Determining the voltage polarity of the independent current source

[fight_club_alum]

TL;DR Summary

How can I determine the voltage polarity of the current source in the circuit attached?

I am so sorry if I am posting this in the wrong forum; it is just not a homework problem, and I can't find the right place - it's more of a study help question.

 

[phinds]

Summary: How can I determine the voltage polarity of the current source in the circuit attached?

Just add up the voltages around the loop and make it zero by assigning the needed voltage to the current source.

 

[fight_club_alum]

Just add up the voltages around the loop and make it zero by assigning the needed voltage to the current source.

Thank you so much for replying
I did so and got -5 V (with that number how can I determine the positive and negative ends of the current source's voltage?)

 

[fight_club_alum]

Thank you so much for replying
I did so and got -5 V (with that number how can I determine the positive and negative ends of the current source's voltage?)

It would make a difference if I am calculating, for instance, the power developed and dissipated

 

[phinds]

Thank you so much for replying
I did so and got -5 V (with that number how can I determine the positive and negative ends of the current source's voltage?)

You just did.

 

[phinds]

It would make a difference if I am calculating, for instance, the power developed and dissipated

No, it would not. Why is that true?

 

[fight_club_alum]

No, it would not. Why is that true?

if I am going through the positive terminal of a component, I will write P = +V I
if I am going through the negative terminal of a component, I will write P = - V I
Isn't that correct?

 

[phinds]

if I am going through the positive terminal of a component, I will write P = +V I
if I am going through the negative terminal of a component, I will write P = - V I
Isn't that correct?

No. Power does not have a polarity. Think about it this way: if you have a single 10 ohm resistor drawn vertically and you connect +10 volts to the top and ground the bottom, the power dissipation is 10 watts. If, on the other hand, you connect +10 volts to the bottom and ground the top, the power dissipation will be 10 watts.

@fight_club_alum another way to look at it is that P = I*2 x R so clearly it doesn't matter which way the current is flowing.

 

[cabraham]

The polarity matters. The current source has a terminal voltage of -5 volts, minus on top, plus on bottom. Since positive current exits the positive terminal, the current source is delivering power.
If the terminal voltage were reversed plus on top, then the current source is receiving power.
An example is a car battery. When starting engine, positive current exits pos terminal. Battery delivers power to crank the starter. After engine starts, pos current from alternator enters battery pos terminal. Battery here is receiving power from alternator, ie getting recharged.

Claude Abraham
EE 41 years
PhD candidate

 

[phinds]

Good point. I was forgetting that it's an active component so my analogy with a passive component is not a good one.

 

[anorlunda]

It would make a difference if I am calculating, for instance, the power developed and dissipated

Yes it makes a difference. But remember your example with only ideal voltage sources and ideal current sources is very unrealistic. There is no real life circuit that behaves like that example.

It is easy to create contradictory examples using ideal components. For example an ideal voltage source with a short circuit that predicts infinite current. Ideal components are extremely useful to help analyze real world circuits which never contain infinities. It is silly to apply them to unrealistic circuits.

 

[anorlunda]

No. Power does not have a polarity. Think about it this way: if you have a single 10 ohm resistor drawn vertically and you connect +10 volts to the top and ground the bottom, the power dissipation is 10 watts. If, on the other hand, you connect +10 volts to the bottom and ground the top, the power dissipation will be 10 watts.

That is a bit overstated. Your example is true only true for a resistor. If power had no polarity it wouldn't matter if power flows from power plant to customer or from customer to power plant.

 

[phinds]

That is a bit overstated. Your example is true only true for a resistor. If power had no polarity it wouldn't matter if power flows from power plant to customer or from customer to power plant.

Yes, see post #10

 

[anorlunda]

Good point. I was forgetting that it's an active component so my analogy with a passive component is not a good one.

We all trip over our own words once in a while. I know you know this, but it is not even true for every passive circuit.

The correct statement is that P=VI does have polarity, and can be positive and negative. But in a passive resistor, when V changes sign, so does I. When the circuit includes passive L or C, that is no longer true.

 

[phinds]

We all trip over our own words once in a while. I know you know this, but it is not even true for every passive circuit.

The correct statement is that P=VI does have polarity, and can be positive and negative. But in a passive resistor, when V changes sign, so does I. When the circuit includes passive L or C, that is no longer true.

 

[sophiecentaur]

another way to look at it is that P = I*2 x R so clearly (??) it doesn't matter which way the current is flowing.

True but (for a passive device) that doesn't take care of the P = VI sign. If you are measuring relative to the Positive terminal you will get +V and +I; if you measure relative to the - terminal you get a -V and -I so VI is positive either way.
Kirchoff II tells us that the emf's subtract from the IR's and the signs also behave properly according to whether Power is going in or going out.

 

[anorlunda]

True but (for a passive device) that doesn't take care of the P = VI sign.

Did you miss #13?

 

[sophiecentaur]

I think the dementia is kicking in.