Resistors in Series and Parallel

Set a supply voltage and two or three resistances, then wire them one after another, side by side, or one in front of a parallel pair. The page works out the equivalent resistance, what flows through each branch, and where the power ends up.

Worked example — Resistors in Series and Parallel

Equivalent resistance	1,410.67 Ω
That total ÷ the smallest resistor in it	2.666673
Current out of the supply	0.1630 A
Power the network dissipates	37.500 W
The branch powers added up	37.500 W
Its share of the total power	62.50%

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Lesson

The theory — Resistors in Series and Parallel

Two resistors are in series when the same current has to go through both of them one after the other, and in parallel when each has the full supply voltage across it and the current divides between them. Series resistances add, R = R₁ + R₂ + R₃. Parallel ones add as reciprocals, 1/R = 1/R₁ + 1/R₂ + 1/R₃. The reciprocal of a resistance is its conductance, in siemens, and in that currency the second rule is the first one: conductances in parallel add exactly as resistances in series do.

What each symbol means

R: resistance of one component, in ohms. R_eq is what the whole network looks like from the supply's two terminals
G: conductance, 1/R, in siemens. This is what adds when branches are in parallel
V: voltage, in volts. Shared equally by parallel branches, divided between series ones in proportion to R
I: current, in amperes. Identical through every resistor in series, divided between parallel ones in proportion to G
P: power, in watts. VI, and equally I²R or V²/R — which form to reach for depends on which quantity the branches have in common

Assumes: Steady direct current, perfect wires, a supply with no resistance of its own, and resistances that hold still. The bulb figures lean on that last one harder than anything else here. A tungsten filament is a much better conductor cold than hot, so a 60 W bulb running at 23 W sits far below its rated temperature and its resistance drops below the 881.67 Ω used above. The exact 37.500 W is a textbook number rather than a bench measurement. Which bulb glows brighter survives it, because both filaments cool and the 60 W one keeps the larger resistance either way.
Breaks when: On anything that is not a resistor. A diode has no resistance to combine, only a curve relating its voltage to its current, so none of these rules reach it. Capacitors follow rules of the same shape with the words exchanged — they add in parallel and as reciprocals in series — but only for alternating current at one frequency, and RLC Resonance is where that begins. The rules also run out on their own ground: a Wheatstone bridge is five resistors that no amount of series and parallel reduction will simplify, which is the case Kirchhoff's node and loop equations were written for. Springs do the mechanical version, getting weaker end to end exactly as parallel resistors do, and Hooke's Law works that out in its own units.

Two bulbs in series, and the 60 W one is brighter 🖖

A rating is the power a bulb draws when it has the whole 230 V to itself, so it fixes a resistance: 529 Ω for the 100 W bulb and 881.67 Ω for the 60 W one. The dimmer bulb is the bigger resistor. Put them in series and there is one current through both, and power is I²R, so the bigger resistance takes the bigger share: 23.437 W against 14.062 W. The pair draws 37.500 W between them. Press The same two in parallel and the total is 160.000 W, the sum on the boxes, with the 100 W bulb back on top. Same two bulbs, same supply, 4.27 times the power.

Adding a branch in parallel can only take the total down 🖖

A parallel branch is another path, never a blockage, so the total can only fall — and it falls towards a floor it never reaches, the smallest branch. Ten ohms beside a kilohm comes to 9.90 Ω. Watch the row that divides the total by the smallest resistor in it: in parallel that row is below 1 whatever you type, and at this setting it reads 0.990099. The second panel is the same fact drawn. Put 100 Ω and 200 Ω in and sweep the third: a 1 MΩ branch moves that row from 0.666667 to 0.666622, while the identical sweep takes the series total from 300.10 Ω to 1,000,300.00 Ω.

The share swaps ends and the number stays put 🖖

Press 10 Ω beside 1 kΩ: the 10 Ω takes 99.01% of the power. Press Those two in series: 99.01% again, now held by the 1 kΩ. With two resistors that is exact, not close. Series splits the power as R₂/(R₁+R₂) because the current is shared and power is I²R; parallel splits it as G₁/(G₁+G₂) because the voltage is shared and power is V²/R. Write the second one out in resistances and it collapses into the first. The bulbs do it too, at 62.50% each way.

Problem solved in full

Three bulbs in one string, and the smallest rating wins 6 steps
A 100 W, a 60 W and a 40 W bulb, all rated for 230 V, are wired in series across a 230 V supply. Work out what the string draws and which bulb is the brightest, before reading on. Press 100 W and 60 W in series to watch the two-bulb version on the tool.

Open this problem in the tool
1. A rating is a promise about one bulb on its own at 230 V, so turn each one into a resistance first: 529.00, 881.67 and 1,322.50 Ω. The bulb with the smallest rating is the largest resistor.
  
  $P 100 W 60 W 40 W R = V^{2} / P 529.00 Ω 881.67 Ω 1, 322.50 Ω$
2. In series resistances add, and nothing else about the bulbs matters: 2,733.17 Ω across the three of them.
  
  $R_{eq} = 529.00 + 881.67 + 1, 322.50 = 2, 733.17 Ω$
3. One loop, one current, and every bulb carries the same 0.084151 A.
  
  $I = \frac{V}{R _{eq}} = \frac{230}{2 , 733 . 17} = 0.084151 A$
4. Power in each bulb is I²R, so the shares follow the resistances: 3.746 W, 6.244 W and 9.365 W. The 40 W bulb is taking 48.39% of the total.
  
  $R_{i} 529.00 Ω 881.67 Ω 1, 322.50 Ω P_{i} = I^{2} R_{i} 3.746 W 6.244 W 9.365 W P_{i} / \sum P 19 . 35% 32 . 26% 48 . 39%$
5. The three boxes add up to 200 W and the string draws 19.355 W, which is 9.68% of what is printed on them. Every bulb added to a series string makes the whole string dimmer, including the bulbs already in it.
  
  $\frac{19 . 355}{100 + 60 + 40} = \frac{19 . 355}{200} = 0.0968$
6. Now drop the resistances out. Substitute R = V²/P into the series sum and the supply voltage cancels from both halves, leaving 19.355 W built from the three ratings alone:
  
  $(\frac{1}{P _{1}} + \frac{1}{P _{2}} + \frac{1}{P _{3}})^{- 1} = (\frac{1}{100} + \frac{1}{60} + \frac{1}{40})^{- 1} = 19.355 W$
Answer

19.355 W from three bulbs whose boxes add up to 200, and the 40 W bulb is the brightest of the three at 9.365 W. The closed form in the last step is general: rated powers of bulbs in series combine exactly the way resistances in parallel do, with no resistance and no voltage left in the expression. It is the same reciprocal rule seen from the other side, and it says what a string of n identical bulbs does: the string draws a fraction 1/n of one bulb's rating, and each bulb in it gets 1/n² of what it was built for.

$(\frac{1}{100} + \frac{1}{60} + \frac{1}{40})^{- 1} = 19.355 W$

Learning path

What Ohm’s law leaves out

References (4)

The misconception this page is built against, measured on 145 students before and after instruction. Current being 'used up' along a circuit, and a second bulb in parallel changing the first, are both documented here: L. C. McDermott and P. S. Shaffer, "Research as a guide for curriculum development: An example from introductory electricity. Part I: Investigation of student understanding." American Journal of Physics 60(11), 1992, pp. 994-1003. DOI 10.1119/1.17003
The combination rules themselves, and the three forms of resistive power the branch table uses. The same edition is cited on Ohm's Law and RLC Resonance, deliberately: one circuits text across the chapter. J. W. Nilsson and S. A. Riedel, Electric Circuits, 10th ed. Pearson/Prentice Hall, 2015. ISBN 978-0-13-376003-3.
Where the lesson's `breaks` points next. A Wheatstone bridge cannot be reduced by series and parallel combination at all, and these are the equations written for the circuits that cannot: G. Kirchhoff, "Ueber den Durchgang eines elektrischen Stromes durch eine Ebene, insbesondere durch eine kreisförmige." Annalen der Physik 140(4), 1845, pp. 497-514. DOI 10.1002/andp.18451400402
The temperature dependence the lesson's `assumes` turns on, tabulated element by element. Tungsten is the entry behind the warning that a filament running below its rating is not the resistor its rating describes: P. D. Desai, T. K. Chu, H. M. James and C. Y. Ho, "Electrical Resistivity of Selected Elements." Journal of Physical and Chemical Reference Data 13(4), 1984, pp. 1069-1096. DOI 10.1063/1.555723

By the 42Math team · Reviewed against published sources. Corrections welcome.