Question

determine the resistance of a 250 - w lightbulb connected to its proper source voltage of 110 v. express your answer to two significant figures and include the appropriate units. part d determine the current through a 250 - w lightbulb connected to its proper source voltage of 110 v. express your answer to two significant figures and include the appropriate units.

Explanation:

Step1: Recall the power formula \(P=\frac{V^{2}}{R}\)

We know that power \(P = 250\space W\) and voltage \(V=110\space V\). Rearranging the formula \(P=\frac{V^{2}}{R}\) for \(R\), we get \(R=\frac{V^{2}}{P}\).
Substitute \(V = 110\space V\) and \(P=250\space W\) into the formula: \(R=\frac{110^{2}}{250}=\frac{12100}{250}=48.4\space\Omega\approx48\space\Omega\) (to two significant figures)

Step2: Recall the power formula \(P = VI\)

We know that power \(P = 250\space W\) and voltage \(V = 110\space V\). Rearranging the formula \(P=VI\) for \(I\), we get \(I=\frac{P}{V}\)
Substitute \(P = 250\space W\) and \(V=110\space V\) into the formula: \(I=\frac{250}{110}\approx2.27\space A\approx2.3\space A\) (to two significant figures)

Answer:

For resistance: \(R = 48\space\Omega\)
For current: \(I=2.3\space A\)