Question

25.35 • light bulbs. the power rating of a light bulb (such as a 100 w bulb) is the power it dissipates when connected across a 120 v potential difference. what is the resistance of (a) a 100 w bulb and (b) a 60 w bulb? (c) how much current does each bulb draw in normal use?

Explanation:

Step1: Recall power - resistance - voltage 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 find $R=\frac{V^{2}}{P}$. The voltage $V = 120$ V.

Step2: Calculate resistance of 100 W bulb

For a 100 W bulb, substituting $P = 100$ W and $V=120$ V into $R=\frac{V^{2}}{P}$, we get $R_1=\frac{120^{2}}{100}=\frac{14400}{100}=144\ \Omega$.

Step3: Calculate resistance of 60 W bulb

For a 60 W bulb, substituting $P = 60$ W and $V = 120$ V into $R=\frac{V^{2}}{P}$, we get $R_2=\frac{120^{2}}{60}=\frac{14400}{60}=240\ \Omega$.

Step4: Recall power - current - voltage formula

The power formula is also $P = VI$, where $I$ is current. We can re - arrange it to find $I=\frac{P}{V}$.

Step5: Calculate current of 100 W bulb

For a 100 W bulb, substituting $P = 100$ W and $V = 120$ V into $I=\frac{P}{V}$, we get $I_1=\frac{100}{120}=\frac{5}{6}\approx0.83$ A.

Step6: Calculate current of 60 W bulb

For a 60 W bulb, substituting $P = 60$ W and $V = 120$ V into $I=\frac{P}{V}$, we get $I_2=\frac{60}{120}=0.5$ A.

Answer:

(a) $144\ \Omega$
(b) $240\ \Omega$
(c) For 100 W bulb: $\frac{5}{6}\text{ A}\approx0.83$ A; For 60 W bulb: $0.5$ A