A 50-watt lamp, a 20-watt, a 75 watt lamp, and a 100-watt lamp are connected in parallel to a 110-volt circuit. Calculate a) the current through each lamp; b) the resistance of each lamp; c) the total resistance of the circuit.

That's a neat problem. I like it.

Do you have any idea how to solve it?

Do you know the formulae you need to know? For example, for the first part, you need P = IV

When you know this, you should use the formula from Ohm's Law, and apply your knowledge of how the total resistance in a circuit is affected when the components are in parallel.

This is what I have so far..
50 W/110V=.45A
20W/110W=.18A
75W/110V=.68A
100W/110V=.91A

For the second part
110V/.45A=244 Ohms
110V/.18A=611 Ohms
110V/.68A=162 ohms
110V/.91A=121 ohms

Now for the third part I know I would set it up like this
1/244 +

A slight detail I'd like to point out, is that when you use the rounded off numbers, you are losing the accuracy of your results. If you have a storage facility in your calculator, I advise that you use it.

For the resistances, for example, if you take all the numbers, you get:

1. 110/0.4545... = 242 ohms
2. 110/0.1818... = 605 ohms
3. 110/0.6818... = 161 ohms
4. 110/0.9090... = 121 ohms

Only the third one needed rounding off, but all the rest were exact resistances.

Then, for finding the total resistance, I get:

\frac{1}{242} + \frac{1}{605} + \frac{1}{161} + \frac{1}{121} = 0.0202

Then, R_T = \frac{1}{0.0202...} = 49 \Omega

to two sig fig.

The 0 is significant. Get in the habit of using it. So, 68 becomes 0.68. In certain classes, namely physics, I got points knocked off from that many of times.

We also had to do all calculations paying attention to significant figures and error bars.

See, the problem wasn't that hard.

And in case you wonder why I chose to use 2 significant figures for the last part, it's because the values you got are all to 2 significant figures. You just cannot get more accurate answers than the values you used at first, right? ;)

But isn't the final part asking to find total resistance of a parallel circuit? So I thought I had to use the other equation

Yes, you have to use:

\frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2} + ...

And that is what I used. Oh, and avoid using the 'reply' button. There is limited space there, and I have to see the thread to know that you replied to me. This is a feature that should be removed... The admins of the site are still working on it.