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In the following circuit, one could use a simple KVL to calculate that the voltage across the resistor equals 10V, and therefore the current flowing in the circuit is 1A, from the right side to the left side of the resistor.

This then shows that the power delivered by the right voltage source (using the passive sign convention) is P = -VI = -(20)(1) = -20W = 20W delivered.

However, when moving to the left voltage source, and still applying the passive sign convention, since the current enters the positive terminal, the power of the source is P = VI = (10)(1) = 10W = 10W absorbed.

We have faced this situation often in my current circuit analysis course. My question is - how is it possible that a voltage source has a net absorption of power? This just doesn't make any logical sense to me. If its a source, why is it absorbing power?

schematic

simulate this circuit – Schematic created using CircuitLab

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    \$\begingroup\$ You have some good answers already. I just want to add that the behavior of an ideal voltage source is only to constrain the voltage between two points to a certain value. Current can flow either way through it. It may provide power or absorb power. Be careful not to confuse an ideal voltage source with a real voltage source. \$\endgroup\$ Aug 24 '20 at 21:08
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Imagine V1 is a rechargeable battery. If you put a resistor across it, it delivers power. If you connect it through a resistor to a higher voltage, you will charge the battery.

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  • \$\begingroup\$ Does this apply to non-rechargeable batteries i.e. what will happen if V1 physically can't absorb any power? Also, do rechargeable batteries recharge at the same voltage they supply? \$\endgroup\$
    – Gary Allen
    Aug 24 '20 at 12:04
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    \$\begingroup\$ If the battery can’t absorb the energy it has to go somewhere, typically it shows up as heat. As an NiMH battery approaches full charge, the cell temperature rises and if charging continues excess hydrogen gas may be vented, damaging the cell. You need at least a bit more voltage than the battery voltage to get energy to flow into the battery. Lead acid batteries produce about 12V but charge with about 13.8V. \$\endgroup\$ Aug 24 '20 at 12:26
  • \$\begingroup\$ This begs the question @SpehroPefhany , what defines the ESR and thus current flow as a function of cell voltage vs charger or load \$\endgroup\$ Aug 24 '20 at 21:51
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Ideal voltage sources are bidirectional.

Even non-rechargeable batteries can be charged at low currents , just not efficiently.

Generators being charged with force turn into motors.

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It's a hypothetical perfect voltage source of course.

But take the real, slightly imperfect lead acid battery as an example to prove that voltage sources can, in principle, absorb power.

Some designs of real voltage source can't, of course, for reasons like diodes getting in the way.

But others - not just rechargeable batteries, but also generators can absorb power too. Of course, when a generator absorbs power, we call it a motor...

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Power supplies can have one, two, or four quadrants.

A single quadrant power supply has a pass element from high voltage to the output. It will regulate the pass element to give the desired output. This is a typical linear DC supply. This supply can not sink current.

A dual quadrant supply has all the single quadrant stuff but also has a second pass element. This pass element connects the output to common. It has a large heat sink on it. When the voltage is greater than the regulated set point it will burn off the excess with this second element. This type of power supply is typically used with battery testing because it can both charge and discharge batteries.

A four quadrant supply has all the dual quadrant supply stuff but also a mirror image to a negative voltage. This type of supply can be used with AC. It can sink or source both positive and negative voltage and current.

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If we want to attempt understanding this at the atomic-electron level, imagine this illustration. You have two sources; one which wants to output 10 electrons / second and another which outputs 20 electrons / second. You connect these with a 10Ω resistor. Because there is 10 electron/sec difference in "voltage", coupled by a 10Ω resistor, the current that flows is 1A. But this current flows from the bigger supply, through the resistor, into the smaller supply - even though it is a "supply."

So yes, energy is being forced into the 10V supply from the 20V supply, even though they both are intended to be outputs. Such effect might be called "back-feeding" and generally is not a good idea.

What happens to that energy, where does it go? Well, it almost certainly is dissipated in the 10V power supply as heat.

There are many different topologies of power supply, and each has various strengths and weaknesses. In general through, inside both supplies, components regulate the voltage and current output to safe limits. Depending on how robustly the supply was engineered, this safety margin may be small ("cheap" supplies), or larger ("expensive" supplies.)

  • A supply you find on eBay, Amazon, Aliexpress for $10 is a "cheap" supply. These are typically designed on the low-side of the robustness spectrum. They can work fine for their intended purpose, but are not very tolerant of abuse such as back-feeding. This type of supply has every conceivable cost-cutting measure taken, such as reducing or eliminating heat-sinks. Without a suitable heat-sink to dissipate extra power introduced by back-feeding, such a supply could fail quickly.
  • An "expensive" supply is generally much more costly. These are more expensive because a) the additional engineering work done to make them more robust, and b) the addition of extra features/components such as larger heat-sinks, physically bigger components (to allow for better heat-sinking), thermal monitoring, etc.

In short, a more expensive supply is generally able to handle back-feeding better than a cheap supply. This usually isn't much of a problem, because system designers understand that back-feeding a supply should be avoided.

Incidentally, this is one reason why bench power supplies for electronics engineers are expensive - they are designed to be robust against several types of transient and continuous events, including accidental back-feeding.

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    \$\begingroup\$ Good answer, however, electrons per second is a description of current rather than voltage. This makes the first paragraph somewhat nonsensical, as it makes it sound like you are treating the supplies as current sources. \$\endgroup\$ Aug 24 '20 at 14:35
  • \$\begingroup\$ Thanks, feel free to edit it or write a new answer. \$\endgroup\$
    – rdtsc
    Aug 24 '20 at 15:45

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