# How Joule/s and V*A both equal to same 1 Watt?

In physics, Watt equals to Energy per second [J/s]. But also, Volt times Amps also equals to Watt. How can one tell they both are the same Watt? How can one measure it?

For example, in motors. The efficiency calculation is based on ratio of electrical power / mechanical power.

How can we be sure that 1 W for 1 second in electrical power can do (if it is 100℅ efficiently used) J of work? What is the reference?

• That's from the definition of the volt. The volt is the electric potential difference that yields one joule of energy difference per Coulomb of charge. So multiplying by an Ampere (one Coulomb per second) gives one joule per second. – knzhou Aug 14 '16 at 8:59
• So, that's how you can trust that a battery that says '2 Volts' can deliver up to $IV$ in power. The other question is how we know that battery is 2 volts in the first place, i.e., how do we define the unit of the volt to begin with? To do this, put your batter in a tank of water, short the ends together, and measure the total current and how much the water heats up. This gives you a conversation between electrical and mechanical energy. – knzhou Aug 14 '16 at 9:01
• @knzhou: That's hardly a practical way to measure the voltage on a battery and, worst of all, it will give you the wrong number to boot. A real voltmeter is, indeed, measuring by how much the energy of a charge changes. – CuriousOne Aug 14 '16 at 9:05
• @CuriousOne I didn't mean to give a realistic setup, I just wanted to make a point that somehow or other, one has to make the link between 'electrical energy' and 'mechanical energy'. Back in the 19th century, one couldn't just measure the energy change of a single charge. – knzhou Aug 14 '16 at 9:08
• @knzhou: The definition of [Volt] does not depend on microscopic charges and in practice very few instruments use single electrons to measure voltages. The classic 19th century "electrometer" does convert a potential into a change of potential energy of the charge that it deposits on the electrometer, i.e. it is a true mechanical potential measurement. The great confusion about the function of voltmeters doesn't happen until a high school teacher sells a galvanometer with a series resistor as a voltmeter to students... which is akin to defining a force with a spring force gauge... ugh! – CuriousOne Aug 14 '16 at 9:12

One ampere is an elementary unit of current $I$ in the SI units. One coulomb is defined as one ampere second – the charge that the current one ampere transfers in one second $Q=It$.
One joule is defined as twice the kinetic energy that one kilogram gets after being accelerated to one meter per second, $2E=mv^2$, so one joule is really defined as ${\rm kg}\cdot{\rm m}^2/{\rm s}^2$. The factor of $2$ or $1/2$ comes from integration.
One watt is defined as one joule per second, the power needed to produce one joule of energy as quickly as in one second. So one watt is defined as ${\rm J}/{\rm s}$.
One volt is defined as the voltage at which one ampere current has the power of one watt. Equivalently (because of the multiplication by $t$ explained in the first paragraph), one volt is defined as the voltage at which one coulomb of charge gains the energy of one joule. So $$1\,{\rm V} = \frac{1\,{\rm J}}{1\,{\rm C}} =\frac{1\,{\rm J}/{\rm s}}{1 \,{\rm C}/{\rm s}}= \frac{1\,{\rm W}}{1 \,{\rm A}}$$ Here, the seconds cancelled in the numerator and the denominator.
If there are $K$ names of units, one obviously needs $K$ definitions. Those are basically equivalent to the statements you described as "equivalent" except that you reverted the other one. A more conventional way to talk is to say that "one watt is one joule per second" and "one volt is one watt per ampere". These are just two definitions. But these two statements are equivalent to the two statements you made.