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Can someone answer me without using an analogy or answering me only with maths (j/c * c/s = j/s), I want to know the logic behind that, thanks.

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  • \$\begingroup\$ I'm not convinced this can be answered fully without the math. At some point you have to realize that the energy lost per second is the power. \$\endgroup\$ Oct 20, 2016 at 2:04
  • \$\begingroup\$ Do you understand why mechanical power is force times distance over time? \$\endgroup\$ Oct 20, 2016 at 3:31
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    \$\begingroup\$ Without maths or analogy, there's not much left. \$\endgroup\$
    – Chu
    Oct 20, 2016 at 7:29
  • \$\begingroup\$ You don't want an analogy but I'm unclear whether you want maths or not? \$\endgroup\$
    – Andy aka
    Oct 20, 2016 at 10:35

6 Answers 6

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You know that "watt" actually means energy-flow, right? "Joules per second" is renamed the Watt. This can be a little confusing to beginners, since at the start we should be saying "joules per second" this, and "joules per second" that. Giving lots of examples of energy quantity and energy flow, but without mentioning Watts, so "Joules" and "Joules per second" become solidly known and obvious. We come to see that Joules per second is a rate, it's not like a "stuff." We cannot have a bucket of watts, since a bucket cannot contain a rate. But the joules, they ARE like a stuff. We can have a bucket of energy. Energy isn't created or destroyed; just moved around.

You know that "ampere" actually means charge-flow, right? "Coulombs per second" is renamed the Ampere. It's a little confusing to beginners, since at the start we should be saying "coulombs per second" this, and "coulombs per second" that. Giving lots of examples of charge quantity and charge flow, but without mentioning Amperes, so "Coulombs" and "Coulombs per second" become solidly known and obvious. We'll come to see that coulombs per second is a rate, it's not like a "stuff." We cannot have a bucket of amps, since a bucket cannot contain a rate. But the coulombs, they ARE like a stuff. We can have a bucket of charge. Charge isn't created or destroyed; just moved around.

So, delete this "watts" and "amperes" stuff, and instead deal with the concepts concealed behind them: quantities of energy, and quantities of charge.

Your question then reads: "Why is the amount of energy determined by multiplying the charge times the voltage?"

Simple answer: voltage is a way of measuring e-fields. So, if we perform work by lifting a quantity of charge against an e-field, then the electrical energy stored is equal to the work needed to lift the charge. (The energy was stored in altered fields and moved charges, as with charging a capacitor.) And, when we allow an e-field to perform work on a charge by pulling it along, the energy which appears is equal to the work performed by the moving charge. (The electrical energy came out as charges moved "downhill," as when we discharge a capacitor.)

So, if we double the voltage, we double e-field intensity, so we double the force, which doubles the amount of energy stored when we move a certain amount of charge. Volts times coulombs gives joules. Or, if we double the amount of charge, that must double the force as before, then when we move it through an e-field, it gives double the amount of energy. Double both of them at once, and we'd get four times the force, therefore four times the energy when we move the charge. Obviously it's just a simple product, volts times coulombs. (DOH, gravity-field analogy: expending energy to lift a rock! Higher gravity requires more force so more energy, but higher-mass rock also requires more force so more energy.)

Finally, we can perform the same work over a short time or a long time (different joules per second but total transferred energy not changing.) And, we can transport charge over a short time or a long time (different coulombs per second but total moved charge not changing.) That gets us back to joules/sec equaling coulombs/sec times volts.

Last question would be... why isn't energy the square of the coulombs, or the square of the volts? (It's because doubling the force will double the work performed over distance. Not quadrupling the work.)

Also, note that I'm not going deeply into e-field intensity versus the planes of equipotential spread over a certain distance. Length and e-field and voltage is all in there somewhere, mixed up, but I was keeping length constant, so volts then is proportional to e-field intensity, and we can temporarily see them as the same thing. Peeling it all apart is really a separate question.

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    \$\begingroup\$ Very good! Shows how the underlying concepts lead to the commonly-used shorthand notation. \$\endgroup\$
    – Dave Tweed
    Oct 20, 2016 at 11:55
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Explaining without math is complicated, since you can just go by the definitions of each thing to understand why voltage times current equals power.


1) Power is just how much energy you have per second

$$ \mbox{1 Watt} = \frac{\mbox{1 Joule}}{\mbox{1 second}} $$

2) Voltage is how much energy you have per charge

$$ \mbox{1 Volt} = \frac{\mbox{1 Joule}}{\mbox{1 Coulomb}} $$

3) Current is how much charges you have per second

$$ \mbox{1 Ampere} = \frac{\mbox{1 Coulomb}}{\mbox{1 second}} $$

If you multiply voltage and current, you get:

$$ \mbox{1 Volt} \times \mbox{1 Ampere} = \frac{\mbox{1 Joule}}{\mbox{1 Coulomb}} \times \frac{\mbox{1 Coulomb}}{\mbox{1 second}} = \frac{\mbox{1 Joule}}{\mbox{1 second}} = \mbox{1 Watt} $$

It just happens to be like this because of the way those three things are defined, that's it. There is no magical logic behind it.

Measuring current is like looking at a section of a wire and counting how many charges are going by each second, and measuring power is like measuring current, but instead of just counting charges, you are counting each charge's energy.

You can think of the product of voltage and current as an "indirect" way to measure power, since you have no tools to count energy per second, but you have multimeters which can measure current and voltage.

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Voltage is a measurement of the potential energy in each electron (qV). A single electron requires energy to be forced to a higher voltage. Similarly, an electron which travels from a higher voltage to a lower voltage loses energy (for example as as heat in a resistor). The greater the change in voltage the electron experiences, the more energy which it releases.

Now, current is just the sum total of all the electrons moving. So it is the same example as above for each electron. This is just multiplication per number of electrons per second.

Each electron provides an amount of energy (qV). The current is proportional to the number of electrons per second, so together they form the total energy per second. This is what the scientific community has decided will be called Watts (Joules/second or Volt-Amps; the units are defined in such a way that they are normalized to each other: i.e. V=IR and P=VI and F=ma and F=-kx and Q=CV and B=Li -- almost all basic equations lack any constants because the units are normalized themselves)

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Watts is an instantaneous measure of power that varies from microsecond to microsecond. Watts voltage times current. Current is the number of electrons moving, and volts is the potential energy of each electron (pressure?). Tell me if I'm getting close to helping?

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  • \$\begingroup\$ Not even close. You're just parroting definitions. What is the thinking behind those particular definitions? Why are they useful? \$\endgroup\$
    – Dave Tweed
    Oct 20, 2016 at 11:51
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What's a watt?

It has been theoretically analyzed, tested and proven that the real product of the voltage "potential" times (x) the rate of charge flow per second or current yields this quantity.

It is a Law of nature, formalized by the measurement standards for each quantity and can be converted to any other form of power and adopted internationally.

The Watt is defined by International System of Units (SI), named after the Scottish engineer James Watt (1736–1819).

The unit of 1 Watt is also defined as 1 joule per second and can be used to express the rate of energy conversion or transfer with respect to time.

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Sadly, very basic math is unavoidable, but I'll try to keep it to a minimum.

The voltage \$V_{AB}\$ across two points A and B in a circuit is, by definition, the potential energy change per unit charge that an electrical charge undergoes when it moves from A to B. Therefore the voltage unit can be expressed as: volt = joule / coulomb.

The current intensity \$I\$ is, by definition, the charge that flows across a given section of a conductor in the unit time, so its unit can be expressed as: ampere = coulomb / second.

Therefore, if you have a given charge quantity Q that moves between A and B during a time \$\Delta t\$ you know that its energy changes by \$ E = Q \cdot V_{AB}\$. Since the power (i.e. the "wattage") is, by definition, the energy per unit time, that means that the power exchanged in the process is:

$$ P = \frac {E}{\Delta t} = \frac{V_{AB} \cdot Q} {\Delta t} = V_{AB} \cdot \frac{Q} {\Delta t} = V_{AB} \cdot I $$

That means that the same formula holds for the unit of those quantities:

$$ watt = volt \times ampere $$

To express the same concepts in a more informal way: each time a charge "packet" moves between A and B it loses/gains energy. If you multiply by the number of charges, you get the total energy exchanged in the process. If you divide that by the time in which the process happens, you get the (average) power of the process. You get the same result if you multiply voltage by current because the quantities are the same, but they are "mixed differently" in the formulas (as I showed before).

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