A charge of 54 C passes through a circuit every 6 s. If the circuit can generate 162 W of power, what is the circuit's resistance?

2 Answers
Jun 3, 2016

$R = 2 \text{ } \Omega$

Explanation:

$I = \frac{\Delta Q}{\Delta t} \text{ Current of circuit}$

$\Delta Q = 54 C \text{ ; } \Delta t = 6 s$

$I = \frac{54}{6} = 9 \text{ } A$

$P : \text{Power}$

$R : \text{Resistance}$

$P = {I}^{2} \cdot R$

$162 = {9}^{2} \cdot R$

$R = \frac{162}{81}$

$R = 2 \text{ } \Omega$

Jun 3, 2016

I found $2 \Omega$

Explanation:

You know that current $I$ is the charge passing a section of conductor each second.
In your case you have:
$I = \frac{Q}{t} = \frac{54}{6} = 9 A$
Power is the rate of transformation (work) energy with time:
$P = \frac{E}{t} = V \cdot I$
Using Ohm's Law we get:
$P = {I}^{2} R$
rearranging:
$R = \frac{P}{I} ^ 2 = \frac{162}{{9}^{2}} = 2 \Omega$