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Electronics College class help

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New Member
Hi folks, I'm new here (first post). I have a bit of a problem. I lost my job recently and am back at school after 30 years of printing newspapers. I am in retraining and one of my classes is "The World of Electricity". Problem is the instructor has a heavy accent and sometimes I don't grasp what he is saying untill 10 minutes later. My dilema is I need to figure out which mathematical formulas to plug into the problems he gives us. (Math is my weak area). Right now I need the formula for finding the voltage drop in 2 parallel resistors and voltage drop in 3 of more parallel resistors. I know this might sound simple for most of you, but I am a hands on learner. I have a proto board and a bunch of resistors, DMM and stuff to carry out labs at home. Thanks so much in advance for any help...BTW I did use the search button, but after 20 minutes of looking I think I needed to just ask.
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Well-Known Member
First make sure you are not meaning 'series' instead of 'parallel' in your question.

In parallel the voltage drop would be the same on each resistor.

In series, you have to calculate the current flowing along them all and apply Ohm's law; where individual voltage across each resistor = individual resistance value times current circulating in such resistor.


New Member
Thanks Externet. The instructor is using the Seiman's Step2000 Basic electricity 1&2 book from the internet with his own notes and problems and homework and labs added in.
I have the parallel/series thing down, but I have a couple of formulas. Series is Σ1+Σ2+Σ3+Σn=Σt (Kirchoff voltage law)
Is the parallel formula (Kirchoff's Current Law) 1/R1+1/R2+1/R3+1/Rn=Rt
So if I have R1=5Ω, R2=10Ω and R3=20Ω then Rt=1/5+1/10+1/20...find the lowest common denominator which is 20 so 4/20+2/20+1/20=7/20=.35Ω
If 2 resistors are used then it is Rt=R1×R2/R1+R2
I think I have that part down. Now the instructor gives us the problem R1is a series resistor that feeds into a parallel circuit R2 and R3
R1=25.25Ω,R2=45Ω,R3=3.35Ω Σ=100V
So if I grasp the concept right then R1+R2×R3/R2+R3=Rt 25.25+45×3.35/25.25+3.25=25.25+150.75/28.5=30.539Ω
NOW to get the voltage drop I use Ohm's Law Σ=I×R SO I have to figure out the current first...I=Σ/Rt...I=100v/30.539Ω=3.274A
THEN 3.27A/30.529Ω=.107V
This answer doesn't look correct :confused:


Well-Known Member
Most Helpful Member
jackofalltrades said:
It is actually: 1/R1+1/R2+1/R3+1/Rn=1/Rt

If you make that change, your numbers will make more sense.



New Member
Thanks, I'll try that! I'll also build it with a proto board and actually see it work. I am a very hands on learner! Give me a hunk of metal and a shop drawing and I'll build it...what I can't see is what I have trouble understanding.
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