Analyzing Power from Current Source

Thread Starter

stoltz1986

Joined Sep 21, 2015
7
Hello,

I am having trouble understanding how the power is analyzed for the current source in the attached circuit.
Capture.PNG
I have solved the circuit using the node voltage method and determined v0 to be +4 V with the polarities like this:
Capture.PNG .
I then calculated the power dissipated by the resistors as follows:

R20 = (.2A)²(20Ω) = +.8 W,

R80 = (.2A)²(80Ω) = +3.2 W,

R25 = (4V)²/(25Ω) = +.64 W,

with the total power dissipated by the resistors as +4.64 W.
Then for analyzing the power for the sources, I have as follows:

VS = -(24V)(.2A) = -4.8 W,

IS = -(-4V)(.04A) = +.16 W.

Adding all of these up I end up with total power dissipated at +4.8 W and total power produced at -4.8 W, which is great. The problem I am having is I know the current source is absorbing power but I do not know if I am calculating it correctly. I am assuming the voltage would be negative for the current source because the voltage is reversed from the voltage of the rest of the circuit. Is the way I did it above correct or would it be calculated with one of these options:

IS = -(4V)(-.04) = +.16 W or,

IS = (4V)(.04) = +.16 W?

Thanks in advance for any help, it is greatly appreciated.
 

Attachments

DGElder

Joined Apr 3, 2016
351
Is that 25 ohm resistor supposed to be part of the current source? I can't read the subscript you have for r. You need to know that to determine the CS power dissipation.

Assuming it is not part of the current source then your answer looks correct to me, the current source is acting as a current sink.
 
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MrAl

Joined Jun 17, 2014
11,389
Hi,

If you can replace the source with a passive device (resistor) and everything remains the same, then the source must have been absorbing power. The resistor value would have to be R=V/I or else it doesnt work. If doing that changes anything, then it must have been delivering power. If doing that does not change anything, then the source must have been absorbing power.
 

Thread Starter

stoltz1986

Joined Sep 21, 2015
7
Hey DGElder. Thanks for the response. Yeah, the 25 ohm resistor is not part of the current source. The subscript next to the 25 ohm resistor is v sub 0, with positive polarity at the top of the circuit and negative polarity at the bottom.

Thanks MrAl. I was not aware of that technique.

I know that the current source is absorbing power but what I'm wondering is what is the proper sign assignment for the values in determining the power for the current source. I know with passive sign convention that if a source is absorbing power that power will be positive and vice versa. So I guess what my question is is what equation is the correct one for determining the power for the current source?
1) -(-4V)(.04A) = .16W
2) -(4V)(-.04A) = .16W
3) (4V)(.04A) = .16W
I was thinking the first one with voltage being negative because of the way I assigned polarities in the circuit.
Also, here is a larger picture with my assigned polarities.Capture.JPG
 

DGElder

Joined Apr 3, 2016
351
Your math is wrong in equation 1), but that aside...
I wouldn't assign a voltage polarity to a current source. It's superfluous, adds confusion ( to which you can attest) and is potentially contradictory if you goof the signs (which you have done). The voltage polarity is defined across the 25 ohm and the current polarity is defined by the CS arrow symbol; that is all you need, so...
P = 4V *.040A
 
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MrAl

Joined Jun 17, 2014
11,389
Hey DGElder. Thanks for the response. Yeah, the 25 ohm resistor is not part of the current source. The subscript next to the 25 ohm resistor is v sub 0, with positive polarity at the top of the circuit and negative polarity at the bottom.

Thanks MrAl. I was not aware of that technique.

I know that the current source is absorbing power but what I'm wondering is what is the proper sign assignment for the values in determining the power for the current source. I know with passive sign convention that if a source is absorbing power that power will be positive and vice versa. So I guess what my question is is what equation is the correct one for determining the power for the current source?
1) -(-4V)(.04A) = .16W
2) -(4V)(-.04A) = .16W
3) (4V)(.04A) = .16W
I was thinking the first one with voltage being negative because of the way I assigned polarities in the circuit.
Also, here is a larger picture with my assigned polarities.View attachment 106049

Hi again,

Well, i have no idea why you would want to assign the plus (+) and minus (-) signs to the current source in that way. That does not show the true voltage across the current source. The voltage polarity for the current source must be the same as for the resistor in parallel with it, and since you have determined that the upper node is positive and the lower negative, then that must be the same for the current source. If that current source was a load that would be the same situation.
The arrow in the current source is correct because that shows the direction of the current in the current source, but the polarity must be the same as across the parallel resistor. As i mentioned before, if the current source is replaced by a resistor just to test our idea, that's the voltage polarity we would see again.

I am still a little curious though to find out why you assigned 'that' voltage polarity to the current source. That is the polarity when the current source is supplying power to another part of the circuit. The voltage polarity of the current source has to come from the external circuit components not from the source itself unless it is the only source, and of course if it was like that then it would have been found that it was supplying power to the circuit.
 

DGElder

Joined Apr 3, 2016
351
To head off possible confusion, let me rephrase my last sentence....
The voltage polarity is defined across the 25 ohm, Vo, and the current direction is defined by the CS arrow symbol; that is all you need, so...
P = Vo *0.04A
P = 4V *0.04A
 

Thread Starter

stoltz1986

Joined Sep 21, 2015
7
Ok. That makes sense. Yeah, I'm not exactly sure why, but I was under the impression that current sources were supposed to be marked that way. I'm learning this on my own and I often find myself over complicating simple problems. Thanks again to both of you.
 

WBahn

Joined Mar 31, 2012
29,978
I know that the current source is absorbing power but what I'm wondering is what is the proper sign assignment for the values in determining the power for the current source. I know with passive sign convention that if a source is absorbing power that power will be positive and vice versa. So I guess what my question is is what equation is the correct one for determining the power for the current source?
There are a couple variants of the passive sign convention. First, realize that it is only that -- a convention. You could assign a current direction to each device by flipping a coin and then assign a voltage polarity to each device by flipping another coin. As long as you are consistent, you will be fine -- but if you do it this way you will find that being consistent is easier said than done.

This is where the passive sign convention comes in -- it helps you be consistent.

The convention you seem to be trying to use is the one that says that positive current through a device symbolically flows into the positive voltage terminal and out the negative voltage terminal. This applies whether the device is a source or a load and if the power computed for the device turns out to be positive then the device is absorbing power, otherwise it is supplying power.

This is a perfectly fine convention, but it does have the weakness that it forces us to routinely work with negative powers and humans are not very good dealing with negative numbers -- we make a lot more mistakes compared to when we work with positive numbers. So a very common variant of the passive sign convention is that for devices that we expect to behave as sources, we have the current flow OUT of the positive terminal so that the device has positive power if it is delivering power to the circuit.

You can use either one, you have to use it consistently. Of course, other people will have no idea what you are using, so this underscores the importance of defining both the location and the polarity of the various terms you use.
 

MrAl

Joined Jun 17, 2014
11,389
Hello again,

The way i understand it is that the passive sign convention is an across the board convention meaning it works for both passive (like resistors) and active devices (sources). It's actually defined by power flow though and for an active device like a voltage source it can be negative. If the source is supplying power then it must be negative power or else an audit of the power in a circuit would show up as being twice as high as what we are actually putting into it.

For a quick example, a 1v source and a 1 ohm resistor, the resistor absorbs 1 watt and the source supplies 1 watt, so the power in the resistor is positive and the power in the source is negative. The power levels balance out so all is well :) If we call both powers positive, then the total power is 2 watts, which doesnt seem to have any practical appliciation.

For energy though it is just the opposite, the source supplies so much (positive) energy and the load absorbs that energy (negative), but the convention is based on 'power' not energy.

If we had a circuit that had 1 watt going in and 0.9 watts coming out, that would mean that the circuit itself was absorbing 0.1 watts so the efficiency is 90 percent. The efficiency is a ratio but if we only knew the power absorbed by the circuit (0.1 watts) we could subtract that from the input power (1 watt) and get the output wattage (0.9 watts) and then we'd be able to calculate the efficiency. This is backwards but works because it's really based on energy.
If we knew the output power (0.9 watts) and the input power (1 watt) then we'd be able to calculate the power absorbed by the circuit by subtracting 1-0.9=0.1 so that has a lot of practical significance too.

But a main point here is that we have to know what is a passive element and what is an active element, and then proceed accordingly. If the element is active then it is supplying power, and if the element is passive then it is absorbing power, and the convention covers both cases. Positive current entering the most positive voltage node means the element is passive, positive current leaving the most positive voltage node means the element is active. Unless there is only one source the most positive voltage node has to be solved for when considering a current source, and for a voltage source we have to solve for the direction of the current, into or out of the sources most positive node (in=absorbs power, out=supplies power).

I realize this is a little confusing at first. Just like anything else though, once a few problems are worked it starts to made a lot of sense. Just like learning negative numbers for the first time we have to learn all the rules like with multiplication where two signs result in a positive sign :)
 

WBahn

Joined Mar 31, 2012
29,978
For a quick example, a 1v source and a 1 ohm resistor, the resistor absorbs 1 watt and the source supplies 1 watt, so the power in the resistor is positive and the power in the source is negative. The power levels balance out so all is well :) If we call both powers positive, then the total power is 2 watts, which doesnt seem to have any practical appliciation.
It all depends on using whichever convention you choose to consistently.

If you blindly apply the same assignment to all components, then you have to consistently use the relationship that the sum of all powers has to be zero. Perfectly acceptable.

But another perfectly acceptable and reasonable relationship is that the sum of all power supplied by all sources must equal the sum of all power dissipated by all loads. To be consistent with this, the assignments on all sources must be such that they are the same sign (universally chosen to be positive) when supplying power as the sign on loads when they are absorbing power.

You just need to be consistent.

For energy though it is just the opposite, the source supplies so much (positive) energy and the load absorbs that energy (negative), but the convention is based on 'power' not energy.
Since power is, by definition, the time-rate-change of energy, they could only be "opposite" if time is negative. If you want to claim that a source supplies positive energy, then that same source HAS to supply positive power. That is, unless you either want to declare that P = - dE/dt or you want to join the electron flow crowd and start sprinkling magical mystery minus signs throughout your work.
 

DGElder

Joined Apr 3, 2016
351
You guys are making this way too complicated. By the Passive Sign Convention: for ANY device that is dissipating power, P (power consumed), is positive; for ANY device that is delivering power, P (power consumed) is negative. That's it.

A voltage source is just that, a source of voltage - not neccesarily power. A current source is just that, a source of current - not neccesarily power. Don't take "source" too literally as both can act as sinks or sources.

One can invent other ways of bookkeeping but one would only succeed in confusing other people trying to understand their circuit analysis and probably confuse themselves. The PSC is universally accepted by EEs so if one wants to be part of that world then one must conform - resistance is futile.
 
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WBahn

Joined Mar 31, 2012
29,978
You guys are making this way too complicated. By the Passive Sign Convention: for ANY device that is dissipating power, P (power consumed), is positive; for ANY device that is delivering power, P (power consumed) is negative. That's it.
As I pointed out above, that is ONE interpretation of the passive sign convention -- there is more than one way to interpret the intent of the term "passive" in that name. It is not the ONLY interpretation (although it is arguably the purest one) and, to the best of my knowledge, it is not even the most widely used interpretation in the practical world where people are far more interested in minimizing mistakes than they are in adhering dogmatically to something the way that people that have a shaky/inexperienced grasp of the underlying concepts often need to.
 

DGElder

Joined Apr 3, 2016
351
It's not open to nor dependent on the interpretation of the word "passive". The convention is unambiguously defined and followed by EEs in academia and industry. Suggesting they/we are inexperienced, impractical, with a shaky grasp of concepts is ridiculous. The next time you are out driving your car, if you don't want to slavishly adhere to the sign convention of a one-way street sign then don't complain after your head on collision with a truck.
 
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MrAl

Joined Jun 17, 2014
11,389
Hello again,

WBahn:
Not sure what you are saying in total there. I am coming from the point of circuit analysis, which does not obey 'energy' laws in the same way that a physics forum might. In physics, energy is either supplied or absorbed, and it is really always positive. Energy in equals energy out when we have 100 percent efficiency. But in circuit analysis, the power can be negative because that helps us understand certain things more clearly, or should i say mathematically even though they are not really physical quantities we can really measure, such as negative power. We dont measure negative power, but we use that concept anyway in circuit analysis itself. In other words, we divide the elements into the two groups, suppliers and absorbers, and make the absorbers positive. We dont have to do this, but it makes sense to do so. We can also call the energy in positive and the energy absorbed positive, but then we have to keep track of a third thing: how the power is actually flowing, when we dont have to keep track of that when we use signs because the sign says what it is and carries through.
I know what you mean about being consistent through, i also believe that is the most important thing.

An example might similar to what DGElder was talking about. If we had three sources, V1, V2, and V3, and maybe resistors connecting them, we dont know what is happening if we say that:
V1 power is 10 watts,
V2 power is 15 watts,
V3 power is 5 watts.

The question is, which source(s) is/are absorbing power? If we say that V2 supplies 15 watts, then V1 and V3 are absorbing, but if we say that V2 is absorbing, then V1 and V3 are supplying.

But now let's ask that same question including the signed power:
V1 power is +10 watts,
V2 power is -15 watts,
V3 power is +5 watts.

Which source(s) is/are absorbing power? We know right away that V1 and V2 are absorbing because we are using the passive sign convention where when current enters the most positive voltage node the power is absorbed, and we usually call this positive.

Again consistency, as you say, is the most important so we want to keep that in mind too :)

Just a side note:
Voltage sources dont really have just one sign, they have two signs.

"To be consistent with this, the assignments on all sources must be such that they are the same sign (universally chosen to be positive) when supplying power as the sign on loads when they are absorbing power."

Not sure i understand that part. You mean the 'assumed' sign, before the analysis is done?
Like when we call a battery voltage Vbatt, and it is positive up and negative down, then later we find out that is was discharged and charged in reverse so the sign reversed so Vbatt=-1.5 for example instead of the assumed +1.5v or so.

Capacitors can absorb or deliver power, first absorbing then delivering.
 
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MrAl

Joined Jun 17, 2014
11,389
It's not open to nor dependent on the interpretation of the word "passive". The convention is unambiguously defined and followed by EEs in academia and industry. Suggesting they/we are inexperienced, impractical, with a shaky grasp of concepts is ridiculous.
Hi,

Yes, that is the way the passive sign convention is defined. There is no interpretation here because it's a set way of doing things, and is determined by the CIRCUIT itself NOT by the way we assign polarities.

The passive sign convention is a convention just like "conventional current flow" is from the positive to the negative, which is not subject to interpretation it's an agreed upon standard.

The other convention would be called something else, not passive sign convention.
So in other words, this is not correct:
"We use a passive sign convention",

but this is correct:
"We use the passive sign convention".

The difference is 'a' and 'the'. There's only one passive sign convention.
 

WBahn

Joined Mar 31, 2012
29,978
Hello again,

WBahn:
Not sure what you are saying in total there. I am coming from the point of circuit analysis, which does not obey 'energy' laws in the same way that a physics forum might.
They better, otherwise you are working with a mathematically inconsistent system.

If we have an amount of energy that is stored in anything, be it a battery, a capacitor, or an inductor, then our definition of power needs to be consistent with that concept.

Clearly if a black box contains an amount of energy E, then positive power flowing into that box better result in the amount of energy in that box increasing. Similarly, positive power leaving that box better result in the energy in that box decreasing. This of course applies to the total of all energy flow across the box's boundary.

Just a side note:
Voltage sources dont really have just one sign, they have two signs.
They have a single polarity. If they had "two signs", then you could assign one sign to one end and then arbitrarily assign a sign to the other end -- so you could assign + to both ends or - to both ends. They have a single polarity -- End A is either more positive than End B or the other way around.

"To be consistent with this, the assignments on all sources must be such that they are the same sign (universally chosen to be positive) when supplying power as the sign on loads when they are absorbing power."

Not sure i understand that part. You mean the 'assumed' sign, before the analysis is done?
Yes -- that is the "symbolic" polarity. Using the extremely common underlying relationship that the power supplied by the sources equals the power absorbed by the loads, you may (though hardly anyone would) assign voltages and currents such that power supplied by a source is negative as long as you also defined power absorbed by a load as being negative. All that matters is that these two situations have the same sign. It is no different than using a very common interpretation of KCL that states that the sum of the currents entering a node equals the sum of the currents leaving a node. This is equivalent to the interpretation that all currents at a node must sum to zero, except in this case the underlying assumption is that all currents are symbolically entering the node (or all are exiting). The assigned symbolic directions and resulting positive/negative results differ between these two cases, but as long as you are internally consistent in assigning polarities according to your interpretation, everything works out just find.

Like when we call a battery voltage Vbatt, and it is positive up and negative down, then later we find out that is was discharged and charged in reverse so the sign reversed so Vbatt=-1.5 for example instead of the assumed +1.5v or so.
Huh?

Vbatt would (should) always be taken to be the positive terminal relative to the negative terminal. If the battery was discharged the terminal voltage would start out at something less than 1.5 V, but it would NOT be negative. The current would have the opposite sign of what it would have if the battery were sourcing power to the circuit, hence the sign of the power would be the opposite as well (according to either interpretation since both interpretations are mathematically consistent).

Capacitors can absorb or deliver power, first absorbing then delivering.
Sure. So what? The instantaneous power absorbed by the capacitor is either positive or negative depending on whether the capacitor is actually absorbing or delivering power at that moment in time. The same is true for inductors or any other device.
 

WBahn

Joined Mar 31, 2012
29,978
It's not open to nor dependent on the interpretation of the word "passive". The convention is unambiguously defined and followed by EEs in academia and industry. Suggesting they/we are inexperienced, impractical, with a shaky grasp of concepts is ridiculous. The next time you are out driving your car, if you don't want to slavishly adhere to the sign convention of a one-way street sign then don't complain after your head on collision with a truck.
I have several EE textbooks that refute your claim.

It very much does depend on the interpretation of the word passive. One interpretation uses the word in the sense of it being a sign convention for passive components. The other interpretation uses the word in the sense that the once the polarity of one quantity is assigned that the polarity of the other quantity is passively implied.

Unlike a one-way street sign, whose interpretation is set by a regulatory body having the legal authority to do so, conventions are not. If you think they are, then please provide the name of the regulatory body having authority to dictate the one true definition of "passive sign convention," a reference to the legal charter that granted them that authority, and the body that made that grant.

There are plenty of arbitrary concepts that have such legal underpinnings -- for instance, the definition of one ampere or one volt -- but (to the best of my knowledge), the passive sign convention is not among them.
 

WBahn

Joined Mar 31, 2012
29,978
We can certainly agree to disagree. That is probably the best course since you seem to believe that Wikipedia is an authoritative source.
 
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