power

[lilkrazyrae]

Given two light bulbs with the following ratings: 75.0W 120VAC and 100W 120VAC. They are connected in series across a potential difference of 120. VAC. Assuming that the operating resistances of the bulbs are not affectd by the series connection, calculate (a) the power produced in each bulb, and (b) the total power produced in the series circuit.

Using I=P/v gives (a) 75W & 100W and (b) 175W

is this correct?

[turbo]

We can't do your homework for you. Have you reviewed your course-work and textbooks, and do you have any questions related to your understanding of them? That would be the place to start.

[lilkrazyrae]

Actually I was reviewing for a test and working the odd problems. I was checking to see if I did them right. You guys always seem to assume the worst!!!!

[krab]

Your answer (a) is wrong because those powers only apply to case when there is 120V across each light bulb, like if they were in parallel. That's not the case here.

[lilkrazyrae]

Even though it says to ignore the series connection?

[turbo]

Even though it says to ignore the series connection?You are can ignore the changes in the operating resistances (printed on the bulbs) imposed by the circuit values caused by the series connection, but that does not leave you free to ignore the effects of the series connection on the calculated voltages and powers. They are simplifying the question by allowing you to assume that the resistive values of each bulb is constant regardless of applied voltage. Does that help?

[chroot]

If a bulb uses 75 W of power when connected across 120 VAC, you can directly calculate its resistance


\begin{gathered}
P = VI = V\left( {\frac{V}
{R}} \right) = \frac{{V^2 }}
{R} \hfill \\
R = \frac{{V^2 }}
{P} \hfill \\
R = \frac{{\left( {120\,{\text{V}}} \right)^2 }}
{{75\,{\text{W}}}} = x\,{\text{ohms}} \hfill \\
\end{gathered}


Use the same process to find the resistance of the 100 W bulb.

Finally, find the current through a resistance equal to the sum of the bulb's resistances. Then use P=VI to find the power dissipated.

As a check, consider that resistances in series add together to form a larger resistance, so the power dissipated by the two bulbs in series cannot possibly be larger than the power dissipated by either alone.

- Warren