Engineering Determining if voltage or current is positive or negative with sources

[Johnstonator]

Homework Statement

What value of alpha is required to make this a valid interconnection?
For this value of alpha, find the power associated with the current source.

Relevant Equations

v= alpha * i
P= i v

So I'm confused on a few things about this circuit. I know my math is right (since I got the absolute values correct), but the signs are off since I can't tell when the voltage, current or if the general power equation is positive or negative.

A) So my first question is, where is the current going? Is it going CW like what I drew in on here (based off of the current source) or is it going the opposite direction because of iDelta? My confusion stems from the other two questions below

B) The current iDelta in the first equation should be negative 15mA because it's pointed opposite from the current source right?

C) Is the voltage negative (-6V in the first equation)? The dependent voltage source polarity is different than the 6V polarity so I guess I can see that but is that something I'll have to remember rather than understand?

D) Finally, is the general power equation (P = i v) negative or positive? Since positive current entering a positive terminal is positive power, and in this case the current (dictated by iDelta) enters a negative terminal, I want to say it's negative. But if I use P = - i v, and current is negative, and voltage is negative, my power is negative (when the power is really positive).

 

[BvU]

Hello @Johnstonator , !

I had to look up the meaning of the diamond symbol , but other than that it is clear:

B) You are correct.
C) Again, correct ! Good thing too: the difference between + side and - side is -6V

So for A) -- which follows from B): clockwise ! That's what current sources do for a living

For D) I am slightly lost too, so you have to see for yourself if you agree with my reasoning:

The diamond side with the - sign is at +6 V, so no current from the top line flows in the central branch. That means the 15 mA is supplied by the Voltage source and the current source dissipates the 90 mW. In other words: I agree with you.​

However, that means we both disagree with the first answer in another forum !

I hope a real expert will come along and agrees (or proves us wrong ).
$\$

 

[gneill]

Homework Statement:: What value of alpha is required to make this a valid interconnection?
For this value of alpha, find the power associated with the current source.
Relevant Equations:: v= alpha * i
P= i v

So I'm confused on a few things about this circuit. I know my math is right (since I got the absolute values correct), but the signs are off since I can't tell when the voltage, current or if the general power equation is positive or negative.

A) So my first question is, where is the current going? Is it going CW like what I drew in on here (based off of the current source) or is it going the opposite direction because of iDelta? My confusion stems from the other two questions below

B) The current iDelta in the first equation should be negative 15mA because it's pointed opposite from the current source right?

Right. The $i_{\Delta}$ and arrow is just a label on the circuit diagram, and essentially just instructs you how to interpret the current flowing though that wire. It has no effect on the actual current there. So your interpretation that $i_{\Delta}$ will be $-15\;mA$ is correct.

C) Is the voltage negative (-6V in the first equation)? The dependent voltage source polarity is different than the 6V polarity so I guess I can see that but is that something I'll have to remember rather than understand?

It's straightforward to understand. If a source has to have a particular polarity in order to be correct in the circuit, then if the source has been "hardwired" to a given direction in the schematic, it follows that the sign of the value that source is assigned needs to compensate for any mismatch with "reality". A negative value for voltage or current sources is perfectly fine.

D) Finally, is the general power equation (P = i v) negative or positive? Since positive current entering a positive terminal is positive power, and in this case the current (dictated by iDelta) enters a negative terminal, I want to say it's negative. But if I use P = - i v, and current is negative, and voltage is negative, my power is negative (when the power is really positive).

The general equation p = iv complete in and of itself. It's up to us to define conventions as to what directions or polarities are to be associated with the i and v values. Typically this is done by defining a general two-lead component and the potential polarity and current direction for its leads. Think of a resistor with a potential drop across it. Current flows into its "+" lead.

We most often tend to think of "positive power" being associated with components that absorb or dissipate power. Things like resistors which dissipate power as heat. With this convention, components that are injecting power into a circuit are associated with negative power. (But that is not the only choice for a definition)

Yes, it can be a tad confusing if the convention to be used is not made explicit. I'm not fond of questions that simply state " find the power associated with (some component)" without making the definition clear. If the course material states a convention to be used, then follow that.

Things to keep in mind:

When a voltage source is pushing current out of its positive lead, it's injecting power into the circuit. By the common convention it would thus be associated with a negative power, even though it is in fact "generating" power. If current is flowing into the positive lead then it is absorbing power and would then be associated with a positive power value. If it was a battery, it would be being "charged up".

When a current source is pushing current out of its more positive lead (moving charge from the lower to higher potential across itself) it is also "generating" power and is thus associated with a negative power. On the other hand if it's pushing current out of its lower potential terminal then it is absorbing energy, and is then associated with a positive power value.

Going by this power convention, what signs would you associate with the sources for this circuit?

 

[DaveE]

Just to reinforce what @gneill said above:

There is a concept known as the "passive sign convention" which helps with polarity definitions. It basically boils down to define V and I signs as if the components were simple resistors which can only absorb power. So once you have chosen a current polarity, the voltage polarity is defined as if the current caused the voltage drop, as in a resistor. This means that the current always flows into the terminal with + voltage polarity. A component defined this way is generating power if P=VI is negative.

However, there is often some ambiguity in what people expect, some assumption in the language about whether people are talking about power generated or power dissipated. I see this often with photodiodes, for example. This doesn't prevent you from solving the circuit, but you may have to translate between your definitions and theirs.

You don't have to use the passive sign convention, but it can add some clarity or common structure to your approach. There are other common conventions, like current always flowing clockwise in a loop, or current always flowing into a node. You can't always apply everyone, there is a lot of personal preference at play here.

It is simplest, IMO, to just define a set of voltage and current polarities at the beginning without regard to the actual circuit solution. I would draw these on the schematic, like you did in the OP. Then you solve the equations without much concern over whether the resulting values are positive or negative. A negative value just means that your original definition was backwards. There is no need to guess correctly at the beginning, indeed sometimes that is nearly impossible. The important thing is to document your polarity choices and stay consistent. It is the consistency that is crucial, since sign errors are really easy to make in complex networks.

 

[Johnstonator]

Right. The iΔ and arrow is just a label on the circuit diagram, and essentially just instructs you how to interpret the current flowing though that wire. It has no effect on the actual current there. So your interpretation that iΔ will be −15mA is correct.

So the main takeaway is that whenever I see iΔ I just have to know that only applies to the dependent voltage source equation (v = alpha* i), and then ignore that for everything else?

Going by this power convention, what signs would you associate with the sources for this circuit?

For the 6V Voltage source, the power would be negative. P = -i v, where i = 15mA, and v = 6V.

For the dependent voltage source, the power would be positive (since you said iΔ has no effect on actual current). P = i v, where i = 15mA, and V= -6V?

For the independent current source, the power would be positive (since I would be ignoring iΔ ). P = i v, where i =15mA, v = 6V. A part of me wants to say the voltage is negative because the current I drew in purple goes from 6V, to 0V (I know this is wrong since it is positive). Since I drew the polarity of the current source based on where the 6V and 0V were, along with the current direction, the voltage cannot be negative across this element right? Or did I get the right answer with the wrong reasoning?

 

[BvU]

That way you would have 15 mA in all three branches from the top central node ?

 

[gneill]

So the main takeaway is that whenever I see iΔ I just have to know that only applies to the dependent voltage source equation (v = alpha* i), and then ignore that for everything else?

Well, in this particular circuit its only function is to relate the controlled source to the current in that particular wire, so yes.

For the dependent voltage source, the power would be positive (since you said iΔ has no effect on actual current). P = i v, where i = 15mA, and V= -6V?

You'll have to explain how you arrived at a current of 15 mA through the controlled (dependent) source. Did you find a value for $\alpha$ and what was your reasoning?

 

[Johnstonator]

You'll have to explain how you arrived at a current of 15 mA through the controlled (dependent) source. Did you find a value for $\alpha$ and what was your reasoning?

$\alpha$ is 400 V/A.

Vs =$\alpha$ iΔ , where, Vs = -6V, iΔ= -15mA, so $\alpha$ = 400V/A.

So it sounds like I can't say iΔ (-15mA) is the value of the current going through the dependent voltage source then?

 

[gneill]

$\alpha$ is 400 V/A.

Vs =$\alpha$ iΔ , where, Vs = -6V, iΔ= -15mA, so $\alpha$ = 400V/A.

Good!

So it sounds like I can't say iΔ (-15mA) is the value of the current going through the dependent voltage source then?

Both the independent and dependent sources are to be considered ideal and are in parallel in this circuit. So how might you decide how much of the (fixed) available current passes though each?

 

[Johnstonator]

Both the independent and dependent sources are to be considered ideal and are in parallel in this circuit. So how might you decide how much of the (fixed) available current passes though each?

I think I see it. iΔ only applies when I have Vs =$\alpha$ iΔ. I forgot to think about that iΔ can be ignored.

I want to say use KVL/KCL here but that doesn't seem right to me?

 

[gneill]

Think of the situation where you have two identical voltages sources in parallel supplying current to a load (something that may be approximated by having two batteries in parallel).

How much of the total current would you attribute to each battery?