# How much power is produced if a voltage of 9 V is applied to a circuit with a resistance of 3 Omega?

Mar 11, 2016

I found: $27 \text{Watts}$

#### Explanation:

Consider that your voltage is telling you the quantity of energy, in joules, carried by each Couomb of charge; in our case every coulomb of charge is carring $9$ Joules of energy. But how many coulomb are passing by to give you the energy in, say, $1$ second?
Ohm tells us that in our wire of resistence $3 \Omega$ passes a current given as:
$i = \frac{V}{R} = \frac{9}{3} = 3 A$ but this current tells us that we have $3$ Coulombs each second passing by!!!

So we can join volts and amperes together to say that the power $P$ (equal to energy/time) will be:

$P = V i = \text{joule"/"Coulomb"*"Coulomb"/"time"="Joule"/"time"="Watts}$
YES!!!

so:

Power$= 3 \cdot 9 = 27 \text{Watts}$

Mar 11, 2016

$= 27 W$

#### Explanation:

Power P=${V}^{2} / R$ , where V = voltage and R = resistance
$\therefore P = {V}^{2} / R = {9}^{2} / 3 = 27 W$