Recent content by Elysium

No I haven't. it's in my previous post. Am I complicating things?

That means: \int \vec{E} \cdot d\vec{S} = \frac{Q_{enclosed}}{\varepsilon_\varnothing} \int E \cdot dS cos(0) = \frac{Q_{enclosed}}{\varepsilon_\varnothing} E \cdot 4 \pi R^2 = \frac{Q_{enclosed}}{\varepsilon_\varnothing} Charge is intact, so... \int E = \frac{1}{4\pi\varepsilon_\varnothing}...

Hi, I'm currently doing a practice problem and I need help in solving it. Ok first the problem: a)Ok, the potential energy of the electron is: U = -eV_R To find V of R we need to refer to electric field: V_f - V_i = -\int_{i}^{f} \vec{E} \cdot d\vec{R} V(R) - V(\infty) =...

Too late, sent it out. I spent a lot more than an hour on the problem, but I did solve it (I feel extremely awful about the time it required to solve it, bu I did learned from the experience). I got an answer in the magnitude of 10^(-2) for I_2. Eh.

Yea... I know. I first started with Cramer's Rule because of that stupid textbook, but then the linear algebra knowledge kicked in at the last minute... :cry: Yea, I know about this. I was just surprised that the current of the 10.0V voltage source was occurring in the in the opposite way of...

I'm getting a negative value for I_5 (At the 10V voltage source in the direction of the diagram. I didn't label it before the scan) . Did I make a mistake?

ARGH! I've been wasting my time. :cry: I'll simply row-reduce this matrix and call it quits. \left(\begin{array}{cccc}0&40&30&15\\20&0&30&10\\1&1&-1&0\end{array}\right) Zat ok?

I'm at a loss. How is it that the top resistor at I1 has only 5.0V? Can you explain? So I only need to use Cramer's Rule for I2, I3 and I4? (I4's polarity is wrong in the diagram) Thanx

Whoops again, I have to divide that number by the determinant. :redface: I get: I1 = 0.5 A This is hard doing it manually. :s

for I1. I get... 13000? That can't be right...

That would make sense since I would know the current comining in and out after. Ok then, thank you for helping me out. :)

Whoops, I4 was backwards, but not I2 though. So there's no I5 and I6 for each of the voltage sources?

Supernode? Sorry I'm not familiar with that term. :(

Here's a link to the image. Sorry it took so long. [img=http://img234.imageshack.us/img234/2566/a5p4600010jx.th.jpg]

I don't really know how to use Maple either. :-p It'll ceratinly make my life easier finding the determinant manually. But is this system of equation compliant with the attached circuit drawing?