Recent content by s_a

In using this method, you are treating the wye resistors & delta resistors as a 3 terminal black box, and you're determining the resistance seen at any two of these terminals with the remaining third terminal hanging and equating the resistances. This is useful for determining the wye values...

The Thevenin resistance is easy to find, you disable the current (turn it into an open circuit) and voltage sources (turn it into a short circuit), and then find the equivalent resistance across RLOAD (removing the resistor RLOAD itself of course). So you get Rt = R2 // R3 = (20^-1 + 45^-1)^-1...

For b) you need to take a circular Amperian loop of radius 6mm, and the enclosed current is I2 - I1. So if you apply Ampere's Law, the answer is mu*(I2 - I1)/(2*pi*0.006).

Because in the case of V1 = Vcm = 0, there is no potential difference across r1, hence no current through it. So whatever the value of r1, no current flows through it, so it has no effect on the output Vo. Likewise, in the case of V2 = Vcm = 0, there is no potential difference across r3, r4...

A and B are vectors which are both tangential (but not necessarily perpendicular to each other) to the surface of the hemisphere. A x B is a vector which is perpendicular to both A and B (and hence NORMAL to the hemispherical surface). The magnitude of A x B is the area of the parallelogram...

What I'd do is determine the potential energy (U) stored in the springs for any value of q (using Hooke's Law), and then use the fact that the force F = -dU/dq. From this you can determine the acceleration in terms of the position q and hence form a simple 2nd order DE to solve for the equation...

Firstly, this line integral is incorrect. The circle may be parametrised as (x, y, z) = (cosФ, sinФ, 0) d/dФ (x, y, z) = (-sinФ, cosФ, 0) So the line integral is: Int{0 -> 2[FONT=Arial]π} (y, -x, z) . (-sinФ, cosФ, 0) dФ = Int{0 -> 2[FONT=Arial]π} (sinФ, -cosФ, z) . (-sinФ, cosФ, 0) dФ =...

z = 1000 - 0.01x^2 - 0.02y^2 grad(z) = (-0.02x, - 0.04y) at (50,80,847), grad(z) = (-0.02*50, - 0.04*80) = (-1,-3.2) (this vector points in the direction of the steepest uphill slope) a) The unit vector for "south" is (0,-1), so (-1,-3.2) . (0,-1) = 3.2 > 0. So a walk south from the...

Use the following facts: E = V/L, (E = electric field, V = voltage, L = separation) Q = CV -> dQ/dt = C * dV/dt -> I = C * dV/dt, C = εA/d (provable using Gauss' Law)

I'm assuming you want to find the magnetic field where the black dot is in the diagram. Working out the field due to the straights part of the wire is easy, it is as if there's one infinitely long wire carrying current I and distance r from the point where you want to find the B field. Working...

Suppose you have a straight wire carrying current I, of length 2L, and the midpoint M of the wire is a perpendicular distance D from a point O where you wish to find the magnetic field. Suppose OM is the line going from point O to point M (the line OM meets the wire at 90 degrees). Suppose...

Actually, no. The wire with the current going up contributes the same magnetic field at point O as the wire going down (use the right hand grip rule). So determine the magnetic field due to the wire with the current going up (or down), double it, and add it to the magnetic field due to the...

Almost there. Remember E . dE/dt = 1/2 d/dt |E|^2 and B . dB/dt = 1/2 d/dt |B|^2 (sorry I don't know Latex). Use these results in your last equation, and not surprisingly you should get the expression for the power per unit volume of the electric and magnetic fields.

I'd just assume the length is the distance between the rails, otherwise there wouldn't be much point in telling us the distance between the rails (actually the length of the metal rod isn't relevant anyway, only the length of it exposed to the magnetic field)

In this case there are two forces acting on the current carrying conductor, perpendicular to each other. Weight force (down), and magnetic force (left). The vector sum of these two forces points in the direction of the two supporting wires, diagonally down at an angle @ (theta) to the vertical...