a lightbulb has a resistance of 195 ω and carries a current of 0.62 a. the power rating of the lightbulb, to…

Explanation:

Step1: Recall power - resistance - current formula

The formula for power $P$ in terms of resistance $R$ and current $I$ is $P = I^{2}R$.

Step2: Substitute given values

Given $R = 195\ \Omega$ and $I=0.62\ A$. Substitute into the formula: $P=(0.62)^{2}\times195$. First, calculate $(0.62)^{2}=0.62\times0.62 = 0.3844$. Then, $P = 0.3844\times195=74.958$.

Step3: Round to nearest whole number

Rounding $74.958$ to the nearest whole number gives $75$.

Answer:

$75$