# Calculating parallel resistances

Problem: Parallel resistances can be are hard to understand because your dealing with a ratio not a sum. If you had two 10 ohm resistors connected in parallel, the total resistance accross the two would be 5ohms – this much is fairly easy to grasp because both resistors would share the current equally. But as soon as the values are different, each resistor handles a different share of the current.

Solution: To work out the total resistance accross 2 or more resistors connected in parallel you need this formula:

Rtotal = 1 / ( (1/R1) + (1/R2) + (1/Rn) )

Note: (1/Rn) is repeated for each additional resistor in parallel, so for 4 resistors the formula would be:

Rtotal = 1 / ( (1/R1) + (1/R2) + (1/R3) + (1/R4) )

Example: The other day I was asked if an engine temperature sender could be wired in parallel to a resistor to make a miss-matched temperature gauge read properly (short answer: NO). So lets substitute some real world values from this example problem to see how the above formula works:
R1 is the resistance of the sender at a given temp, R2 is the resistor to be used in parallel. So substitute 400ohms for R1 and (for arguments sake) 39ohms for R2:

1) Rtotal = 1 / ( (1/400) + (1/39) )
2) Rtotal = 1 / ( (0.0025) + (0.0256) )
3) Rtotal = 1 / 0.0281
4) Rtotal = 35.6 ohms

If you play about with some values you’ll soon notice two things:
– Rtotal will never be higher than the smallest Resistor used. i.e with setup above, no matter what resistance the sender reads, Rtotal will never be over 39ohms.
– Rtotal will always be at most half of the value of the highest resistor used. With above info, whatever you choose for R2, Rtotal will always be 200ohms or less.

This could get complicated if we go into WHY the person wanted to change the working range of the sender. This post if just intended to give a real world example of parallel resistance. Please question, correct or comment below if you want 🙂

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