Verify that the units of a volt-ampere are watts, as implied by the equation \(P = IV\).

The power of a process is the amount of some type of energy converted into a different type divided by the time interval \(\Delta t\) in which the process occurred:

\(P = \frac{{\Delta E}}{{\Delta t}}......(1)\)

The SI unit of power is the watt\((W)\).

\(1\) Watt is \(1\) joule/second.

The potential difference \(\Delta V\) between the two points is

\(\Delta V = \frac{{\Delta U}}{q}......(2)\)

Here\(q\)is the change and\(\Delta U\)is the potential energy.

The unit of electric potential is joule/coulomb \((J/C)\) and is called the volt \((V)\).

The magnitude of the electric current passes through the wire

\(I = \frac{{|q|}}{{\Delta t}}......(3)\)

Where\(q\)is the charge, and\(\Delta t\)is the time interval.

The unit of current is the ampere \(A\), equivalent to \((C/s)\).

According to Equation\((1)\), the unit of watt is equivalent to joule/second. So:

\((W) = \left( {\frac{J}{{s}}} \right)\)

According to Equation\((2)\), the unit of volt is equivalent to joule/coulomb, and according to Equation\((3)\), the unit of the ampere is equivalent to coulomb/second. So:

\(\begin{aligned}{}(V.A) &= \left( {\frac{J}{C}\cdot\frac{C}{s}} \right)\\ &= \left( {\frac{J}{s}} \right)\end{aligned}\)

Therefore, the units of a volt-ampere are equal to watts.