Question

Show that the units \(1\;{V^2}/\Omega = 1\;W\) , as implied by the equation \(P = {V^2}/R\).

Short answer

The units \({V^2}/\Omega \) and \(W\) are equivalent.

Step-by-step solution

Definition of Power

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\((1W = 1J/s)\) .

Definition of Potential difference

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 \((V)\).

Ohm's law

Current through the electric circuit is determined by,

\(I = \frac{V}{R}......(3)\)

Where \(V\) is the potential difference, and \(R\)is the resistance.

Definition of Electric current

The magnitude of the electric current passes through the wire

\(I = \frac{{|q|}}{{\Delta t}}......(4)\)

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

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

Verifying the unit

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

\((W) = \left( {\frac{J}{s}} \right)......(5)\)

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

\(\begin{aligned}{}\left( {\frac{{{V^2}}}{\Omega }} \right) = \left( {\frac{{\frac{{{J^2}}}{{{C^2}}}}}{{\frac{V}{A}}}} \right)\\\left( {\frac{{{V^2}}}{\Omega }} \right) = \left( {\frac{{{J^2}.A}}{{{C^2}.V}}} \right)\end{aligned}\)

Final result

Again, the unit of volt is equivalent to joule/coulomb, and according to Equation\((4)\), the unit of the ampere is equivalent to coulomb/second:

\(\begin{aligned}{}\left( {\frac{{{V^2}}}{\Omega }} \right) = \left( {\frac{{{J^2}.A}}{{{C^2}.V}}} \right)\\ = \left( {\frac{{{J^2}.\frac{C}{s}}}{{{C^2}.\frac{J}{C}}}} \right)\\\left( {\frac{{{V^2}}}{\Omega }} \right) = \left( {\frac{J}{s}} \right)......(6)\end{aligned}\)

From eq.\((5)\)and\((6)\), itcan be concluded that the units\({V^2}/\Omega \)and\(W\)are equivalent.

Therefore, the units \({V^2}/\Omega \) and \(W\) are equivalent.