This video is what I am watching.

R2. is 36 Ohms and R3. is 12 Ohms I know I do not just take 36 + 12.

I will do this 1. over 36 = .0277

Then I do this 1. over 12 = .0833

Then I do 1. over .0607 = 16.474 Ohms.

I know there is a lot of ways to do this but is this right?

Am I right then I take .0277 + .0833 = .0607

 

1/36= 0.0277
1/12=0.0833
0.0833 + 0.0277 = 0.1111
1/0.1111=9

 

R2 and R3 are in parallel. You add conductance, not resistance.
R for resistance
G for conductance

Conductance = 1 / Resistance

 

1/12 + 1/36 = 3/36 + 1/36 = 4/36 = 1/9

Don’t do division on a calculator. Add with common denominator.

 

For two resistors it is product over sum.

\( \cfrac{36 \Omega\times 12\Omega}{36\Omega+12\Omega}\;=\;\cfrac{432}{48}\;=\;9\Omega \)

 

Thanks.

 

I will make one up now just to see if I have it.

If I have two resistors in Parallel and R1. is 5. Ohms and R2. is 25. Ohms I do this.

1. over 5. is 0.2
1. over 25. is 0.04

then I Add 0.2 + 0.04 = 0.24

Now I do 1. over 0.24 = 4.1666 Ohms.

Do I have it?

 

I will make one up now just to see if I have it.

If I have two resistors in Parallel and R1. is 5. Ohms and R2. is 25. Ohms I do this.

1. over 5. is 0.2
1. over 25. is 0.04

then I Add 0.2 + 0.04 = 0.24

Now I do 1. over 0.24 = 4.1666 Ohms.

Do I have it?

Check:

\( \cfrac{25\times5}{25+5}\;=\;\cfrac{125}{30}\;=\;4.17\Omega \)

N.B. The result will always be less than the smaller of the two parallel resistors.

 

See my post #4.
The answer with 5Ω in parallel with 25Ω is 25/6 Ω = 4 1/6 Ω.
I did not need to use a calculator.

 

1/12 + 1/36 = 3/36 + 1/36 = 4/36 = 1/9

Don’t do division on a calculator. Add with common denominator.

That's fine, when it's apparent that the numbers are going to work out nice and neat. But that is the rare exception in real life.

R1 = 470 Ω and R2 = 91 Ω

 

Check:

\( \cfrac{25\times5}{25+5}\;=\;\cfrac{125}{30}\;=\;4.17\Omega \)

N.B. The result will always be less than the smaller of the two parallel resistors.

While the smaller of the two is the upper bound, half of the smaller is the lower bound.

So if you have a 68 Ω resistor in parallel with a 91 Ω resistor, you know immediately that

34 Ω < Reg < 68 Ω

Any answer outside this range is guaranteed to be wrong.

Another useful point is that if the larger is three times the smaller, the equivalent resistance if 3/4 of the smaller (i.e., half way between the upper and lower bounds).

These bounds are very useful in quickly estimating effective resistance and determining if the estimate is good enough to not even have to do the actual calculation.

 

That's fine, when it's apparent that the numbers are going to work out nice and neat. But that is the rare exception in real life.

R1 = 470 Ω and R2 = 91 Ω

I agree with you. I am trying to indicate to the TS two things.

1) It is better to learn principles rather than trying to remember formulae.
Learn that you add conductance when in parallel. This reinforces the concept of current flow in parallel circuits.

2) You don’t always have to use a calculator.

(TS used a calculator and still got it wrong.
12 in parallel with 36 must be lower than 12. Hence 16.474 cannot be correct.)

Now, suppose one did not have a calculator, can one arrive at an estimate using mental math?
475 in parallel with 95 is 475/6 = 79
Hence the correct answer for 470//91 would be slightly less than 79.