# Can I put multiple resistors in series to get higher resistance?

If I put 3 1MΩ resistors in series, is that the equivalent of one 3MΩ resistor?

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this question seems awfully simple... do you have some context that makes this answer less obvious? –  vicatcu Dec 17 '10 at 16:51
You should read an introduction to electronics. –  starblue Dec 17 '10 at 21:12
@vicatu: It may be simple to YOU, but everyone has to start somewhere - and the best start anyone interested in a subject can make is to ask questions. –  Linker3000 Dec 19 '10 at 18:21

The short answer: Yes. It's the same.

The current through resistors in series stays the same, but the voltage across each resistor can be different. The sum of the potential differences (voltage) is equal to the total voltage. To find their total resistance:

By Wikipedia

But this the ideal value. In the real word, resistors have tolerances and you have take in count. E.g., tolerance of 10%. The final value can vary between 2.7MΩ to 3.3MΩ.

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yeah, the tolerance part makes sense because it is multiplied as well. thanks for the answer! –  Matt Williamson Dec 17 '10 at 19:32

If you place resistors in series the total resistance is the sum of the resistances of all resistors in the chain. There are two main reasons why you would do this.

1. to obtain a value you don't have in you box. For instance a 30k resistor is not an E12 series value, but you can obtain it by putting two 15k resistors in series
2. to allow a higher voltage. 0603 resistors have a working voltage of 50V, So if you want to use them in a 70V circuit you'll have to place at least two of them in series. Care should be taken if you pick unequal values. Divide the voltage by the total resistance to get the current flowing through the resistors, and multiply this current by each of the resistances in turn to get the voltage over this resistor. In the 0603 example none of the voltages should exceed 50V.
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3. More power dissipation before the resistors burn. –  pingswept Dec 17 '10 at 18:32

Yes of course, and to add some marginally more interesting content to my answer, you can also put resistors in parallel to get a lower resistance, though the effective resistance is not quite as straightforward as the simple sum for series resistance. Parallel resistance is calculated as (R1 * R2) / (R1 + R2), which equals R/2 for R = R1 = R2...

More formally, for n resistors in series: R_effective = SUM(i=1, i=n, R_i)

...and for n resistors in parallel: R_effective = 1 / SUM(i=1, i=n, 1/R_i)

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