Question

determine the resistance of a 75 - w lightbulb connected to its proper source voltage of 110 v. express your answer to two significant figures and include the appropriate units. r = value units submit request answer part b

Explanation:

Step1: Recall power - voltage - resistance formula

The power formula is $P=\frac{V^{2}}{R}$, where $P$ is power, $V$ is voltage and $R$ is resistance. We can re - arrange it to solve for $R$, getting $R = \frac{V^{2}}{P}$.

Step2: Substitute given values

Given $P = 75\ W$ and $V=110\ V$. Substitute these values into the formula: $R=\frac{(110)^{2}}{75}=\frac{12100}{75}\ \Omega$.

Step3: Calculate and round

$R=\frac{12100}{75}\approx161.33\ \Omega$. Rounding to two significant figures, $R = 160\ \Omega$.

Answer:

$R = 160\ \Omega$