A charge of $40 C$ $40 C$ passes through a circuit every $8 s$ $8 s$ . If the circuit can generate $42 W$ $42 W$ of power, what is the circuit's resistance?

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Answer 1 · 2016-02-10T11:05:06

$R = 1.68 Omega$

The current, $I$ , is the charge passing through per unit time ( $Q/C$ ).

So,

$I = frac{40"C"}{8"s"} = 5"A"$

The power arises from the fact that charges are moving over a potential difference.

$P=IV$

From Ohm's Law, we know that $V=IR$ for an ideal resistance.

Therefore,

$P = I^2R$

$42"W" = (5"A")^2R$

$R = 1.68 Omega$

Answer 2 · 2016-02-10T11:08:01

$1.68 Omega$

We'll need 2 formulae to answer this. namely:

$Q=It$ and $P = I^2R$

Where Q is charge (40C), t is time (8s), P is power (42W), I is current and R is resistance.

Begin by using the first formula to obtain the current:

$I = Q/t = 40/8 = 5A$

Now use the second formula to get the resistance:

$R = P/I^2=42/5^2=42/25=1.68 Omega$

A charge of 40 C40C40 C passes through a circuit every 8 s8s8 s. If the circuit can generate 42 W42W42 W of power, what is the circuit's resistance?