A charge of $35 C$ $35 C$ passes through a circuit every $5 s$ $5 s$ . If the circuit can generate $21 W$ $21 W$ of power, what is the circuit's resistance?

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Answer 1 · 2018-05-03T12:53:28

$\frac{3}{7} \approx 0.43$ $3/7~~0.43$ ohms

We first find the current produced in five seconds. Current is given by the equation,

$I = \frac{Q}{t}$ $I=Q/t$

$Q$ $Q$ is the charge in coulombs

$t$ $t$ is the time in seconds

So, we get:

$I=(35 \ "C")/(5 \ "s")$

$=7 \ "A"$

Now, power is related by the equation,

$P=I^2R$ , since $P=IV$ and $V=IR$ (Ohm's law)

$I$ is the current in amperes

$R$ is the resistance in ohms

$P$ is the power in watts

Rearranging for resistance, we get:

$R=P/I^2$

Plugging in our given values, we get:

$R=(21 \ "W")/((7 \ "A")^2)$

$=(21 \ "W")/(49 \ "A"^2)$

$=3/7 \ Omega$

A charge of 35 C35C35 C passes through a circuit every 5 s5s5 s. If the circuit can generate 21 W21W21 W of power, what is the circuit's resistance?