# what would be the unit of power

1. if a test question says 'what is the power dissipated by the dependent source?' but the dependent source voltage is negative , than should I write down the power in negative value?

2. if resistor unit is kiloohm and current unit is mA, what is the power unit? is it kW or W or mW?

• Really? Hint: kW W and mW are all the same unit. – R Drast Oct 17 '17 at 9:38

The sign used with power is an indication of the direction of energy flow. A positive sign usually indicates power supplied by a source and absorbed by a load. A negative sign usually indicates energy absorbed by a source and supplied by a load, the reverse of the "normal" direction of energy transfer. For this convention to work properly, care must be taken it the definitions of sources and loads, voltage polarities and current directions.

• Hmm... that isn't confusing at all is it.. Not a convention I have ever been happy with. – Trevor_G Oct 17 '17 at 1:33
• As in, if the load is driving the source.. then the load becomes the source and the source becomes the load.... in the end it all goes out the window as heat. – Trevor_G Oct 17 '17 at 1:37
• It doesn't really go out the window, the signs reverse when the energy flow reverses. It can be difficult to keep track of what is going on, but in real systems, you generally know if a reverse in power flow can be anticipated. If there are energy flow reversals in a system, it essential to have a system to keep track of them. – Charles Cowie Oct 17 '17 at 1:52
• Yup but I have always preferred the more definitive method of stating the battery absorbed 90W and the resistor dissipates 90W. The concept of something having negative energy simply goes against my grain. But I will admit I may be missing some key point. – Trevor_G Oct 17 '17 at 2:01
• I'm also pretty sure we are just confusing the OP. Perhaps some illustrations would help – Trevor_G Oct 17 '17 at 2:02
1. if a test question says 'what is the power dissipated by the dependent source?' but the dependent source voltage is negative , than should I write down the power in negative value?

Dissipated power is always positive. Voltage polarity is only relative to some other point so does not matter. In this case you use the absolute values.

1. if resistor unit is kiloohm and current unit is mA, what is the power unit? is it kW or W or mW?

It will be in whatever units you want to convert it to.

But in natural units you have from

$I^2R = mA * mA * k\Omega$ $=> 10^{-3}A* 10^{-3}A * 10^{+3}V/A = 10^{-3}VA => mW$

• Power does not have a direction. Yes it does. Passive sign convention dictates that positive is dissipated, and negative is generated. – Matt Young Oct 17 '17 at 1:22
• @MattYoung .. hmm.. not going there.... – Trevor_G Oct 17 '17 at 1:25
• @MattYoung reworded to appease those that adhere to that convention. – Trevor_G Oct 17 '17 at 2:31

if a test question says 'what is the power dissipated by the dependent source?' but the dependent source voltage is negative , than should I write down the power in negative value?

No because if you put a resistor across a battery's terminals then the resistor will heat up as much as if you reverse the battery's terminals.

Remember that $P=V×I$ and $V = I×R$, substitute and get $P=\frac{V^2}{R}$. Then the sign of the $V$ doesn't matter. Because $\frac{(-V)^2}{R}=\frac{(V)^2}{R}$.

if resistor unit is kiloohm and current unit is mA, what is the power unit? is it kW or W or mW?

Let me rephrase your question, if resistor is $1000Ω$ and current is $0.001A$, what is the power dissipated in the resistor?

$P = V×I = I^2×R=(0.001A)^2×1000Ω=0.001W=1$mW

• I had to check up on what was meant by "dependent source", and I learned from wikipedia that the power dissipated in one, depends on things, as I understand the matter, of no intrinsic relevance to the power dissipated in a load. Your example is 1mA in the load dissipating 1mW. If I understand OK, then if the dependent source has a low Rinternal, say 1ohm, then power dissipated in the dependent source would be 1uW, a microwatt. The 1mA of your example would be the power dissipated in the load, and load Pdiss wasn't the question. I myself stand to be educated here. – Stan H Oct 17 '17 at 5:49
• @StanH This question is not the best formatted so if I misunderstood it then so be it. - This is what I began hating about the university towards the end. You had the knowledge, you knew your stuff, but sometimes it's just difficult to understand the words so you need to look them up in wikipedia to even understand what the question even is. - I did not look up on wikipedia this time, hence my misunderstanding. – Harry Svensson Oct 17 '17 at 6:31
• I remember that aspect of universities. It was Charles Cowie's answer that alerted me to the fact that something about the question wasn't as it appeared to be on first reading. His answer covered subleties that made me go look for the definition. Happy to sit back now and continue my learning curve from other comments or answers. – Stan H Oct 17 '17 at 7:09
• Just discovered this question. Will look at answers soon. electronics.stackexchange.com/questions/264721/… – Stan H Oct 17 '17 at 7:12