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

$8 \frac{J}{s} \mathmr{and} 8 W$

Explanation:

$P = I V = {I}^{2} R = {V}^{2} / R$

What are given:
$V = 4 V$
$R = 2 \setminus \Omega$

Therefore,
$P = {V}^{2} / R = {\left(4 \left(k g \setminus \cdot {m}^{2} \setminus \cdot {s}^{-} 3 \setminus \cdot {A}^{-} 1\right)\right)}^{2} / \left(2 \left(k g \setminus \cdot {m}^{2} \setminus \cdot {s}^{-} 3 \setminus \cdot {A}^{-} 2\right)\right) = 8 \left(k g \setminus \cdot {m}^{2} \setminus \cdot {s}^{-} 3\right) = 8 \frac{J}{s}$

For the determination of units,
$V = E d = \left(\frac{F}{q} _ 0\right) \left(d\right) = \frac{m a d}{{q}_{0}} \implies \frac{k g \cdot \frac{m}{s} ^ 2 \setminus \cdot m}{C} \implies \frac{k g \cdot \frac{m}{s} ^ 2 \setminus \cdot m}{A s}$

And

$\setminus \Omega = \frac{V}{A} = \frac{k g \cdot \frac{m}{s} ^ 2 \setminus \cdot m}{{A}^{2} s}$

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

$V = E d = \left(\frac{F}{q} _ 0\right) \left(d\right)$
$\setminus \Omega = \frac{V}{A}$