Question

A charge of `9 C` is passing through points A and B on a circuit. If the charge's electric potential changes from `29 J` to `5 J`, what is the voltage between points A and B?

Answer 1

THe voltage is `=2.67V`

Explanation:

The charge is `Q=It`

We apply Ohm's Law

`U=RI`

The power is

`P=UI`

And the energy is

`E=UIt`

`E=UQ`

and

`DeltaE=QDeltaU`

The voltage is

`DeltaU=(DeltaE)/Q=(29-5)/9=24/9=2.67V`