Question

How much power is produced if a voltage of `4 V` is applied to a circuit with a resistance of `2 Omega`?

Answer 1

`8 J/s or 8 W`

Explanation:

`P = IV = I^2 R=V^2/R`

What are given:
`V=4 V`
`R = 2\Omega`

Therefore,
`P = V^2/R =(4 (kg\cdotm^2\cdot s^-3\cdotA^-1))^2/(2(kg\cdot m^2 \cdot s^-3 \cdot A^-2))=8 (kg\cdot m^2 \cdot s^-3 )=8 J/s`

For the determination of units,
`V = Ed= (F/q_0)(d) = (mad)/(q_0) => (kg *m/s^2 \cdot m)/C => (kg *m/s^2 \cdot m)/(As)`

And

`\Omega=V/A = (kg *m/s^2 \cdot m)/(A^2s)`

*Take note that the equation `P = I^2 R=V^2/R` can be played using `P=IV` using the different fundamental equations of electricity.

`V = Ed =(F/q_0)(d)`
`\Omega=V/A`