- #1
Discover the Output Voltage Vo for Figure Attached
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Discussion Overview
The discussion revolves around determining the output voltage Vo in a circuit involving a current source and two resistors. Participants explore various interpretations of the circuit configuration and the implications for the voltage measurement across the resistors, specifically focusing on the role of the small resistor (r) versus the larger resistor (R).
Discussion Character
- Debate/contested
- Technical explanation
- Mathematical reasoning
Main Points Raised
- Some participants propose that Vo can be expressed as Vo = -I * r, based on the direction of current flow and the assumed polarity of the voltage measurement.
- Others argue that removing the small resistor (r) does not affect the output voltage, suggesting that the voltage remains the same regardless of its presence.
- A participant provides specific values for the resistors and current, calculating the voltage across the small resistor as 30 volts, while also noting the voltage across the larger resistor and the current source.
- There is a discussion about the apparent configuration of the circuit, with some participants questioning whether the resistors are in series or parallel and how this affects the voltage readings.
- One participant expresses confusion about the voltage measurements across different branches of the circuit, leading to further exploration of the circuit's behavior.
- Another participant suggests that rearranging the circuit into a more familiar form can clarify the relationships between the components and their voltages.
- Some participants challenge the interpretation of the circuit, questioning the conditions under which the resistors are considered in series or parallel.
Areas of Agreement / Disagreement
Participants do not reach a consensus on the correct interpretation of the circuit configuration or the resulting output voltage. Multiple competing views remain regarding the roles of the resistors and the implications for the voltage measurement.
Contextual Notes
There are unresolved assumptions about the circuit configuration, particularly regarding the series versus parallel arrangement of the resistors, which affects the interpretation of the output voltage. Participants express varying degrees of understanding and visualization of the circuit.
- #2
psparky
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I X R?
Tough one.
Tough one.
- #3
vk6kro
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If the head of the Vo arrow is supposed to be the positive end, then Vo = minus I times little r = -(I * r)
This is because the current is flowing upwards in the right hand side of the loop, so the bottom of the resistor r is positive.
This is because the current is flowing upwards in the right hand side of the loop, so the bottom of the resistor r is positive.
- #4
cabraham
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-I*r = Vo.
- #5
psparky
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I see what you guys are saying...but if you remove little r...the voltage is still the same!
Little r has no bearing on v out in my opinion.
Little r has no bearing on v out in my opinion.
- #6
jim hardy
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psparky said:
looks to me like if you open little r you have opened the load on a current source, so the voltage will become the maximum the source can produce.
Infinite, for "ideal" source .
So it's little r not big R determines Vo in that circuit.
- #7
vk6kro
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It helps to put some real sizes on the components:
Let R = 20 ohms, r = 30 ohms and the current source = 1 amp.
Like this:
http://dl.dropbox.com/u/4222062/current%20gen.PNG
So, the voltage across the resistor "r" = Vo = (1 amp * 30 ohms) = 30 volts (Negative 30 volts relative to the arrow)
So, the voltage across the resistor "R" = (1 amp * 20 ohms) = 20 volts
With this load, the voltage across the current source is 50 volts (1 amp * (30 ohms + 20 ohms)) and you can get the same result if you regard the resistors as part of a voltage divider.
Let R = 20 ohms, r = 30 ohms and the current source = 1 amp.
Like this:
http://dl.dropbox.com/u/4222062/current%20gen.PNG
So, the voltage across the resistor "r" = Vo = (1 amp * 30 ohms) = 30 volts (Negative 30 volts relative to the arrow)
So, the voltage across the resistor "R" = (1 amp * 20 ohms) = 20 volts
With this load, the voltage across the current source is 50 volts (1 amp * (30 ohms + 20 ohms)) and you can get the same result if you regard the resistors as part of a voltage divider.
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- #8
psparky
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Oh ya...I see what you are saying once again!
The circuit appears to be in parallel at first glance...but apparentlly is not!
The voltage across is clearly not the same. (edit...yes, voltage across is the same)
Still though, If I slide a volt meter across one branch to the other branch I get two different voltages?
How can this be?
Perhaps I can answer this...I can see how the current device has 50 volts across it.
I guess it drops 20 volts across R...leaving 30 volts across each branch.
Must be since the voltmeter is not going to change branch to branch.
The circuit appears to be in parallel at first glance...but apparentlly is not!
The voltage across is clearly not the same. (edit...yes, voltage across is the same)
Still though, If I slide a volt meter across one branch to the other branch I get two different voltages?
How can this be?
Perhaps I can answer this...I can see how the current device has 50 volts across it.
I guess it drops 20 volts across R...leaving 30 volts across each branch.
Must be since the voltmeter is not going to change branch to branch.
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- #9
vk6kro
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Perhaps I can answer this...I can see how the current device has 50 volts across it.
I guess it drops 20 volts across R...leaving 30 volts across each branch.
Yes, that's right, the 50 volts across the current source is partially canceled by the 20 volts across the 20 ohms resistor (because they have opposite polarities in the left hand leg of the loop), so this leaves 30 volts between the top and bottom of the diagram.
I guess it drops 20 volts across R...leaving 30 volts across each branch.
Yes, that's right, the 50 volts across the current source is partially canceled by the 20 volts across the 20 ohms resistor (because they have opposite polarities in the left hand leg of the loop), so this leaves 30 volts between the top and bottom of the diagram.
- #10
M. next
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vk6kro thank you!
- #11
psparky
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This is a great problem. I have stumped 5 out of 6 elec engineers with this problem at work...including myself.
It is much easier to see when when you re-arrange the circuit in a basic series arrangement. Add the two resistors in series and you get 50 ohms...or 50 volts. Obviously there is 50 volts across the 50 ohm resistor...or current source.
Lesson learned...always re-arrange circuits in a more familiar form...regardless if you are in college...taking state exams or just fooling around on this forum or whatever.
Psparky out...thank you for schooling me as well.
Excellent question M. next...keep em coming.
It is much easier to see when when you re-arrange the circuit in a basic series arrangement. Add the two resistors in series and you get 50 ohms...or 50 volts. Obviously there is 50 volts across the 50 ohm resistor...or current source.
Lesson learned...always re-arrange circuits in a more familiar form...regardless if you are in college...taking state exams or just fooling around on this forum or whatever.
Psparky out...thank you for schooling me as well.
Excellent question M. next...keep em coming.
- #12
Ratch
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psparky,
Well, I just don't get it. It looks to me like a problem whose solution can be discerned almost as fast as it can be read. I mean, a current source in series with two resistors? What's the big deal?
Ratch
Well, I just don't get it. It looks to me like a problem whose solution can be discerned almost as fast as it can be read. I mean, a current source in series with two resistors? What's the big deal?
Ratch
- #13
lazyaditya
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Are those resistance really in series ?
- #14
psparky
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lazyaditya said:
Yes! Redraw the circuit leaving out the vout tails.
You have a current source in series with two resistors...simple as that.
- #15
lazyaditya
- 176
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i think answer should be IR !
- #16
lazyaditya
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suppose there are terminals in at the ends of r resistance , that would'nt be in series then and it did not asked for the voltage drop at r it asked for V(0) which would be equal to current in small r when small r and capital R are parallel!
- #17
NascentOxygen
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lazyaditya said:
I'm not able to picture what you mean. Can you draw it?
- #18
- #19
vk6kro
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Here is the same circuit as above but redrawn in a more familiar way:
http://dl.dropbox.com/u/4222062/series%202.PNG
The question asked for the voltage ACROSS the 30 ohm resistor, "r". That is, what voltage would you measure with a voltmeter?
http://dl.dropbox.com/u/4222062/series%202.PNG
The question asked for the voltage ACROSS the 30 ohm resistor, "r". That is, what voltage would you measure with a voltmeter?
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- #20
lazyaditya
- 176
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ya i got it ! i was confused there ! thanks !
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