- #1

- 382

- 0

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter M. next
- Start date

- #1

- 382

- 0

- #2

psparky

Gold Member

- 884

- 32

I X R?

Tough one.

Tough one.

- #3

vk6kro

Science Advisor

- 4,081

- 40

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

- 1,176

- 84

-I*r = Vo.

- #5

psparky

Gold Member

- 884

- 32

Little r has no bearing on v out in my opinion.

- #6

jim hardy

Science Advisor

Gold Member

Dearly Missed

- 9,847

- 4,888

Little r has no bearing on v out in my opinion.

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

Science Advisor

- 4,081

- 40

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 [Broken]

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 [Broken]

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.

Last edited by a moderator:

- #8

psparky

Gold Member

- 884

- 32

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 accross 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 accross R.....leaving 30 volts across each branch.

Must be since the voltmeter is not going to change branch to branch.

Last edited:

- #9

vk6kro

Science Advisor

- 4,081

- 40

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 cancelled 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

- 382

- 0

vk6kro thank you!!

- #11

psparky

Gold Member

- 884

- 32

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

- 315

- 0

This is a great problem. I have stumped 5 out of 6 elec engineers with this problem at work....including myself.

Psparky out....thank you for schooling me as well.

Excellent question M. next....keep em coming.

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

- 176

- 7

Are those resistance really in series ?

- #14

psparky

Gold Member

- 884

- 32

Are those resistance really in series ?

Yes! Redraw the circuit leaving out the vout tails.

You have a current source in series with two resistors.....simple as that.

- #15

- 176

- 7

i think answer should be IR !

- #16

- 176

- 7

- #17

NascentOxygen

Staff Emeritus

Science Advisor

- 9,244

- 1,073

I'm not able to picture what you mean. Can you draw it?

- #18

- 176

- 7

I'm not able to picture what you mean. Can you draw it?

here it is

- #19

vk6kro

Science Advisor

- 4,081

- 40

Here is the same circuit as above but redrawn in a more familiar way:

http://dl.dropbox.com/u/4222062/series%202.PNG [Broken]

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 [Broken]

The question asked for the voltage ACROSS the 30 ohm resistor, "r". That is, what voltage would you measure with a voltmeter?

Last edited by a moderator:

- #20

- 176

- 7

ya i got it ! i was confused there ! thanks !

Share:

- Replies
- 5

- Views
- 6K