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

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  • #1
goonking
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Homework Statement

[/B]
upload_2016-3-29_10-8-31.png


Homework Equations




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?
 

Answers and Replies

  • #2
donpacino
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Homework Statement

[/B]
View attachment 98161

Homework Equations




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?

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
BvU
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Funny. I thought the 6 ##\Omega## 'sees' 3 V0. Whereas the 2V0 voltage source 'sees' 10 ##\Omega##
 
  • #4
goonking
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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.
so would it be different if the one of the current sources was replaced with a voltage source?
 
  • #5
donpacino
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so would it be different if the one of the current sources was replaced with a voltage source?
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.
 
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  • #6
donpacino
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Funny. I thought the 6 ##\Omega## 'sees' 3 V0. Whereas the 2V0 voltage source 'sees' 10 ##\Omega##
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
BvU
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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
cnh1995
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But now I don't understand the V0 sign indication at the 4 Ohm resistor
I believe it's the voltage drop across 4Ω resistor.
Vo=4*(10+2Vo)..
 
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  • #9
donpacino
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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
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.
 
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  • #10
goonking
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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
NascentOxygen
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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?
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. :cool:
 
  • #12
goonking
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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. :cool:
but isn't there a direction to the current?
 
  • #13
cnh1995
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but isn't there a direction to the current?
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
goonking
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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.
what happens when two current sources are in series, and directed towards each other?
 
  • #15
cnh1995
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what happens when two current sources are in series, and directed towards each other?
It is invalid to connect two current sources in series(and two voltage sources in parallel).
 
  • #16
NascentOxygen
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but isn't there a direction to the current?
Yes. Current flows out. That's why it is called a constant current source.

Inverted, it can be called a current sink.
 
  • #17
BvU
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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
gneill
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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## ?

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.
 
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  • #19
BvU
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Thanks gneill and donpacino. I learned something from this one.
 
  • #20
goonking
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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.
 
  • #21
gneill
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another question, but are 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.
The polarity of the Vo label has nothing to do with the current. It simply tells you how to read whatever potential difference is there: where to (conceptually) connect your positive and negative meter leads. The actual potential difference that the current produces across that resistor may well have the opposite polarity, in which case Vo will be read as a negative value.
 
  • #22
goonking
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The polarity of the Vo label has nothing to do with the current. It simply tells you how to read whatever potential difference is there: where to (conceptually) connect your positive and negative meter leads. The actual potential difference that the current produces across that resistor may well have the opposite polarity, in which case Vo will be read as a negative value.
I was told resistors don't have polarity, but in the diagrams, they do. Are the polarities on resistors only used when applying Kirchoff's Voltage Law?
 
  • #23
gneill
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I was told resistors don't have polarity, but the diagrams, they do. Are the polarities on resistors only used when applying Kirchoff's Voltage Law?
Resistors have no inherent polarity. Any polarity attributed to them is either an artifact of an assumed current causing a potential drop (KVL) or because the circuit designer is telling you how to interpret a voltage reading across that resistor.
 
  • #24
goonking
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Resistors have no inherent polarity. Any polarity attributed to them is either an artifact of an assumed current causing a potential drop (KVL) or because the circuit designer is telling you how to interpret a voltage reading across that resistor.
If a circuit diagram has multiple loops and the resistor polarities aren't indicated, is it still possible to apply Kirchoff Laws? Or do we have to assume flow of current and then polarity of resistors according to our assumption?
 
  • #25
gneill
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If a circuit diagram has multiple loops and the resistor polarities aren't indicated, is it still possible to apply Kirchoff Laws? Or do we have to assume flow of current and then polarity of resistors are then labeled accordingly?
It's up to you to choose assumed currents for your analyses and label the resulting voltage drops. It's very rare indeed that you'll find a circuit diagram with potential drops pre-labeled for you. Kirchhoff's laws always hold; they are a matter of the physics and independent of any labeling. But you will frequently find particular currents and voltages associated with controlled sources to be indicated.
 
  • #26
goonking
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It's up to you to choose assumed currents for your analyses and label the resulting voltage drops. It's very rare indeed that you'll find a circuit diagram with potential drops pre-labeled for you. Kirchhoff's laws always hold; they are a matter of the physics and independent of any labeling. But you will frequently find particular currents and voltages associated with controlled sources to be indicated.
Would it be possible to apply KVL on the right loop in this case (the loop with 2 current sources)?
 
  • #27
gneill
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Would it be possible to apply KVL on the right loop in this case (the loop with 2 current sources)?
Applying KVL with current sources is tricky because they don't have an inherent potential difference due to the current. The potential drop they exhibit depends on the rest of the circuit: they will produce any potential required in order to maintain their fixed current. So you either need to introduce and solve for another variable to represent that potential difference (which means finding another equation including it somehow), or choose another solution method. Nodal analysis is the usual way to go in this case.

Eventually you will cover a method called mesh analysis, which uses KVL, and includes a techniques called a supermesh that can handle current sources. It provides a method to obtain the required extra equation without explicitly introducing any new variables.
 
  • #28
goonking
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Applying KVL with current sources is tricky because they don't have an inherent potential difference due to the current. The potential drop they exhibit depends on the rest of the circuit: they will produce any potential required in order to maintain their fixed current. So you either need to introduce and solve for another variable to represent that potential difference (which means finding another equation including it somehow), or choose another solution method. Nodal analysis is the usual way to go in this case.

Eventually you will cover a method called mesh analysis, which uses KVL, and includes a techniques called a supermesh that can handle current sources. It provides a method to obtain the required extra equation without explicitly introducing any new variables.
I see, what about this case, a circuit consisting of only one loop containing a current source of 10 A and a resistor of 5 ohms.

since P = IV = I2R
I2 R = (10)2 (5) = 500 Watts
therefore 500 watts = IV = (10) (V)
V = 50, then we can say the voltage drop across... something is 50 volts?

Is this even right?
 
  • #29
gneill
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I see, what about this case, a circuit consisting of only one loop containing a current source of 10 A and a resistor of 5 ohms.

since P = IV = I2R
I2 R = (10)2 (5) = 500 Watts
therefore 500 watts = IV = (10) (V)
V = 50, then we can say the voltage drop across... something is 50 volts?

Is this even right?
Well, the current source will produce 50 V in order to drive 10 A though 5 Ohms. So the potential difference across each component is the same: 50 V. KVL would be satisfied for the loop.
 
  • #30
goonking
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Well, the current source will produce 50 V in order to drive 10 A though 5 Ohms. So the potential difference across each component is the same: 50 V. KVL would be satisfied for the loop.
Are there cases where applying KCL is required but KVL isn't?
 
  • #31
gneill
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Are there cases where applying KCL is required but KVL isn't?
Both are fundamental "laws" that always hold. In terms of approaching a given problem sometimes one is more straightforward or advantageous in terms of the mathematical complexity. For example, there might be three loops but only two essential nodes in a circuit, so using KVL loops would require three equations to find the loop current, but only two node equations to find the node voltages. However, neither method would work at all if both KVL and KCL didn't hold for the circuit.
 
  • #32
monski24
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what is the answer for this problem. can you give us the steps how to deal with this
 
  • #33
Tom.G
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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.
Just a clarification here.

The Controlled Current Source (CCS) V0 is sensing the voltage across the 4 Ohm resistor. This is shown by the V0 symbol below the resistor. This is a differential measurement, i.e. fully floating with respect to any Ground, and conceptually uses two wires.

The wires are typically not drawn in the schematic to reduce clutter. The "+" and "-" symbols at the resistor show the polarity connection of the two sense wires for the CCS.

Hope this helps!
Tom
 
  • #34
boo
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Forget Vo for a minute. Just apply KCL at that top node (junction of the 2 current sources and the 4 ohm resistor). Take the bottom as the reference node. We then have (V - 0)/(4 + 6) -10 - 2*Vo =0. Now Vo = 4*(0 - V)/(4 + 6). Substitute for Vo and solve for V. Finally, since the current is entering the negative terminal of the dependent current source the power absorbed will be equal to - (V)*(2*Vo). (Hint Vo will actually be negative in this case.)
 
  • #35
boo
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Let me also point out that the main reason you are getting stuck on this problem is because you are taking Vo as the voltage at that top node which it is not. The left going current is neither Vo/4 nor Vo/(4 + 6). The correct (left going) current is simply V/(4 + 6) where V, as above, is simply the voltage at that essential node on top.
 

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