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I am wondering if the energy loss in a voltage drop over an entire circuit is equal to the energy transferred from the circuit (e.g. light/sound/mechanical energy/heat).

If this is true, wouldn't that make it so that the voltage drop in a short circuit (with the power supply being a battery) would be 100% since all of the electrical energy must be converted into something else?

What is voltage drop with explanation of the example given?

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    \$\begingroup\$ The voltage drop in a short circuit is ~100%. A short, by definition, has no potential across it. \$\endgroup\$ – Shamtam Feb 14 '13 at 16:59
  • \$\begingroup\$ If the short has no potential across it, then it drops 0 V, not 100% of the voltage. But in reality there is no perfect short (nor any perfect voltage source)...so if you want to talk about short circuits your model has to include more detail than just perfect voltage sources and perfect wires. \$\endgroup\$ – The Photon Feb 14 '13 at 20:03
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I am wondering if the energy loss in a voltage drop over an entire circuit is equal to the energy transferred from the circuit (e.g. light/sound/mechanical energy/heat).

Yes. Energy is conserved. Any electrical energy that is lost by the circuit is transformed into some other kind of energy.

It is often heat, for example in a resistor or a diode.

It could be something like chemical energy, in a battery charger.

If this is true, wouldn't that make it so that the voltage drop in a short circuit would be 100% since all of the electrical energy must be converted into something else?

Yes, the energy lost from the circuit will be converted to heat. The wire forming the short circuit will heat up (because of its resistance, even though that resistance is too small to matter in more useful circuits), as will the battery (because of its internal resistance).

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  • \$\begingroup\$ Heating the battery is one reason you shouldn't drop a common rectangular 9V battery into a pocket full of loose change or keys. The battery will get hot. You will notice. Other people will notice you jumping around with a hot pocket. If you aren't burned, you will still be mocked.... ;-) \$\endgroup\$ – RBerteig Feb 15 '13 at 0:02
  • \$\begingroup\$ @RBerteig I "learned" this by leaving a pair of D-cells short circuited when I went to lunch. Came back, picked up the battery, and gave myself a nice burn on the hand. \$\endgroup\$ – The Photon Feb 15 '13 at 0:37
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All of this can be understood by two equations:

\$ E = IR \$
Ohm's law

\$ P = IE \$
a combination of Ohm's law and Joule's law

To know how much heat/light/mechanical/chemical/whatever energy is being created as a consequence of the electrical energy being used, you need to know the voltage drop and current. The electrical power, as given by the product of voltage and current, is exactly the rate of energy conversion from electrical to something else.

If you consider what happens if you put an ideal short (zero resistance) across an ideal battery (perfect voltage source), you get a very different universe than the one we have. By Ohm's law, a short can't have a voltage drop:

\$ E = I\cdot0\Omega = 0A\$

But an ideal battery has always exactly the same, non-zero voltage drop. The contradiction means it's impossible. To consider why, think of what happens if you connect a resistor across a battery. As the resistor gets smaller, more current flows, and the power increases. As the resistance approaches \$0\Omega\$, power approaches \$\infty\$. Since we don't have infinite energy available to make infinite power, this can't happen.

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