making combinations of series and/or parallel circuits. What are these 4 ways, and what is the net resistance in each case?

### 5 Answers

240/3 = 80Ω (all 3 in parallel)

240×3 = 720Ω (all 3 in series)

240/2+240 = 360Ω (2 in parallel with third in series)

240×2 = 480

480×240/(240+480) = 160Ω (2 in series all in parallel with third)

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240 Ohm Resistor

Right, you can have them connected in straight off series (3 in a row) giving R=240+240+240

—-===—–===—-===–

Or 3 in parallel in which case

1/R = 1/240 + 1/240 + 1/240

—===—

–===–/

-===-/

Or you could have two in parallel, which is then attached to another.

Giving R=240+r where 1/r = 1/240+1/240

–===—-===–

-===-/

Finally you could have a situation where there’s 2 in series, in parallel with another

This would give 1/R = 1/240 + 1/480

—===–===—-

—–===——-/

Do the maths 🙂

All 4 in series, 2 in series with 2 in parallel in series, 1 in series with 3 in parallel, all 4 in parallel. If you can think of any other possible configurations, you tell me! I say 4.

all parallel, all series, two in series parallel with another, two in parallel in series with another.

equivalent resistance in parallel = 1/R1 + 1/Rn (add reciprocals)

equivalent resistance in series = R1+Rn (just add’em up)