Series and Parallel Resistor Calculator
Add your resistors, pick series or parallel, and see the total straight away with the working shown. Advanced does capacitors, inductors, power and more.
Capacitors add up in parallel and use 1 ÷ C in series, the other way round to resistors.
Tap a unit to change it; the number converts for you. Empty boxes are skipped.
Find the resistor to add
Current, voltage and power in each resistor
Voltage divider (two resistors in series)
The power table and the voltage divider are for resistors. Switch back to Resistors to use them.
How your total is worked out
Add them, or add their 1 ÷ values and flip it back.
- 1
List the parts
100 Ω, 220 Ω,330 Ω3 resistors
- 2
Pick the rule
In series:R1 + R2 + R3= add them up
- 3
Work it out
100 + 220 + 330= 650 Ω
- 4
Check it
More than thelargest (330 Ω)= looks right
Read more: the formula, worked examples, capacitors, inductors and standard values
Parts joined in series sit end to end, so the same current flows through each one in turn. Parts joined in parallel sit side by side between the same two points, so each gets the same voltage and the current splits between them.
The formula
resistors in series: total = R1 + R2 + R3 + …
resistors in parallel: total = 1 ÷ (1 ÷ R1 + 1 ÷ R2 + 1 ÷ R3 + …)
two in parallel: total = R1 × R2 ÷ (R1 + R2)
capacitors in parallel: total = C1 + C2 + …
capacitors in series: total = 1 ÷ (1 ÷ C1 + 1 ÷ C2 + …)
inductors: the same rules as resistors
Worked example
Three resistors of 100 Ω, 220 Ω and 330 Ω in series: 100 + 220 + 330 = 650 Ω. That is more than the largest one (330 Ω), as a series total always is.
The same three in parallel: 1 ÷ 100 = 0.01, 1 ÷ 220 = 0.004545 and 1 ÷ 330 = 0.0030303. Using 3,300 as the common bottom number they are 33/3300 + 15/3300 + 10/3300 = 58/3300, so the total is 3,300 ÷ 58 = 56.90 Ω, less than the smallest one (100 Ω), as a parallel total always is.
Mixing units
Change everything to the same unit before you add. 4.7 kΩ + 470 Ω + 1 MΩ is 4,700 + 470 + 1,000,000 = 1,005,170 Ω, or about 1.005 MΩ. This calculator does that for you, so each part can have its own unit.
Capacitors and inductors
100 nF and 220 nF capacitors in parallel make 100 + 220 = 320 nF. In series they make 1 ÷ (1/100 + 1/220) = 100 × 220 ÷ 320 = 68.75 nF. Inductors work like resistors (when they are not coupled to each other): 10 µH + 22 µH = 32 µH in series and 10 × 22 ÷ 32 = 6.875 µH in parallel.
Finding the part to add
to go lower (resistor in parallel): Rx = R × target ÷ (R − target)
to go higher (resistor in series): Rx = target − R
To bring 1 kΩ down to 750 Ω: 1,000 × 750 ÷ (1,000 − 750) = 750,000 ÷ 250 = 3,000 Ω, a standard E24 value (3.0 kΩ). Check: 1,000 × 3,000 ÷ 4,000 = 750 Ω. For capacitors it is the other way: add one in parallel to go up and one in series to go down.
Voltage, current and power in each resistor
series: current I = V ÷ total; each resistor: V = I × R, P = I² × R
parallel: each resistor gets the full V; I = V ÷ R, P = V² ÷ R
On 12 V, 100 + 220 + 330 Ω in series draw 12 ÷ 650 = 18.46 mA. The 330 Ω one drops 0.01846 × 330 = 6.092 V and makes 0.01846² × 330 = 0.1125 W, under half of a 1/4 W rating. In parallel the 100 Ω one gets the full 12 V and makes 12 × 12 ÷ 100 = 1.44 W, more than five times a 1/4 W rating, so it needs a 2 W part. Picking a rating at least twice the heat is common practice, not a rule.
Voltage divider
Vout = Vin × R2 ÷ (R1 + R2)
R2 for a wanted Vout = R1 × Vout ÷ (Vin − Vout)
12 V with R1 = 10 kΩ and R2 = 4.7 kΩ: 12 × 4.7 ÷ 14.7 = 3.837 V, with 12 ÷ 14,700 = 0.8163 mA flowing. A load across R2 sits in parallel with it: a 10 kΩ load makes R2 act like 4.7 × 10 ÷ 14.7 = 3.197 kΩ, and Vout drops to 2.907 V.
Standard values
Resistors and capacitors are sold in the IEC 60063 preferred values: E12 (10, 12, 15, 18, 22, 27, 33, 39, 47, 56, 68, 82 in each decade, for ±10% parts) and E24 (24 values, for ±5%). 56.90 Ω is closest to 56 Ω in both, 1.6% low. 650 Ω is not a standard value, but 470 Ω + 180 Ω in series gives exactly 650 Ω from two E12 parts.
These are ideal values. Real parts have a tolerance (±5% is common), and resistance changes with temperature. Capacitor values are often ±10 to ±20%. Stay safe around mains voltage.
Common questions
How do you add resistors in series?
Add them up. 100 Ω, 220 Ω and 330 Ω end to end make 100 + 220 + 330 = 650 Ω. The same current flows through each one, and the total is always more than the largest resistor.
How do you work out resistors in parallel?
Add up 1 ÷ R for each resistor, then take 1 ÷ that sum. For 100 Ω, 220 Ω and 330 Ω: 1/100 + 1/220 + 1/330 = 58/3300, so the total is 3300 ÷ 58 = 56.9 Ω. For just two resistors you can use product ÷ sum: 1,000 × 2,200 ÷ 3,200 = 687.5 Ω.
Why are capacitors the other way round?
Capacitors side by side give more plate area, so in parallel their values add up: 100 nF + 220 nF = 320 nF. In series it is like a thicker gap, so you use the 1 ÷ rule: 100 nF and 220 nF in series make 100 × 220 ÷ 320 = 68.75 nF. Inductors follow the same rules as resistors.
What resistor do I put in parallel to get a lower value?
Use Rx = R × target ÷ (R − target). To bring 1 kΩ down to 750 Ω you need 1000 × 750 ÷ 250 = 3,000 Ω, which is a standard 3.0 kΩ E24 value. The target must be lower than the resistor you have; to go higher, add the difference in series.
What wattage do my resistors need?
In series every resistor carries the same current, so the power in each is I² × R. In parallel each gets the full voltage, so it is V² ÷ R. 100 Ω across 12 V makes 144 ÷ 100 = 1.44 W, far too much for a 1/4 W part. Choosing a rating at least twice the power is common practice.
What are two equal resistors in parallel?
Half of one of them. Two 1 kΩ resistors in parallel make 500 Ω, three make 333.3 Ω. In series they double: two 1 kΩ make 2 kΩ. That is a handy way to get values that are not in the standard series.
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