Question

question 19 of 25
the power in a light - bulb is given by the equation p = i²r, where i is the current flowing through the light - bulb and r is the resistance of the light - bulb. what is the current in a circuit that has a resistance of 30.0 ω and a power of 2.00 w?
a. 0.067 a
b. 0.258 a
c. 15.0 a
d. 3.87 a

Explanation:

Step1: Rearrange the power formula

Given $P = I^{2}R$, we can solve for $I$. First, divide both sides of the equation by $R$: $I^{2}=\frac{P}{R}$. Then take the square - root of both sides: $I=\sqrt{\frac{P}{R}}$.

Step2: Substitute the given values

We are given that $P = 2.00\ W$ and $R=30.0\ \Omega$. Substitute these values into the formula $I=\sqrt{\frac{P}{R}}=\sqrt{\frac{2.00}{30.0}}$.
Calculate $\frac{2.00}{30.0}=\frac{1}{15}\approx0.0667$. Then $I = \sqrt{0.0667}\approx0.258\ A$.

Answer:

B. 0.258 A