Question

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?

Answer 1

`2.5Omega`

Explanation:

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`