Series-Parallel Circuit problem, afraid I'm tripping over my work

[Brolomon]

Homework Statement

I want to solve for the Req, the total I, and total Power first. Then, I want to solve for the V, I, and Power for ALL of the resisters.

Homework Equations

I'm using V = IxR and P = IV for this. Also the formulas for finding Req for series (R1+R2) and parallel(1/R1 + 1/R2).

The Attempt at a Solution

I have solved just about everything except for the 60 ohm resister. I am not sure if my numbers are adding up correctly and if possible I would like some help. By help I mean could someone try solving this as well so I may check my answers and hopefully help me understand this stuff more cleary.

Thank you.

 

[CWatters]

You need to show us some working. What did you get for the total resistance?

 

[Brolomon]

Total resistance was .1217, Req was 41.09, total power is .61

For the 10ohm:
V = 1.22V
I = .1217A
P = .15W

For 20ohm:
V = 1.51V
I = .0756A
P = .11W

For 30ohm:
V = 2.27V
I = .0756A
P = .17W

I attempted the 50 and 40ohm resisters and I am almost certain they're wrong. If someone could show me what they have for all of this and/or check my current answers, that'd be great.

The 40,50, and 60 ohm ones are giving me trouble.

 

[CWatters]

Total resistance was .1217, Req was 41.09, total power is .61

Agreed.
I get Req = 31.09 + 10 = 41.09 Ohms
Power = V²/R = 0.61 W

For the 10ohm:
V = 1.22V
I = .1217A
P = .15W

Agreed.

Then what I would do is calculate the voltage drop across the top group of resistors. Can be done several ways but using the potential divider rule...

V = (5 * 31.09)/(41.09) = 3.783V

Then apply the potential divider rule four times to that voltage to calculate the voltages on the following "four" resistors..

30Ohm
20Ohm
60Ohm
and the 50//40 = 22.222^.Ohm

Knowing the voltage on all the resistors you can calculate I and P for each including the 40 and 50Ohm individually.

For example the voltage drop across the 60 Ohm is...

= (3.783V * 60) / (60 + 50//40)
= (3.783V * 60) / (60 + 22.222)
= 2.83V

 

[Brolomon]

This is very helpful! We haven't learned the divider rule yet, but I'm sure we will. Thank you!

 

[CWatters]

Ah ok.

You can also do it by using Ohms law. for example the voltage drop across the top set is..

V = I * R
= .1217A * 22.222^.

The voltage divider rule typically allows you to avoid working ou the current explicitly. There is a primer here although perhaps skip straight to the section on the Resistive divider...

If you have ever covered "ratios" in other subjects it will be quite easy.

http://en.wikipedia.org/wiki/Voltage_divider

There is also a current divider rule but perhaps get to grips with the voltage divider first.

http://en.wikipedia.org/wiki/Current_divider