# A charge of 36 C passes through a circuit every 9 s. If the circuit can generate 40 W of power, what is the circuit's resistance?

Mar 8, 2016

$2.5 \Omega$

#### Explanation:

The current is

I = Q/t = frac{36"C"}{9"s"} = 4 "A"

The power is given by

$P = {I}^{2} {R}_{\text{eff}}$,

where ${R}_{\text{eff}}$ is the effective resistance of the circuit. So plugging the numbers in,

$40 \text{W" = (4 "A")^2 R_"eff}$

${R}_{\text{eff}} = 2.5 \Omega$

But where did that formula come from? Read on to find out.

In 1 second, the amount of energy generated is

$1 \text{s" xx 40"W" = 40 "J}$

The amount of charge passing through the resistor is

frac{36"C"}{9"s"}xx1"s" = 4"C"

Since $40 \text{J}$ is needed to pass $4 \text{C}$ of charge across a potential, we can find the potential across the resistor using

frac{40"J"}{4"C"} = 10"V"

And by Ohm's law, we can find the resistance by dividing the potential difference across the resistor by the current passing through it.

R_"eff" = frac{10"V"}{4"A"} = 2.5 Omega