Use voltage divison and current division to find

[Color_of_Cyan]

Homework Statement

http://img526.imageshack.us/img526/8817/homeworkprob9.jpg


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.

 

[phinds]

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]

Homework Statement

http://img526.imageshack.us/img526/8817/homeworkprob9.jpg


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.

 

[Color_of_Cyan]

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.

 

[rezeew]

To find Io, we can use current division as you have attempted. However, there is an error in your calculation. The total parallel resistance is not 6Ω, but rather (1/8 + 1/24)^-1 = 6. The total incoming current is 8A, but the current through the 24Ω resistor is not 8A. Instead, it is (1/6)(8A) = 4/3 A. Therefore, the current through the 8Ω resistor is 8A - 4/3 A = 20/3 A. Using this current and the voltage division law, we can find V as follows:

V = (20/3 A)(8Ω) = 160/3 V

To find Vo, we can use voltage division since the 8Ω and 24Ω resistors are in series. Therefore, we can use the formula:

Vo = (24Ω / (8Ω + 24Ω))(160/3 V) = 120 V

Note that in this problem, voltage division and current division can be used interchangeably since the resistors are either in series or in parallel. However, in more complex circuits, it may be necessary to use both methods to find the desired quantities. It is important to remember the formulas for voltage division and current division and to correctly identify the resistors in series and parallel.