Recent content by William Bush

Big thanks goes to "learningphysics"! I was able to solve this problem after another two hours of work. Way to long for me to list the steps but basically I had to use Kirchhoff's junction and loop rules to develope equations that allowed me to solve for current from batteries 1 & 2. Once I...

Thank-you for helping me out...it's past midnight my time and my brain isn't working very well anymore. You have given me a direction to pursue in the morning so hopefully this will make more sense then. If I sleep with my book under my pillow, do you think I will wake up with the solution?!

Also, why is direction important?...I thought that current is constant everywhere in a closed system?

How and when will the 12 ohm resistor come into play?

I have not studied the node voltage method yet. However, I am looking at my textbook and I see that there is a rule that states that the voltage into the node is equal to the voltage coming out of the node. Is this what you are talking about?

I'm not quite sure how to go about finding the voltage at the node joining the 3 resistors. If I use ohms law: I= V/R The current to the node supplied by battery 1 is equal to 10V/40 ohms which gives ".25 amps". The current to the node supplied by battery 2 is equal to 10V/10 ohms...

1. Three resistors and two 10.0 volt batteries are arranged as shown in the circuit diagram. Which of the following entries in the table is correct? 2. P= I^2 X R, P= V^2/R, P= I X V 3. I've been working on this problem for 45 minutes and can't find a way forward. My...

I would have never been able to come up with the derivations that you listed! I didn't know that work is equal to potential energy -0; or that Fd = U. F = qE is in my textbook so I was aware of that one. I'm worried because I don't understand how you came up with those derivations.

What does "U" stand for in your equations?

I spoke to soon...I didn't come out correctly my way! Hope you can bear with my so that I can see if I follow your last post

Okay, I got it!...my mistake was that I was trying to use E=kq/r^2. When I plugged the numbers into E=F/q it came out correct. Big thanks to rootX for the help!

I'm sorry,...but I'm not following. Isn't the work simply equal to 50 V x .1 meters?

Thanks for the quick response! 1. Work = (F)(d); but how does this help? Should I consider the consider the 50 V between points P and Q as the force? 2. When I do so, I come up with: E = (8.99 x 10^9)(50 x 10^-6)/0.1^2 When I plug that into my calculator I get 44950000 or 4.495 x...

Homework Statement P and Q are points within a uniform electric field that are separated by a distance of 0.1 meters as shown. The potential difference between P and Q is 50 V. Determine the magnitude fo this electric field. Homework Equations E=F/q E=(k)(q)/r^2 The Attempt...

Doc, Thanks for the help...I finally got it! Hope you are around in the future because I'm sure I'll be posting again pretty soon. Physics is interesting but it doesn't come easy for me.