Computing consumed power of circuits of multiple voltage sources

WBahn

Joined Mar 31, 2012
26,398
If E1 is a perfect voltage source it will have zero impedance and, therefore, cannot consume power.
By that line of reasoning, a perfect voltage source cannot deliver power, either.

The power is P=V*I where V and I are defined such that V is positive if positive I is entering at the positive terminal. If P is positive then the device is consuming power and if P is negative then the device is producing power.

What actually happens to the power consumed by a voltage source depends on the source. Some will store the power chemically, some will store it mechanically, some will dissipate it as heat, others will do other things. The key thing is that it is removed from the circuit being analyzed.
 

blocco a spirale

Joined Jun 18, 2008
1,546
By that line of reasoning, a perfect voltage source cannot deliver power, either.

The power is P=V*I where V and I are defined such that V is positive if positive I is entering at the positive terminal. If P is positive then the device is consuming power and if P is negative then the device is producing power.

What actually happens to the power consumed by a voltage source depends on the source. Some will store the power chemically, some will store it mechanically, some will dissipate it as heat, others will do other things. The key thing is that it is removed from the circuit being analyzed.
What is the impedance of a perfect voltage source?
 

Thread Starter

Danuuuuu

Joined Jan 2, 2015
3
By that line of reasoning, a perfect voltage source cannot deliver power, either.

The power is P=V*I where V and I are defined such that V is positive if positive I is entering at the positive terminal. If P is positive then the device is consuming power and if P is negative then the device is producing power.

What actually happens to the power consumed by a voltage source depends on the source. Some will store the power chemically, some will store it mechanically, some will dissipate it as heat, others will do other things. The key thing is that it is removed from the circuit being analyzed.
Thanks
 

Thread Starter

Danuuuuu

Joined Jan 2, 2015
3
By that line of reasoning, a perfect voltage source cannot deliver power, either.

The power is P=V*I where V and I are defined such that V is positive if positive I is entering at the positive terminal. If P is positive then the device is consuming power and if P is negative then the device is producing power.

What actually happens to the power consumed by a voltage source depends on the source. Some will store the power chemically, some will store it mechanically, some will dissipate it as heat, others will do other things. The key thing is that it is removed from the circuit being analyzed.
I am sorry, but this doesn't sit with my understanding. If I convert the circuit above using the resistor terminals as A,B I would get a thevenin's equivalent circuit of a thevenin power source of 10 volts in series with the load resistor. Therefor, the power consumed would only be the resistor's of 50[W] and the power delivered would only be the power of thevenin's voltage source of 50[W]. How does your explanation settles with thevenin's theorem ?
 

crutschow

Joined Mar 14, 2008
26,365
The resistor is only absorbing 1/2 the power delivered by V2.
The other half of the power must be absorbed by V1. There's no other place for it to go.
Thevenin's theorem doesn't concern itself the nature of the power source and whether it's absorbing some of the power other than by resistive means.
 

WBahn

Joined Mar 31, 2012
26,398
I am sorry, but this doesn't sit with my understanding. If I convert the circuit above using the resistor terminals as A,B I would get a thevenin's equivalent circuit of a thevenin power source of 10 volts in series with the load resistor. Therefor, the power consumed would only be the resistor's of 50[W] and the power delivered would only be the power of thevenin's voltage source of 50[W]. How does your explanation settles with thevenin's theorem ?
So?

A Thevenin equivalent circuit is ONLY equivalent in terms of the voltage and current relationship at the terminals. You can't tell anything about what is going on inside the portion of the circuit that you took the equivalent of and you certainly can't tell anything about power except power at the terminals. Just consider a Thevenin circuit (voltage source in series with a resistor) and the equivalent Norton circuit (current source in parallel with that same value of resistance). If there is no load, the Thevenin circuit is dissipating now power but the Norton is dissipating maximum power. If the load is short circuited, then the reverse is true.
 
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