Parallel cables

[me1]

Help with calculation please.

If separate cable measure : 10Ω, 5Ω, 5Ω, 4Ω and 2Ω what would it be in parallel?

I get this so far


1 1 1 1 1 1 = What happens next?
10 + 5 + 5 + 5 + 4 + 2

 

[telectrix]

1/Rt = 1/R1 + 1/R2 + 1/R3 +1/R4 + 1/R5., so 1/Rt = ?????, then Rt = ?????

you've got 3 5's there instead of 2.

answer is 0.8. show your workings.

 

[me1]

Sorry there should be two 5s.

I am not sure how you get the result, would you mind showing me please?

Much appreciated

R/1= 1/10+1/5+1/5+1/4+ 1/2 =

 

[DPG]

And don't forget the initial answer will be a reciprocal, ie. 1/r so don't forget the last divide - a lot of people forget this. Tel, I know your answer covers this but thought I'd emphasize it. Daz

 

[i=p/u]

Can you not add the all then do 1 over the lot , ?? I think so

- - - Updated - - -

Starting to forget this stuff

 

[ipf]

Can you not add the all then do 1 over the lot , ?? I think so

- - - Updated - - -

Starting to forget this stuff

Get your common denominator and then...........

 

[matt1386]

Find the common denominator of your resistors

work out how many times each of your resistor values go into your common denominator and add them all together

then divide your common denominator by what answer you get from above

Hope that makes sense lol !

 

[Bick]

Just use a scientific calculator and as been mentioned, don't forget to reciprocate (1/x or x-1 button) after summing the fractions.

 

[Richard Burns]

R1 = 10 Ω
R2 = 5 Ω
R3 = 5 Ω
R4 = 4 Ω
R5= 2 Ω

1/Rt = 1/R1 + 1/R2 + 1/R3 = 1/R4 +1/R5

1/Rt = 1/10 + 1/5 + 1/5 + 1/4 + 1/2

You can do this on a calculator using the M+ function
1/10=M+, 1/5=M+,1/5=M+, 1/4 =M+, 1/2=M+
Mr
1/x

You can do this in your head with decimals
1/10= 0.1, 1/5= 0.2, 1/5= 0.2, 1/4 = 0.25, 1/2 = 0.5
0.1 + 0.2 + 0.2 + 0.25 + 0.5 = 1.25
1/1.25 = 4/5 =0.8

You can add the numbers using the common denominator (20)
To give 2/20 + 4/20 + 4/20 + 5/20 + 10/20 = 25/20
1/ (25/20) =20/25 = 4/5 = 0.8

 

[ipf]

For something as simple as this calc I would do it using the last method......in mi yed.

Then again, it depends on age, capability and learning methods

 

[Engineer54]

That's explained and answered the OP's question in full...

 

[Hawkmoon]

But just to add:

A useful double check after you have calculated is to remember that your answer will always be a value lower than the lowest resistor in any parallel group - in this case < 2 ohms

 

[telectrix]

also useful to remember, although not really related is for capacitors in parallel , you just add them, but in series use the reciprocal method as for parallel resistors.

 

[LeeH]

Good thread, not done this since sparky school.