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?

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Answer 1 · 2016-03-08T02:56:22

$2.5Omega$

The current is

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

The power is given by

$P = I^2 R_"eff"$ ,

where $R_"eff"$ is the effective resistance of the circuit. So plugging the numbers in,

$40"W" = (4 "A")^2 R_"eff"$

$R_"eff" = 2.5 Omega$

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

In 1 second, the amount of energy generated is

$1"s" xx 40"W" = 40 "J"$

The amount of charge passing through the resistor is

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

Since $40"J"$ is needed to pass $4"C"$ 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.

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

A charge of 36 C36 C passes through a circuit every 9 s9 s. If the circuit can generate 40 W40 W of power, what is the circuit's resistance?