By that line of reasoning, a perfect voltage source cannot deliver power, either.If E1 is a perfect voltage source it will have zero impedance and, therefore, cannot consume power.
What is the impedance of a perfect voltage source?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.
ThanksBy 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.
Zero -- you said so yourself.What is the impedance of a perfect voltage source?
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 ?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.
So?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 ?
by Luke James
by Jake Hertz
by Jake Hertz
by Jake Hertz