Recent content by aralbrec

This is correct and is found with Gaussian surface in the interior of the conducting shell. However you cannot draw this conclusion by argument of symmetry. The central charge is being acted on by all charges in space, including the charge induced on the inner and outer part of the shell...

The wire does in fact feel the change in field, which you will see in a moment. The best way to understand it is with Stoke's theorem but first let's take a little detour. One of Maxwell's equations, \nabla \times E = -\partial B / \partial t, says that a changing magnetic field at some point...

The deltas are just a common notation for determinant. It's the system of equations you get from mesh analysis put in matrix form. Problem 4.2 that you posted gives a completely worked out example of this. In mesh analysis, you assume currents I1, I2, ..., In flow around closed loops in the...

Sorry I was pulled away and it was too late to edit. It's an application of Cramer's Rule, maybe processed enough that it's not obvious. Your mesh matrix equation is V=ZI where the entries of the matrix V are the voltage sources in each loop. To find the input resistance at the left of the 7...

That is Cramer's Rule used to solve for an unknown in a matrix equation.

It could be sign-magnitude representation.

It doesn't solve the equation, it introduces another unnecessary unknown. By introducing Io you now have to add another equation by writing KCL at the ground node. You can look up modified nodal analysis to see what is done about voltage sources. Either circuit transformation is done (norton...

That's right, which means it's like they aren't there :) They sample the voltage but don't affect the circuit in any way.

He's doing a nodal analysis. You can't write useful nodal equations at a node connected to a voltage source (the top of V1 or the output of the opamp Vo) and that's what he's saying.

The problem with m→∞ is that the response goes to zero. So you can't let m→∞. Maybe you can say m becomes large enough that it dominates the response. Then I do agree with what you have been saying. You can call the (s+100) pole insignificant as I think you did or you could do this, which is...

it is important. both currents add because they flow in the same direction.

That looks fine except the second step looks to be a typo with the second '+' in the denominator meant to be multiplication. What kind of graphs did you do? Obviously they should be the same whether you use excel or mathematica. Bode plots have increasing frequency on the horizontal axis...

The information you should spot from the Bode Diagram is that there is a single pole (-20dB rolloff at w=400 rad/s) and a single zero (+20dB at w=4k rad/s). There is also a dc gain of 32dB at w=0. The pole and the zero locations will be decided by the two RC products. The thing with the...

With nodal analysis you are solving for voltages and you already have that voltage (Vo) so a useful node equation cannot be written there. Similarly, you can't write a useful node equation at the top of V1. In the case of V1, you could change the combination of V1/R4 into a norton equivalent...

This is because the cout is inside the while loop. Each time an integer is read, that line is executed. Well it's not initially set to NULL :) You're depending on the caller to set it to NULL for you when you should be doing it inside the function. Someone who is unfamiliar with the code...