# Use voltage divison and current division to find

## Homework Statement

http://img526.imageshack.us/img526/8817/homeworkprob9.jpg [Broken]

Find Io, find V, find Vo

## Homework Equations

V = IR

And if these are right:

Voltage division:
(voltage across series resistor) = [ (resistance)/(total series resistance) ](total input V)

Current division (only for two resistors in parallel):
(current through parallel resistor) = [ (OTHER resistance)/(total parallel resistance) ](total incoming current)

## The Attempt at a Solution

I feel like I can't go anywhere else in the problem without first finding the current Io. I'm not sure how to use voltage division law properly. Does it only apply to one wire (one node)?

Same goes for current division law but I think I understand it better:

I simplified all resistors except for the 24Ω resistor. So I am left with 8Ω and 24Ω resistors in parallel, along with the 8 amp total current coming in.

Trying to solve Io with the current division law above, I get:

Io = (8Ω / 6Ω)8A = 10.6A

but it is wrong, supposedly. ( 6Ω in the equation because (1/8 + 1/24)^-1 = 6 , unless I got something wrong. )

Thank you.

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## Answers and Replies

phinds
Science Advisor
Gold Member
How you managed to bring 6 ohms into the split, I do not understand.

You have current splitting between 8 ohms and 24 ohms. What portion will flow through each?

gneill
Mentor

## Homework Statement

http://img526.imageshack.us/img526/8817/homeworkprob9.jpg [Broken]

Find Io, find V, find Vo

## Homework Equations

V = IR

And if these are right:

Voltage division:
(voltage across series resistor) = [ (resistance)/(total series resistance) ](total input V)

Current division (only for two resistors in parallel):
(current through parallel resistor) = [ (OTHER resistance)/(total parallel resistance) ](total incoming current)

## The Attempt at a Solution

I feel like I can't go anywhere else in the problem without first finding the current Io. I'm not sure how to use voltage division law properly. Does it only apply to one wire (one node)?

Same goes for current division law but I think I understand it better:

I simplified all resistors except for the 24Ω resistor. So I am left with 8Ω and 24Ω resistors in parallel, along with the 8 amp total current coming in.
Yes, good so far.
Trying to solve Io with the current division law above, I get:

Io = (8Ω / 6Ω)8A = 10.6A

but it is wrong, supposedly. ( 6Ω in the equation because (1/8 + 1/24)^-1 = 6 , unless I got something wrong. )
Yup. Your current division is not right. It should be: ("other resistor"/"sum of the resistors") x current.

Last edited by a moderator:
Thanks.

Ok so instead of 6 ohms it's 32 ohms then, because 8 + 24 = 32. It's just the mathematical sum instead and not r equiv parallel resistance between (like i thought it was).

So Io is 2A then. It's the right answer too.

So then, V = IR makes V = (2A)(24Ω) = 48V, and that's also correct.

Lastly for Vo, 48V is in parallel with the rest of the circuit so it counts as the (total input V) when using voltage division to find Vo, so:

Vo = (30Ω / 80Ω)(48V) ---> Vo = 18V which is also correct.

Many thanks.