# Circuit question -- Two sources and two resistors...

1. Mar 29, 2016

### goonking

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
Applying KCL at the node , assuming current flows from left to right at the 4 ohm resistor:

Vo/4 + 10 + 2Vo = 0

OR, is it :

Vo/(4+6) + 10 + 2Vo = 0

since the two resistors are in series?

2. Mar 29, 2016

### donpacino

There are two parts to this problem.

1. find vo
2. find power dissipated by the controlled current source.

remember V=IR

you know what R is (4). you need to solve for the current, so you write a KCL.
so...

Vo/4 + 10 + 2Vo = 0 is correct.

think about it... that 6 ohm resistor does not effect Vo at all. because Vo is related to the current through the device and the resistance of the device. the sources are CURRENT sources. so the current will be the same independent of the 6 ohm part.

Now to find the power you'll need to write another equation. there are a few different ways you can do it.

3. Mar 29, 2016

### BvU

Funny. I thought the 6 $\Omega$ 'sees' 3 V0. Whereas the 2V0 voltage source 'sees' 10 $\Omega$

4. Mar 29, 2016

### goonking

so would it be different if the one of the current sources was replaced with a voltage source?

5. Mar 30, 2016

### donpacino

Yes. It would be, because with a voltage source there is constant voltage, so the greater the voltage drop across one resistor, then less drop across the other. with a voltage source it is the relationship between the two resistor that determines the voltage drops.

an ideal voltage source will create any current necessary to make the desired voltage.
an ideal current source will create any voltage necessary to make the desired current.

Last edited: Mar 30, 2016
6. Mar 30, 2016

### donpacino

Im very confused by this statement. there is no voltage source, only current sources.
the current through that 6 ohm resistor will ALWAYS be (10 + 2Vo) amps.
The current sources will produce whatever voltage is necessary to create the desired current.

7. Mar 30, 2016

### BvU

My mistake, I didn't recognize the Voltage meter symbol.

But now I don't understand the V0 sign indication at the 4 Ohm resistor

8. Mar 30, 2016

### cnh1995

I believe it's the voltage drop across 4Ω resistor.
Vo=4*(10+2Vo)..

9. Mar 30, 2016

### donpacino

that diamond is not a voltage meter. it is a dependent current source.

cnh is right when he says that the Vo sign is simply indicating what the voltage drop is. You can think of that as a meter if you are a hands on person.

10. Mar 31, 2016

### goonking

Suppose we had another circuit identical to this except, the resistors are very high, lets say a million ohms. Also, the 2Vo is now a current source of a 100 A, and the 10 A remains the same.

What would happen in this circuit? Would current from the 100 Ampere source just flow through the 10 A source? Would there be no current at all? Would KCL still apply?

11. Apr 1, 2016

### Staff: Mentor

You are forgetting what is meant by a 10A current source. No current flows into it. Precisely 10A flows out of it.

No ifs or buts.

12. Apr 1, 2016

### goonking

but isn't there a direction to the current?

13. Apr 1, 2016

### cnh1995

The currents of 10A source and 100A source will add and flow through the million ohm resistors. Current source provides a constant current, irrespective of the nature of network around it.

14. Apr 1, 2016

### goonking

what happens when two current sources are in series, and directed towards each other?

15. Apr 1, 2016

### cnh1995

It is invalid to connect two current sources in series(and two voltage sources in parallel).

16. Apr 1, 2016

### Staff: Mentor

Yes. Current flows out. That's why it is called a constant current source.

Inverted, it can be called a current sink.

17. Apr 1, 2016

### BvU

Can someone explain the sign language here ? Why is 2V0 a dependent current source ? And 100 A, where does that come from ? And the sign of V0 over the 4 $\Omega$ ?

18. Apr 1, 2016

### Staff: Mentor

A common symbol for a controlled source is the triangle. Controlled sources may be current sources or voltage sources and as the name suggests their output is controlled by some external quantity. Usually this external quantity is a potential difference or current located elsewhere in the circuit. It is presumed (and usually not shown in the schematic) that some circuitry or mechanism exists to monitor and measure the external control signal and feed this information back to the controlled source where it is acted on to control the source's output.

In the present problem there is some hidden mechanism that determines the potential Vo across the 4 Ω resistor which is then used to meter the output of the controlled current source according to the specification I = 2 Vo. Note that it is always implied that the constant multiplying the measured value has units associated with it that make the result come out with the desired units. In this case the "2" should be read as 2 amps per volt.

The Vo label on the 4 Ω resistor is there to inform you how the potential difference across that resistor is to be interpreted in terms of its polarity. Think of it as how you would connect the leads of your voltmeter in order to measure it. Reverse the leads and you would change the sign of measurement value. If you were to change the polarity of the Vo label then the circuit would behave differently: The direction of the current produced by the controlled source would be reversed.

As far as I can tell the 100 A is part of a hypothetical example introduced by goonking in post #7. It's not part of the original problem.

19. Apr 1, 2016

### BvU

Thanks gneill and donpacino. I learned something from this one.

20. Apr 2, 2016

### goonking

another question, but aren't the polarities on the Vo resistor suppose to be switched/flipped ( maybe a typo?) since current can only flow from right to left in this case.