Recent content by nyxsilverjk

I went back and did the entire thing and kept my placing past the thousandths. I also did convert my k ohms and m amps. Then two other students and I compared answers and discussed them, we all came to the same consensus. I'm going to my teacher tomorrow during study hall to confirm some minor...

It's ok if you can't tell me. But here's my answers going in order from left to right the last one is the bottom left resistor. 19.28 W , 8.64 W , 1.59 W , 1.08 W , 41.18 W

Ok, I think I had an awesome Ah-hah! moment and I think I can finish the problem. When I do can I give my answers and then you can give me a yes or no to if they're right?

So Rp = 1.15Ω? Then that number is add to 2.2 and 4.7 to get R[SIZE="1"]total which is equal to 8.1? then from there you can find the current from I = V/R? which I got to equal 2.96A

Ok so I attached a copy of the take home test I've been working on. We are allowed to consult outside sources and work with any other student but for clarity I need you to see the picture of the problem I'm working on it's number 6. To find the quivalent resistance of the parallel resistors it...

So let's say the voltage of the battery is 24 V and the resistance at R1 = 2.2 k ohms. The equation would be P of R1 = (24)^2/2200 ohms? Then when you get to the parallel circuit and go to calculate R2 which equals 1.5 k ohms it would be P of R2 = (24)^2/1500 ohms?

So how do you figure out how much current goes down either parallel path?

Thank you so much for helping me understand this

So I find the voltage drop at r2. I then use the voltage I got at r2 and use it in the equation P = V^2/ R where the R is equal to the resistance for r2? This then finds the overall power for r2?

That's pretty much the problem we're given in that picture you just posted. So the best way to go about this is to find the voltage drop after each resistor and then use that drop to find the next rating?

But don't you need the charge of the plate as well?

I did and I got 49 V, but what is the way you're talking about? Is that where F = K ((q of a)(q of b)/ r2)?

Sorry for the bad wording. I guess I was really just asking if I had to calculate the voltage drop after each resistor in order to find the following resistors watt rating? and would it help to calculate the current? If so do I have to take into account all the resistors or just the first one...

Homework Statement When calculating wattage rating, if you have resistors in both a series and in parallel circuits can you still use P = V2/R to solve for the wattage rating over each resistor if you're given the voltage of the battery or do you have to calculate the voltage drop after each...

Which formula?