Power supply and power dissipated in a circuit

[athrun200]

Can Power supply be smaller than power dissipated?
Because I saw this in one question

There are 3 cells, the power supply of the two 4V cells cancel each other. (one is -4W and another one is 4W)
So the total power supply should be 16W

But in part c, the power dissipated is 4+8+2+6=20W

Power supply < power dissipated?

 

[Simon Bridge]

Can Power [supplied] be smaller than power dissipated?

No. Conservation of energy and all that.

the power supply of the two 4V cells cancel each other

Not exactly - try the problem again, only this time remove the part with the 6ohm resistor.

24W is supplied, but 4W is the rate work is being done against the current - leaving 20W of energy for the current to carry. 20W is dissipated. It balances.
If you built this you'd notice that the "-4W" supply gets hot like all the others.

You can think of this circuit as three PSUs with internal resistances driving a 6ohm load.

 

[Simon Bridge]

Note: treating the bit without the 6Ohm resister as the supply, the Thévenin equivalent would have a voltage of 8V and resistance 2Ohms.

You'll see this still gives 6W dissipated in the 6Ohm resister, and 2W in the Thévenin resister, while 8W is supplied. Again - everything balances.

 

[athrun200]

It seems the Thévenin equivalent should have 7.2V with 1.2 ohm resistance.

 

[Simon Bridge]

Did I mess up?
http://en.wikipedia.org/wiki/Thévenin's_theorem#Calculating_the_Th.C3.A9venin_equivalent

V_AB = open circuit condition.
Ah - that's what went wrong... I just subtracted the voltages.

volt drop across 5Ω is 12-4=8V
thus, volt drop across the 2Ω is 3.2V so V_AB must be 7.2V well done.
Did you take it the rest of the way?

 

[willem2]

Using Thevenin equivalents isn't the easiest way here.

What I did was.

1. replace the 8v and 4v source by a 12 v source, and the 1ohm and 2ohm resistance that are in series with a 3 ohm resistance

2. assume the potential at the bottom is 0, and the potential at the top right is U.

3. work out the currents in the 3 branches: (V-12)/3, (V-4)/2, V/6 (all pointed downwards)

4. the sum of the currents must be 0 (kirchhof)

so U/3+U/2+U/6 - 12/3 - 4/2 = 0, so U=6 volt.

then it's easy to work out the currents with the equations already given.

The answer sheet is entirely correct, but it doesn't give the power in the top 4V source.
This power is not equal to 4W, and it doesn't cancel the power of the other 4V source.
I have no idea why you should think that.

The same current is going through it as through the 8V source, so the power must be half the power of the 8V source.

 

[athrun200]

Thanks everyone, I understand it now.