If I put 3 1MΩ resistors in series, is that the equivalent of one 3MΩ resistor?
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asked Dec 17, 2010 at 15:34
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5\$\begingroup\$ this question seems awfully simple... do you have some context that makes this answer less obvious? \$\endgroup\$– vicatcuDec 17, 2010 at 16:51
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1\$\begingroup\$ You should read an introduction to electronics. \$\endgroup\$– starblueDec 17, 2010 at 21:12
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8\$\begingroup\$ @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. \$\endgroup\$– Linker3000Dec 19, 2010 at 18:21
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2\$\begingroup\$ @Linker3000 nope, the best start anyone interested in a subject can make is to read books and/or articles first, then search for relevant data there, then think about what they read, and then and only then ask questions. You learn by cognitive introspection, by searching for answers in existing sources first - not by mindlessly asking every question you can imagine to everybody. \$\endgroup\$– user20088Mar 26, 2015 at 2:01
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4 Answers
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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:
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.
answered Dec 17, 2010 at 15:38
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\$\begingroup\$ yeah, the tolerance part makes sense because it is multiplied as well. thanks for the answer! \$\endgroup\$ Dec 17, 2010 at 19:32
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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.
- 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
- 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.
answered Dec 17, 2010 at 16:37
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1\$\begingroup\$ 3. More power dissipation before the resistors burn. \$\endgroup\$ Dec 17, 2010 at 18:32
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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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If I put 3 1MΩ resistors in series, is that the equivalent of one 3MΩ resistor?
I don't know exactly whether your question contains a trap.
If you put at most 2 resistors in super cold environment and the remaining one(s) in hotter place, the combined resistance will probably no longer equal to 3 mega ohms.
answered Dec 18, 2010 at 0:45
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3\$\begingroup\$ A question on E&R wouldn't be a trap. If it was, it would be edited by another member to remove the confusion. Questions have to be as clear as possible. \$\endgroup\$– tybluDec 18, 2010 at 0:54
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\$\begingroup\$ +1 for noticing that different operating conditions can affect the values of Rs and make the "equal" R values be unequal in reality... it's a common caveat in power circuit's design, where linear parts laying close to Joule heat sources can (and usually will) exhibit nonlinear behaviour, making it necessary to compensate in some way (e.g. by pairing NTC/PTC parts or by introducing a thermal feedback loop) for thermal value drift. \$\endgroup\$– user20088Mar 26, 2015 at 2:04