuring ranges, load distribution in power supplies

How do you calculate with Resistors in Series and Parallel?

Here is how to work through a typical Resistors in Series and Parallel (Rges = R₁+R₂ ; 1/Rges = 1/R₁+1/R₂) task step by step:

1
Task
Two 60 Ω resistors are connected in parallel. Find R_total.
Solution path
1. $R_{\mathrm{eq}}=\frac{(60\,\Omega)^2}{120\,\Omega}=30\,\Omega$ .
2. For n identical parallel resistors, $R_{\mathrm{eq}}=\frac Rn$ .
2
Task
$R_{1} = 150 \Omega{}$ and $R_{2} = 50 \Omega{}$ in series at $U = 12 V$ . Find R_total and I.
Solution path
1. $R_{\mathrm{eq}}=150\,\Omega+50\,\Omega=200\,\Omega$ .
2. Hence $I=\frac{12\,\mathrm V}{200\,\Omega}=0.06\,\mathrm A=60\,\mathrm{mA}$ .

Understand and apply the equation

Connections, not positions on the page, determine whether resistors are in series or parallel. Common current gives the series sum; common voltage gives the parallel sum of conductances. After reducing the network, recover its individual voltages and currents. A loaded divider and a balanced bridge show why this last step matters.

How do you recognise series and parallel connections?

First mark points joined by ideal wires as one node. An ideal wire has no voltage drop. Two components are parallel only when both terminal nodes match. A series group must have no extra current branch at its shared intermediate node. This check prevents a simple-looking drawing from suggesting the wrong formula. Ohm’s law relates voltage, current and resistance within each ideal ohmic branch; the connections determine which quantities are equal between branches.[1]

Two resistors are in series if their common connection has no additional current-carrying branch: the same current traverses both. They are parallel if both ends connect to the same two nodes, giving equal voltage. Being drawn side by side is insufficient. Crossing lines without a connection are not a node. Ideal wires let each connected wire region be labelled as one node.[1][2]

Calculating equivalent resistance: series and parallel

Equivalent resistance describes behaviour at two external terminals. It does not yet specify voltage division inside the circuit. Keep track of the subgroups while reducing them, then work backwards using common voltage in parallel groups and common current in series groups. Kirchhoff’s junction and loop rules provide the current and voltage balances used here.[1][2]

\begin{aligned}R_{\mathrm s}&=\sum_jR_j\\{}\frac{1}{R_{\mathrm p}}&=\sum_j\frac{1}{R_j}\end{aligned}

Series: U=ΣIRj. Parallel: $I=\sum_j\frac{U}{R_j}$ . Both define the equivalent two-terminal resistance by $\frac UI$ .[1]

\begin{aligned}R_{1\parallel2}&=\frac{R_1R_2}{R_1+R_2}\\{}I_j&=\frac{U}{R_j}\end{aligned}

Positive finite ohmic resistances; the product form applies specifically to two resistors.[1]

Replacement preserves the external current-voltage relation, not all internal quantities automatically. After reduction, expand the equivalent group again. Three arbitrary parallel resistors are not replaced by their product divided by their sum. The reciprocal sum remains the reliable starting point.[1]

Why is parallel equivalent resistance smaller than every branch resistance?

\begin{aligned}R_{\mathrm p}<\min_jR_j\quad(n&\ge2,\ R_j>0)\\{}R_{\mathrm s}&\ge\max_jR_j\end{aligned}

The strict parallel bound requires at least two finite positive branches.[1]

With at least two positive finite resistances, the parallel equivalent is smaller than every branch resistance because positive conductances add. An additional conducting parallel path increases conductance. An open branch contributes zero conductance in the limit; an ideal short between the nodes makes equivalent resistance zero. For n equal resistors R, parallel gives $\frac{R}{n}$ and series gives $nR$ . These describe resistance, not whether a source maintains current when its load changes.[1]

Mixed network, loaded divider and bridge

After recovering branch currents and voltages, electrical power provides a check: in a steady ideal resistor network, power delivered by the source must equal total power dissipated in the resistors. The mixed network in the first example checks this balance.[1][3]

Series resistor followed by parallel branches

Given: Source with $U=9.0\,\mathrm V$ ; $10\,\Omega$ in series with the parallel combination of $15\,\Omega$ and $30\,\Omega$ .
Find: Find all branch currents, voltages and power.

Approach: Reduce the parallel group first, then add the series resistance.
$R_{\mathrm{tot}}=R_0+\frac{R_1R_2}{R_1+R_2}$
First replace only the true parallel group.
$\begin{aligned}R_{\mathrm p}&=\frac{15\cdot30}{15+30}=10\,\Omega\\{}R_{\mathrm{tot}}&=20\,\Omega\end{aligned}$
Find source current and group voltage.
$\begin{aligned}I&=\frac{9}{20}=0.45\,\mathrm A\\{}U_{\mathrm p}&=0.45\cdot10=4.5\,\mathrm V\end{aligned}$
Expand both branches using their common voltage.
$\begin{aligned}I_{15}&=0.30\,\mathrm A\\{}I_{30}&=0.15\,\mathrm A\\{}P_{10}&=2.025\,\mathrm W\\{}P_{15}&=1.350\,\mathrm W\\{}P_{30}&=0.675\,\mathrm W\end{aligned}$

Result: Total current 0.45 A; 4.5 V across each group; load power 4.05 W.

Check: Node: $0.30+0.15=0.45\,\mathrm A$ . Loop: $4.5+4.5=9\,\mathrm V$ . Powers: $2.025+1.350+0.675=9\cdot 0.45$ .

A measuring device loads the divider

Given: Source with $12\,\mathrm V$ ; upper resistor $R_1=3.0\,\mathrm{k\Omega}$ , lower resistor $R_2=6.0\,\mathrm{k\Omega}$ ; load $R_L=10\,\mathrm{k\Omega}$ across $R_2$ .
Find: Compare unloaded and loaded output.

Approach: The load parallels the lower divider resistor; recalculate the divider with that equivalent.
$\begin{aligned}R_{\mathrm{low}}&=\frac{R_2R_L}{R_2+R_L}\\{}U_{\mathrm{out}}&=U_{\mathrm{in}}\frac{R_{\mathrm{low}}}{R_1+R_{\mathrm{low}}}\end{aligned}$
Without a load both divider resistors carry the same current.
$U_{\mathrm{out},0}=12\frac6{3+6}=8\,\mathrm V$
With a load the lower branch is a parallel group.
$\begin{aligned}R_{\mathrm{low}}&=6\parallel10=3.75\,\mathrm{k}\Omega\\{}U_{\mathrm{out}}&=12\frac{3.75}{3+3.75}=6.6667\,\mathrm V\end{aligned}$
Check the branch currents.
$\begin{aligned}I_{\mathrm{upper}}&=1.7778\,\mathrm{mA}\\{}I_{6}&=1.1111\,\mathrm{mA}\\{}I_L&=0.6667\,\mathrm{mA}\end{aligned}$

Result: The load lowers output from 8.0 to about 6.67 V, by 16.7%.

Check: Source current is the sum of both lower currents. $R_L\to\infty$ restores 8 V; $R_L\to 0$ gives zero.

A bridge does not automatically contain the simple groups of the previous examples. The following original case first uses complete node balances to establish the special case of a current-free bridge branch.[1][2]

Simplify a bridge only after comparing potentials

Given: Two branches connect A at 9 V to D at 0 V: A–B is 100 Ω and B–D is 200 Ω; A–C is 150 Ω and C–D is 300 Ω. An additional 50 Ω resistor joins B and C. The source is ideal and resistors positive and ohmic.
Find: Find bridge current and resistance seen between A and D.

Approach: do not assume zero bridge current in advance. Test a common B/C potential in both complete node balances.
$\begin{aligned}\frac{9-V_B}{100}&=\frac{V_B}{200}+\frac{V_B-V_C}{50}\\\frac{9-V_C}{150}&=\frac{V_C}{300}+\frac{V_C-V_B}{50}\end{aligned}$
Both node equations are satisfied at 6 V; equal divider ratios explain the balance.
$\begin{aligned}V_B&=V_C=6\,\mathrm V\\I_{BC}&=\frac{V_B-V_C}{50\,\Omega}=0\\I_1&=\frac{9\,\mathrm V}{300\,\Omega}=0.030\,\mathrm A\\I_2&=\frac{9\,\mathrm V}{450\,\Omega}=0.020\,\mathrm A\end{aligned}$
Only after establishing zero current may the bridge be omitted for this excitation.
$\begin{aligned}I_s&=0.050\,\mathrm A\\R_{\mathrm{eq}}&=\frac{9\,\mathrm V}{0.050\,\mathrm A}=180\,\Omega\\P_s&=0.450\,\mathrm W\end{aligned}$

Result: The balanced bridge carries no current and equivalent resistance is 180 Ω. This is not a general series–parallel rule for arbitrary bridges.

Check: The power sum is $0.030^2\cdot300+0.020^2\cdot450=0.450\,\mathrm W$ . Recovering the node potentials gives 6 V at both. Changing one divider ratio requires solving bridge current again.

Explore further: Unknown branches and measurement loading

\begin{aligned}R_2&=\frac{R_{\mathrm p}R_1}{R_1-R_{\mathrm p}}\\{}0&<R_{\mathrm p}<R_1\end{aligned}

Unknown positive parallel branch; Rges≥R₁ permits no second finite positive branch.[1]

For the original 3 kΩ/6 kΩ divider, $R_{\mathrm{Th}}=(3\parallel6)\,\mathrm{k\Omega}=2\,\mathrm{k\Omega}$ . Reworking the same network gives $\frac{U_{\mathrm{out}}}{U_{\mathrm{out},0}}=\frac{R_L}{R_L+2\,\mathrm{k}\Omega}$ . To limit relative loading error to 5%, RL must be at least 38 kΩ. This is an ideal-model bound; an input with capacitance or frequency-dependent current is not described by one RL.[1]

A load lowers divider voltage

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The calculated curve is

U_L=8\,\mathrm V\frac{R_L}{R_L+2\,\mathrm{k\Omega}}

. The dashed line is the unloaded limit, not an exact value at finite load resistance.[1]

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What units do resistance and conductance use?

\begin{aligned}G&=\frac{1}{R}\\{}[G]&=\mathrm S=\Omega^{-1}\end{aligned}

Conductances add in parallel. Convert all resistances to one unit before summing.[1]

The introductory 100 Ω/300 Ω case gives 400 Ω in series and 75 Ω in parallel. Check $\frac{1}{75}=\frac{1}{100}+\frac{1}{300}$ . Reading a conductance sum as ohms produces a large error. Likewise, kΩ and Ω must not enter one addition as unconverted numerical values.[1]

Solve five topology exercises

Calculate · Foundation

90 Ω parallel to 180 Ω, then 40 Ω in series at 10 V. Find total current.

Show hint

Reduce internally, then add externally.

Learning goal: Reduce a mixed resistor circuit.

Show solution and calculation

Approach: reduce the genuine parallel branches, add the series resistance and recover the shared parallel voltage.
Rparallel=60 Ω; Rtotal=100 Ω; I=0.10 A.
The parallel group has 6 V, giving 0.0667+0.0333=0.10 A.

\begin{aligned}R_p&=\frac{90\cdot180}{90+180}\,\Omega=60\,\Omega\\R_{\mathrm{tot}}&=60\,\Omega+40\,\Omega=100\,\Omega\\I&=\frac{10\,\mathrm V}{100\,\Omega}=0.10\,\mathrm A\end{aligned}

Calculate · Application

Four identical 12 Ω resistors in parallel. Find equivalent and each branch current at 3 V.

Show hint

Equality simplifies the reciprocal sum.

Learning goal: Calculate identical parallel branches and current division.

Show solution and calculation

Approach: four identical parallel branches share voltage and split total current equally.
Rtotal=3 Ω; Itotal=1 A; each branch 0.25 A.
Four branch currents sum to total current.

R_{\mathrm p}=\frac{12}{4}=3\,\Omega

Calculate · Application

In the 9 V example the 15 Ω branch becomes open circuit. Find the new source current.

Show hint

An open branch carries no current.

Learning goal: Remove an open branch topologically.

Show solution and calculation

Approach: an open circuit removes the affected current path; then check the new connections.
Only 10+30=40 Ω remain; $I=\frac{9}{40}\,\mathrm A=0.225\,\mathrm A$ .
Total resistance rises, so current falls at the same supply voltage.

I=\frac{9}{10+30}=0.225\,\mathrm A

Interpret · Application

A schematic has nodes A, B, C. R1=10 Ω is between A and B, R2=20 Ω also between A and B, and R3=5 Ω between B and C. The source is between A and C. Which resistors are parallel? Describe the reduced topology and find equivalent resistance.

Show hint

Parallel means sharing both endpoint nodes.

Learning goal: Use node identity rather than visual proximity.

Show solution and calculation

Approach: R1 and R2 share both endpoints and form the parallel group.
Its equivalent is then in series with R3.
Overall resistance is eleven and two thirds ohms.
R2 and R3 share only one node and are not parallel; visual placement is irrelevant.

\begin{aligned}R_{AB}&=\frac{10\cdot20}{10+20}\,\Omega=\frac{20}3\,\Omega\\{}R_{AC}&=R_{AB}+5\,\Omega=\frac{35}3\,\Omega\end{aligned}

[1]

Error analysis · Application

A 100 Ω resistor is bridged between its terminals by an ideal wire. Someone adds 100 Ω in series to the wire. Find equivalent resistance and resistor voltage. Can a directly connected nonzero ideal voltage source determine finite current?

Show hint

An ideal wire makes both terminals the same potential.

Learning goal: Distinguish an ideal short circuit from an added finite parallel load.

Show solution and calculation

Approach: The bridge is parallel and enforces zero resistor voltage, hence zero resistor current.
Ideal equivalent resistance is zero.
A nonzero ideal voltage source across an ideal short is inconsistent and has no finite current solution.
Internal or wire resistance is needed to calculate physically limited current.

\begin{aligned}R_{\mathrm{eq}}&=100\,\Omega\parallel0\,\Omega=0\\{}U_R&=0,&I_R&=\frac{0\,\mathrm V}{100\,\Omega}=0\end{aligned}

[1]

Where simple reduction ends

Bridges without a genuine series/parallel group need node voltages or mesh currents. Linear resistors and ideal connections define this model. Diodes, heated lamps, active sources, line capacitance and inductive coupling may require other relations. Household sockets are conceptually parallel branches so devices receive the same supply; this alone does not establish allowable total load or protection and installation design.[1][2]

In brief

Check nodes before choosing a formula. Reduce only genuine series and parallel groups, check equivalent-resistance bounds and recover each branch. Adding a load changes the network and often its voltages.

Sources for this explanation

Quanta · Updated October 1, 2026

Editorial responsibility: Quanta Study editorial team

Review scope

This explanation, 3 worked examples and 5 exercises are available in German and English. Structure, source references and mathematical notation undergo automated checks. Calculations, model assumptions and both language versions receive a separate AI review.

An internal human subject review is documented for an earlier release of this page. That confirmation does not apply automatically to the revised explanation and new exercises.

An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

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Change history

October 1, 2026Made topology, equivalent-resistance calculations, the parallel bound and units accessible through direct answers and headings; linked Ohm, Kirchhoff and the existing networks’ power check in context.
October 1, 2026Added two exercises; labelled all five by task type, difficulty and learning goal.
October 1, 2026Added an accessible learning sequence, an additional model example, a quantitative diagram and a closing summary; retained existing examples, exercises and sources.

Cite this reference page

Quanta. Resistors in Series and Parallel. Updated October 1, 2026. https://quanta-study.de/en/formula-sheet/widerstand-reihen-parallel

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Frequently asked questions about Resistors in Series and Parallel

How do you calculate the total resistance of a series circuit?+

First check the connection: the shared node between series resistors must have no additional current-carrying branch. Both then carry the same current. Voltage drops add, and Ohm’s law gives the sum of resistances as their equivalent. The retained $\frac{100 \Omega{}}{300} \Omega{}$ case yields 400 Ω. This equivalent preserves the external current-voltage relation; branch drops are recovered afterwards using the common current. A branching measuring instrument can change the supposed series connection. Actual nodes matter, not the order in which symbols are drawn. Convert all resistor values to the same unit before addition.

How do you calculate the total resistance of a parallel circuit?+

Parallel means that every resistor shares the same two end nodes. Each then has the same voltage, while branch currents add. Add reciprocal resistances first, then take the reciprocal of that sum. The product-over-sum form applies directly only to two branches. The retained $\frac{100 \Omega{}}{300} \Omega{}$ case gives 75 Ω. Check each branch current with the common voltage and compare their sum with source current. A conductance sum has units siemens, not ohms. Open branches contribute no conductance; additional conducting branches reduce equivalent resistance for positive finite resistor values.

Why is the parallel resistance smaller than any single resistance?+

With at least two positive finite resistances, each added parallel branch contributes additional conductance. Total conductance therefore exceeds the conductance of the smallest resistor; its reciprocal is smaller than that resistance. This is a strict bound under those conditions. An open branch leaves the result unchanged in the limit, while an ideal short makes it zero. Parallel 15 Ω and 30 Ω branches, for example, give 10 Ω. Four identical 12 Ω resistors give 3 Ω. This describes the equivalent load; it does not claim that every real source maintains unchanged voltage as its load changes.

How do you approach mixed circuits?+

Label points joined by ideal wires as the same node. Reduce only groups that are genuinely series or parallel. In the original 9 V network, 10 Ω is in series with 15 Ω parallel to 30 Ω. The parallel group gives 10 Ω, total 20 Ω and source current 0.45 A. Expand the equivalent backwards: 4.5 V across both branches gives 0.30 A and 0.15 A. Node current, summed voltage and power balance independently check the result. A bridge without a reducible group instead needs node or mesh equations. Visual proximity or similar numerical values are not valid replacement rules.

Why are household sockets connected in parallel?+

Parallel supply gives devices the same voltage in the idealised model. One device can be switched off without creating a series interruption that stops current through every other device. Branch currents nevertheless add to the total supply current. Independent operation therefore differs from unlimited combined load. Real conductors have resistance, and protection and permissible operating conditions require additional consideration. Series/parallel formulas explain the circuit principle but do not provide complete installation design. A loaded voltage divider illustrates the same topology issue on a smaller scale: a new parallel branch changes the original current and output voltage.

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Subject references and further reading

This list names subject references and further reading. A page or section number appears when it is part of the reference; these are not verbatim quotations.

[1]OpenStax, University Physics Vol. 2, §10.2 Resistors in Series and Parallel

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Resistors in series and parallel

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Symbol

Meaning

Unit

R_{\mathrm{ges}}

Total resistance (equivalent resistance)

R₁, R₂

Individual resistances

Derivation & background

Exam blueprint

Validity range

Derivation steps

Common current adds voltage drops; common voltage adds branch currents.

1Mark electrically identical nodes; drawing proximity alone does not determine series or parallel.
2The same current flows through every resistor in an unbranched series connection.
3The loop rule adds voltage drops: $U=\sum_k U_k=I\sum_k R_k$ .
4Compare with $U=IR_{\mathrm{eq}}$ : in series, $R_{\mathrm{eq}}=\sum_kR_k$ .
5Parallel branches connect the same two nodes. With common voltage, the node rule gives $I=U\sum_k\frac1{R_k}$ .
6Hence $\frac1{R_{\mathrm{eq}}}=\sum_k\frac1{R_k}$ . Reduce only genuine series or parallel groups, then recover individual voltages and currents.

How do you calculate with Resistors in Series and Parallel?

Here is how to work through a typical Resistors in Series and Parallel (Rges = R₁+R₂ ; 1/Rges = 1/R₁+1/R₂) task step by step:

Task

Two 60 Ω resistors are connected in parallel. Find R_total.

Solution path

$R_{\mathrm{eq}}=\frac{(60\,\Omega)^2}{120\,\Omega}=30\,\Omega$ .
For n identical parallel resistors, $R_{\mathrm{eq}}=\frac Rn$ .

Task

$R_{1} = 150 \Omega{}$ and $R_{2} = 50 \Omega{}$ in series at $U = 12 V$ . Find R_total and I.

Solution path

$R_{\mathrm{eq}}=150\,\Omega+50\,\Omega=200\,\Omega$ .
Hence $I=\frac{12\,\mathrm V}{200\,\Omega}=0.06\,\mathrm A=60\,\mathrm{mA}$ .

Understand and apply the equation

How do you recognise series and parallel connections?

Calculating equivalent resistance: series and parallel

\begin{aligned}R_{\mathrm s}&=\sum_jR_j\\{}\frac{1}{R_{\mathrm p}}&=\sum_j\frac{1}{R_j}\end{aligned}

Series: U=ΣIRj. Parallel: $I=\sum_j\frac{U}{R_j}$ . Both define the equivalent two-terminal resistance by $\frac UI$ .[1]

\begin{aligned}R_{1\parallel2}&=\frac{R_1R_2}{R_1+R_2}\\{}I_j&=\frac{U}{R_j}\end{aligned}

Positive finite ohmic resistances; the product form applies specifically to two resistors.[1]

Why is parallel equivalent resistance smaller than every branch resistance?

\begin{aligned}R_{\mathrm p}<\min_jR_j\quad(n&\ge2,\ R_j>0)\\{}R_{\mathrm s}&\ge\max_jR_j\end{aligned}

The strict parallel bound requires at least two finite positive branches.[1]

Mixed network, loaded divider and bridge

Series resistor followed by parallel branches

Given: Source with $U=9.0\,\mathrm V$ ; $10\,\Omega$ in series with the parallel combination of $15\,\Omega$ and $30\,\Omega$ .
Find: Find all branch currents, voltages and power.

Approach: Reduce the parallel group first, then add the series resistance.
$R_{\mathrm{tot}}=R_0+\frac{R_1R_2}{R_1+R_2}$
First replace only the true parallel group.
$\begin{aligned}R_{\mathrm p}&=\frac{15\cdot30}{15+30}=10\,\Omega\\{}R_{\mathrm{tot}}&=20\,\Omega\end{aligned}$
Find source current and group voltage.
$\begin{aligned}I&=\frac{9}{20}=0.45\,\mathrm A\\{}U_{\mathrm p}&=0.45\cdot10=4.5\,\mathrm V\end{aligned}$
Expand both branches using their common voltage.
$\begin{aligned}I_{15}&=0.30\,\mathrm A\\{}I_{30}&=0.15\,\mathrm A\\{}P_{10}&=2.025\,\mathrm W\\{}P_{15}&=1.350\,\mathrm W\\{}P_{30}&=0.675\,\mathrm W\end{aligned}$

Result: Total current 0.45 A; 4.5 V across each group; load power 4.05 W.

Check: Node: $0.30+0.15=0.45\,\mathrm A$ . Loop: $4.5+4.5=9\,\mathrm V$ . Powers: $2.025+1.350+0.675=9\cdot 0.45$ .

A measuring device loads the divider

Given: Source with $12\,\mathrm V$ ; upper resistor $R_1=3.0\,\mathrm{k\Omega}$ , lower resistor $R_2=6.0\,\mathrm{k\Omega}$ ; load $R_L=10\,\mathrm{k\Omega}$ across $R_2$ .
Find: Compare unloaded and loaded output.

Approach: The load parallels the lower divider resistor; recalculate the divider with that equivalent.
$\begin{aligned}R_{\mathrm{low}}&=\frac{R_2R_L}{R_2+R_L}\\{}U_{\mathrm{out}}&=U_{\mathrm{in}}\frac{R_{\mathrm{low}}}{R_1+R_{\mathrm{low}}}\end{aligned}$
Without a load both divider resistors carry the same current.
$U_{\mathrm{out},0}=12\frac6{3+6}=8\,\mathrm V$
With a load the lower branch is a parallel group.
$\begin{aligned}R_{\mathrm{low}}&=6\parallel10=3.75\,\mathrm{k}\Omega\\{}U_{\mathrm{out}}&=12\frac{3.75}{3+3.75}=6.6667\,\mathrm V\end{aligned}$
Check the branch currents.
$\begin{aligned}I_{\mathrm{upper}}&=1.7778\,\mathrm{mA}\\{}I_{6}&=1.1111\,\mathrm{mA}\\{}I_L&=0.6667\,\mathrm{mA}\end{aligned}$

Result: The load lowers output from 8.0 to about 6.67 V, by 16.7%.

Check: Source current is the sum of both lower currents. $R_L\to\infty$ restores 8 V; $R_L\to 0$ gives zero.

Simplify a bridge only after comparing potentials

Given: Two branches connect A at 9 V to D at 0 V: A–B is 100 Ω and B–D is 200 Ω; A–C is 150 Ω and C–D is 300 Ω. An additional 50 Ω resistor joins B and C. The source is ideal and resistors positive and ohmic.
Find: Find bridge current and resistance seen between A and D.

Approach: do not assume zero bridge current in advance. Test a common B/C potential in both complete node balances.
$\begin{aligned}\frac{9-V_B}{100}&=\frac{V_B}{200}+\frac{V_B-V_C}{50}\\\frac{9-V_C}{150}&=\frac{V_C}{300}+\frac{V_C-V_B}{50}\end{aligned}$
Both node equations are satisfied at 6 V; equal divider ratios explain the balance.
$\begin{aligned}V_B&=V_C=6\,\mathrm V\\I_{BC}&=\frac{V_B-V_C}{50\,\Omega}=0\\I_1&=\frac{9\,\mathrm V}{300\,\Omega}=0.030\,\mathrm A\\I_2&=\frac{9\,\mathrm V}{450\,\Omega}=0.020\,\mathrm A\end{aligned}$
Only after establishing zero current may the bridge be omitted for this excitation.
$\begin{aligned}I_s&=0.050\,\mathrm A\\R_{\mathrm{eq}}&=\frac{9\,\mathrm V}{0.050\,\mathrm A}=180\,\Omega\\P_s&=0.450\,\mathrm W\end{aligned}$

Result: The balanced bridge carries no current and equivalent resistance is 180 Ω. This is not a general series–parallel rule for arbitrary bridges.

Check: The power sum is $0.030^2\cdot300+0.020^2\cdot450=0.450\,\mathrm W$ . Recovering the node potentials gives 6 V at both. Changing one divider ratio requires solving bridge current again.

Explore further: Unknown branches and measurement loading

\begin{aligned}R_2&=\frac{R_{\mathrm p}R_1}{R_1-R_{\mathrm p}}\\{}0&<R_{\mathrm p}<R_1\end{aligned}

Unknown positive parallel branch; Rges≥R₁ permits no second finite positive branch.[1]

A load lowers divider voltage

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The calculated curve is

U_L=8\,\mathrm V\frac{R_L}{R_L+2\,\mathrm{k\Omega}}

. The dashed line is the unloaded limit, not an exact value at finite load resistance.[1]

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What units do resistance and conductance use?

\begin{aligned}G&=\frac{1}{R}\\{}[G]&=\mathrm S=\Omega^{-1}\end{aligned}

Conductances add in parallel. Convert all resistances to one unit before summing.[1]

Solve five topology exercises

Calculate · Foundation

90 Ω parallel to 180 Ω, then 40 Ω in series at 10 V. Find total current.

Show hint

Reduce internally, then add externally.

Learning goal: Reduce a mixed resistor circuit.

Show solution and calculation

Approach: reduce the genuine parallel branches, add the series resistance and recover the shared parallel voltage.
Rparallel=60 Ω; Rtotal=100 Ω; I=0.10 A.
The parallel group has 6 V, giving 0.0667+0.0333=0.10 A.

\begin{aligned}R_p&=\frac{90\cdot180}{90+180}\,\Omega=60\,\Omega\\R_{\mathrm{tot}}&=60\,\Omega+40\,\Omega=100\,\Omega\\I&=\frac{10\,\mathrm V}{100\,\Omega}=0.10\,\mathrm A\end{aligned}

Calculate · Application

Four identical 12 Ω resistors in parallel. Find equivalent and each branch current at 3 V.

Show hint

Equality simplifies the reciprocal sum.

Learning goal: Calculate identical parallel branches and current division.

Show solution and calculation

Approach: four identical parallel branches share voltage and split total current equally.
Rtotal=3 Ω; Itotal=1 A; each branch 0.25 A.
Four branch currents sum to total current.

R_{\mathrm p}=\frac{12}{4}=3\,\Omega

Calculate · Application

In the 9 V example the 15 Ω branch becomes open circuit. Find the new source current.

Show hint

An open branch carries no current.

Learning goal: Remove an open branch topologically.

Show solution and calculation

Approach: an open circuit removes the affected current path; then check the new connections.
Only 10+30=40 Ω remain; $I=\frac{9}{40}\,\mathrm A=0.225\,\mathrm A$ .
Total resistance rises, so current falls at the same supply voltage.

I=\frac{9}{10+30}=0.225\,\mathrm A

Interpret · Application

A schematic has nodes A, B, C. R1=10 Ω is between A and B, R2=20 Ω also between A and B, and R3=5 Ω between B and C. The source is between A and C. Which resistors are parallel? Describe the reduced topology and find equivalent resistance.

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Parallel means sharing both endpoint nodes.

Learning goal: Use node identity rather than visual proximity.

Show solution and calculation

Approach: R1 and R2 share both endpoints and form the parallel group.
Its equivalent is then in series with R3.
Overall resistance is eleven and two thirds ohms.
R2 and R3 share only one node and are not parallel; visual placement is irrelevant.

\begin{aligned}R_{AB}&=\frac{10\cdot20}{10+20}\,\Omega=\frac{20}3\,\Omega\\{}R_{AC}&=R_{AB}+5\,\Omega=\frac{35}3\,\Omega\end{aligned}

[1]

Error analysis · Application

A 100 Ω resistor is bridged between its terminals by an ideal wire. Someone adds 100 Ω in series to the wire. Find equivalent resistance and resistor voltage. Can a directly connected nonzero ideal voltage source determine finite current?

Show hint

An ideal wire makes both terminals the same potential.

Learning goal: Distinguish an ideal short circuit from an added finite parallel load.

Show solution and calculation

Approach: The bridge is parallel and enforces zero resistor voltage, hence zero resistor current.
Ideal equivalent resistance is zero.
A nonzero ideal voltage source across an ideal short is inconsistent and has no finite current solution.
Internal or wire resistance is needed to calculate physically limited current.

\begin{aligned}R_{\mathrm{eq}}&=100\,\Omega\parallel0\,\Omega=0\\{}U_R&=0,&I_R&=\frac{0\,\mathrm V}{100\,\Omega}=0\end{aligned}

[1]

Where simple reduction ends

In brief

Sources for this explanation

Quanta · Updated October 1, 2026

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Change history

October 1, 2026Made topology, equivalent-resistance calculations, the parallel bound and units accessible through direct answers and headings; linked Ohm, Kirchhoff and the existing networks’ power check in context.
October 1, 2026Added two exercises; labelled all five by task type, difficulty and learning goal.
October 1, 2026Added an accessible learning sequence, an additional model example, a quantitative diagram and a closing summary; retained existing examples, exercises and sources.

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Quanta. Resistors in Series and Parallel. Updated October 1, 2026. https://quanta-study.de/en/formula-sheet/widerstand-reihen-parallel

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Frequently asked questions about Resistors in Series and Parallel

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