# Calculating Power - Simple DC Circuit

• Engineering
• JosefMTD
In summary: No, the generator-load convention is to make the voltage at the current source be generating power, not absorbing because it's positive.
JosefMTD

## Homework Statement

http://puu.sh/7gYwk.png

Consider the circuit shown above. Find the current Ia flowing through the resistor and the power for each element in the circuit. Which elements are absorbing power?

## Homework Equations

P = V x I = I^2 x R = V^2 / R

## The Attempt at a Solution

I tried to find Ia first, the voltage source with the load produce 2A going the opposite direction from the current source, so I got Ia = 0A and makes all the power = 0W

Is this correct?

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Hi JosefMTD! http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif

Is that a constant current source in your circuit? What do you understand is the behaviour of a "constant current source"?

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I think it's a constant current source.

It's always of the same value and stays on the same direction, does it override all other current source though? I have no idea, I may have missed this during class :-\

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Yes, the constant current source always wins! Everything else must accommodate the dictates of a constant current source.

So you never arrange two where there would be conflict, or there will be fireworks!

Well, thank you very much. I must have fail to listen carefully during class while this happens :P

But then,
Power on resistance = (2A)(2A)(5Ω) = (10 V)(10 V)/(5Ω) = 20 W
Power on voltage source = (10V)(2A) = 20 W
Power on current source = (-10V)(2A) = -20 W

It adds up to 20W, but according to conservation of energy: ∑p = 0 for a closed circuit. My "gut" feeling says the voltage source should have 0 W power, don't know why. So anyways, where did I do wrong ?

Why did you decide to throw in that negative sign when calculating the current source's power output?

Because of the active sign convention.

I did it like I did on the previous question:
http://puu.sh/7h4PK.png

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By the way, can you offer an explaintion of why you believe the current source has a voltage of 10v?

I'm taking reference from the previous question which is shown up there, dunno, I figured it should be the same, or it shouldn't?

Well this one has a resistor in the circuit, so that makes it a bit different.

If you take the voltage at the bottom of the voltage source as V0 = 0v, then work around the circuit and calculate the voltage at each node.
At the node bewtween the voltage source and the resistor, V1 = ?
At the node between the resistor and the current source, V2=?
At the node between the current source and the voltage source V3=?

Oh, I see.
V1 = 10V
V2 according to KCL
∑I = 0, V2-V1/5Ω - 2A = 0
V2 - 10V = 10V, V2 = 20V
V3 according to KVL
V2-V3 = V3-V0
20V = 2V3
V3 = 10V

Damn, I'm confused

How many volts across the constant voltage source?
How many volts across the resistor itself?

I guess that is confusing, when you put it that way.

I thought you would have recognized the following:
V3 = V0 = 0 since they are the same node.
You really do not have to do a KVL calculation on it.

With KVL you can do:
V2-V3 = V2-V0 ie summation of voltages going left from V2 = going right from V2
V2=V2

NascentOxygen said:
How many volts across the constant voltage source?
How many volts across the resistor itself?

1. 10V
2. With using KCL on the node between current source and resistor, I got 10 V across the resistor

I don't know why, but I never had problems in solving these kinds of problems, I think I forgot all of the basics because I had a too long vacation :P

OMG.
@256bits

V2 = 0 V
(V1 - V2)/5 ohm = 2A [KCL]
(10 V - V2) = 10 V
V2 = 10 V - 10 V = 0 V

No voltage across current source.

Do we use the generator-load convention to make the 20 W on voltage source to be generating power not absorbing because it's positive?

Thank you guys so much :D
Finally, I think I'm getting to know the basics again XD

JosefMTD said:
No voltage across current source.

Really? I think maybe you still have a little work to do on the basics.

Also, as nascentoxygen alluded to in post #4, if this were a real world situation and the voltage source was a battery, it would most likely explode. Do you understand why?

Because the current generated from the constant voltage source would always make it 2A to the opposite side of constant current source. Ka boom.

JosefMTD said:
OMG.
@256bits

V2 = 0 V
(V1 - V2)/5 ohm = 2A [KCL]
(10 V - V2) = 10 V
V2 = 10 V - 10 V = 0 V

No voltage across current source.

You already had across the voltage-source 10v, and across the resistor 10v.
so how do you now get 0v?

(V1 - V2)/5 ohm = 2A [KCL]
Try,
(V2- V1)/5 ohm = 2A [KCL]
in the direction of the current.

OH OH OH OH OH LOL
LOL

Yeah, I messed up XD
I realized where I did wrong XD
Thank you guys for bearing with me ;D

(V2 - V1)/5 ohm = 2 A
V2 - V1 = 10 V
V2 = 20 V

V3 = 0 V

Voltage across the constant current source = 20 V
with the active sign convention, the power on constant current source is - 40 W.
with the passive sign convention, the power on constant voltage source is 20 W
the power on resistor is 20 W

Summation of all the power in components is 0. I'm a bit sure that this one is right, now ;D

OK. That's fine.

JosefMTD said:
Because the current generated from the constant voltage source would always make it 2A to the opposite side of constant current source. Ka boom.

Absolutely wrong. Ideal sources always have exactly what they say they have. You cannot HAVE ideal sources in the configuration you show.

My question was why would a real-world BATTERY explode. You seemed to think it would be the current source that would explode.

Also, you did not answer my question about the voltage across the current source. You really DO need to go back and restudy the basics.

phinds said:
Absolutely wrong. Ideal sources always have exactly what they say they have. You cannot HAVE ideal sources in the configuration you show.

My question was why would a real-world BATTERY explode. You seemed to think it would be the current source that would explode.
Not quite. A battery is recharged in this manner, for example.

An ideal voltage source will happily source or sink any amount of current in order to maintain its fixed potential difference. Similarly, an ideal current source will produce any potential difference (positive or negative) required in order to maintain its fixed current.

gneill said:
Not quite. A battery is recharged in this manner, for example.
Well, yeah, but normally at a trickle charge, not 2 amps !

An ideal voltage source will happily source or sink any amount of current in order to maintain its fixed potential difference. Similarly, an ideal current source will produce any potential difference (positive or negative) required in order to maintain its fixed current.

Right. I was wrong about the circuit having a contradiction in the ideal case, since the 10v source will just let -2 amps flow through it. Thanks for that correction.

I do agree I need to read the basics again.

"OK. That's fine"
I'll assume that equations I brought up is correct then. Thank you all for helping me to solve this problem. ;D

## 1. How do I calculate power in a simple DC circuit?

To calculate power in a simple DC circuit, you need to know the voltage and current in the circuit. You can use the formula P = VI, where P is power in watts, V is voltage in volts, and I is current in amps. Simply multiply the voltage by the current to get the power.

## 2. What is the unit of power in a DC circuit?

The unit of power in a DC circuit is watts. This is the same unit used to measure electrical power in other systems, such as light bulbs and appliances.

## 3. Can power be negative in a DC circuit?

Yes, power can be negative in a DC circuit. This occurs when the current and voltage are in opposite directions, resulting in a negative value. Negative power can indicate that energy is being absorbed or consumed by the circuit, rather than being produced.

## 4. How does power relate to voltage and current in a DC circuit?

Power is directly proportional to both voltage and current in a DC circuit. This means that as the voltage or current increases, the power will also increase. Similarly, if the voltage or current decreases, the power will decrease as well.

## 5. Can I use the power formula to calculate power in an AC circuit?

No, the power formula P = VI is only applicable to DC circuits. In AC circuits, the power formula is P = VIcosϕ, where ϕ is the phase angle between voltage and current. This is because in AC circuits, the voltage and current are constantly changing and may not be in phase with each other, resulting in a different calculation for power.

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