Yes, when they are in series, that means they are in a row. Think of it like the electricity is a group of lazy people, the wire is a hallway, and the resistors are hurdles. Everybody has to go over the same hurdles because they've got no other direction to go.
Rtotal = R1 + R2 + R3.....
Parallel resistors, the electricity can flow in more that one path. So now there are several hallways, with different heights of hurdles. Most of the lazy people are going to go for the short hurdles, but some of them, because they don't want to wait in line are going to go for the high ones. Not every body takes the same path.
In parallel, to figure out the total amount of hurdles jumped
1/Rtotal = 1/R1 + 1/R2 +1/R3....
2007-06-15 10:15:45
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answer #1
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answered by JC DDS 2
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Yes! 100 ohm in series with 100 ohm is equivalent to 200 ohm
2007-06-15 09:48:05
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answer #2
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answered by Babar 1
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Yes, although it is less accurate putting two resistors in series than two in parallel. You would do better to put two 400 ohm resistors in parallel.
2007-06-15 09:51:39
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answer #3
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answered by ooorah 6
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sequence connection of means resistors would not enable greater means distribution previous what one million might. to lessen contemporary, you may desire to be certain what voltage drop you want, i think of you're working it out incorrect. additionally, parallel connection of means resistors will multiply means dissipation with the aid of the style of resistors used, however the ohmic fee may well be greater. have you ever learnt the 10A will fall with the aid of fact the battery is charged? in case you rather need to lessen contemporary, i might hook up an ammeter in sequence, connect the charger, then with the aid of fact the battery fees, and the present drops to the utmost you want to apply, verify the terminal voltage. presented it grew to become into relatively 14V at 10A initially, subtract the hot voltage from 14, and that's the voltage parent you utilize at your needed contemporary, to artwork out a resistance fee, then means dissipation score. a pair of means transistors may well be used to offer a drop, utilising a potentiometer under pressure out in the backside circuit. That way, the present could be continuously replaced at will. in step with probability this may well be a greater effective decision?
2016-12-13 03:57:50
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answer #4
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answered by ? 4
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Yes.
In series resistors add.
Rtotal = R1 + R2.
In parallel, they do this..
1/Rtotal = 1/R1 + 1/R2.
2007-06-15 09:45:37
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answer #5
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answered by anotherbsdparent 5
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Yes.
2007-06-15 17:05:17
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answer #6
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answered by Scott S 4
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yes
2007-06-15 09:38:17
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answer #7
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answered by csmithballsout 4
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yes!
2007-06-15 09:44:00
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answer #8
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answered by Anonymous
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Why not?
2007-06-15 10:22:22
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answer #9
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answered by Anonymous
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