Find Vo in the circuit / current direction / potential direction

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SUMMARY

The discussion centers on calculating the output voltage Vo in a circuit with a known voltage drop Va of 8V across a 4Ω resistor. Participants utilize fundamental electrical engineering principles such as Ohm's Law (V = IR), Kirchhoff's Voltage Law (KVL), and voltage division to analyze the circuit. The conclusion reached is that Vo is determined to be 48V, contingent upon the assumption of the direction of potential drop across the resistors. The importance of clearly stating assumptions regarding current and voltage directions is emphasized for accurate problem-solving.

PREREQUISITES
  • Understanding of Ohm's Law (V = IR)
  • Familiarity with Kirchhoff's Voltage Law (KVL)
  • Knowledge of voltage and current division techniques
  • Ability to analyze series and parallel resistor configurations
NEXT STEPS
  • Study advanced circuit analysis techniques using Kirchhoff's Laws
  • Learn about the implications of resistor configurations on voltage and current
  • Explore the concept of dependent and independent voltage sources in circuits
  • Practice solving circuit problems with unspecified voltage and current directions
USEFUL FOR

Electrical engineering students, circuit designers, and anyone involved in analyzing or troubleshooting electrical circuits will benefit from this discussion.

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Homework Statement




http://imageshack.us/a/img29/9056/homeworkprob10.jpg


Given: Va = 8V. Find Vo in the circuit.


Homework Equations


V = IR,

KVL, KCL

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)/(sum of resistors) ](total incoming current)


The Attempt at a Solution



I think you can redraw it like this:

http://img24.imageshack.us/img24/8245/homeworkprob10redraw.jpg

And since 8V = Va across the 4Ω resistor, V = IR means the current in the whole right node means that I = V/R, and then I = 8V / 4Ω

I = 2A in the right branch right?


Then I = 2A also in the 8Ω resistor in series with it,

so then V = IR for the 8Ω resistor means, V = (2A)(8Ω), V = 16V over the 8Ω resistor

Adding these, V = 24V for the entire right branch.


Left branch resistors is simplified to 4Ω and then you can say V = 24V for it because of being in parallel, right?

Then I = 6A for the left because I = V/R --> 24V/4Ω

then that means 8A is supplied from the middle to both,


but then that means V = (3Ω)(8A) means V = 24V through the middle 3Ω resistor

It can't be more than 24V because it's in parallel right? There was no direction given for Va in the first place and that's confusing. The answer for Vo is given so the only thing that would make sense is if V = -24V for the middle 3Ω resistor and then Vo = 48V (which is also the answer, and just thought of this now).


Is there a better way to tell if either the current or potential in the middle branch is + or - then given only Va first? Again, I just figured it out for the middle branch just now but it's still kind of confusing so, thanks anyway.
 
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Without being given the direction of the potential drop Va, you can't tell which direction the current is flowing. So there will be two possible answers, on for each choice of direction of Va.

Remember that if the central resistor is dropping a certain amount of potential, the voltage source Vo will have to make up the difference so that their sum matches the required branch voltages.
 
Cyan..

Your approach isn't correct. The 2A doesn't flow through the 8 Ohms. I redrew it like this...

PS Vo is on the right.
 

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    redrawn.png
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PPS: Two of the resistors don't effect Vo.
 
CWatters said:
Cyan..

Your approach isn't correct. The 2A doesn't flow through the 8 Ohms. I redrew it like this...

PS Vo is on the right.

Hmm. The way I interpret the problem Va is not a voltage source, it's the potential drop across the 4Ω resistor due to the current flowing though it. The only independent voltage source in the circuit is Vo.
 
Color_of_Cyan said:

Homework Statement


Is there a better way to tell if either the current or potential in the middle branch is + or - then given only Va first? Again, I just figured it out for the middle branch just now but it's still kind of confusing so, thanks anyway.

If you're given a potential drop without its direction being specified, then all you can do is look at the circuit and make deductions/assumptions. Like, looking at the indicated polarity of Vo, if it is assumed that Vo is a positive value then the current can only flow one way through it since it's the only source in the circuit. That in turn tells you which way the current must flow in the branches, hence the polarity of Va.

When these situations crop up on assignments or exams, be sure to explicitly state your assumptions along with your solution.
 
gneill said:
If you're given a potential drop without its direction being specified, then all you can do is look at the circuit and make deductions/assumptions. Like, looking at the indicated polarity of Vo, if it is assumed that Vo is a positive value then the current can only flow one way through it since it's the only source in the circuit. That in turn tells you which way the current must flow in the branches, hence the polarity of Va.

When these situations crop up on assignments or exams, be sure to explicitly state your assumptions along with your solution.

Okay, thanks. And yes, Va was the potential drop across that resistor, and neither the drop direction or current direction were given. It's just given from being positive and then the direction of Vo.
 
gneill said:
Hmm. The way I interpret the problem Va is not a voltage source, it's the potential drop across the 4Ω resistor due to the current flowing though it. The only independent voltage source in the circuit is Vo.

Insert red face.

That makes much more sense now I look at it again. I was fooled by the subscript on Vo.
 

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