I have the equation:
L*d2Q/dt2 + Q/C = Eocos(wt)
And I just realized that the root of my auxiliary equation is r = j/sqrt(LC), which means my complementary solution would be of the form:
Qc(t) = c1*cos(rt) + c2*sin(rt), which is entirely imaginary, correct? In that case, could I neglect the...
You may want to check again for B. I get B = 2*10-6, approximately. Since the magnitude of sin(wt) is at most 1, does a value for B larger than Q make sense?
I have a cosinusoidal voltage source with a series LC circuit which results in an inhomogeneous differential equation that I solved via the method of undetermined coefficients.
I apologize if this isn't relevant to the question, but I'm working on a similar situation where I get the Qocos(wt) term from the particular solution of my differential equation, but why is the complementary solution not included in the entire expression for Q(t)? Specifically, I have the term...
Sorry for the bump, but I'm working on the problem including the series R with L, and can't seem to figure out how to derive the resonance frequency w_o in that case. Wikipedia lists it as w_0 = sqrt((1/LC) - (R/L)^2), where the impedance is only real.
I have the equation Zin = (R +...
You draw the gaussian loop enclosing only half of the inside of the solenoid and the other half s outside. There is no "r" because dl = L when integrated over the length of the solenoid.
Because B does not depend on the radius of the solenoid, the B field inside the solenoid is uniform...
That's correct. The differential length is simply L when integrated, since the magnetic field runs parallel to the axis of the solenoid. The enclosed current Ienc is the current I running through each turn multiplied by the number of turns in the solenoid, or N*I.
Hopefully that helps.
I've started using a coordinate system based with the origin at the left-front corner.
So with the biot-savart law with the bottom front I get:
dl = -Uy*dy
r(hat) = (-sqrt(2)*b*Ux-b*Uz)/(sqrt(3)*b)
r = (sqrt(3)*b)
so then for dl x r(hat) I end up with: ((-sqrt(2)*b*Uz +...
Ok, that helps me visualize it a bit better. I'm still concerned about the magnitude of the side I first calculated though. I cannot see how they can get sqrt(2/3).
Using the RHR, my vectors match up for the first three sides on the base of the cube. However, when I calculate the vector for...
Homework Statement
Find the magnetic field at a point p in the center of the cube with side length 2b.
Homework Equations
Biot-Savart Law
The Attempt at a Solution
I attempted using the biot-savart law, but my answer contradicts what someone told us as a hint. Supposedly the...
For the sequence {arctan2n}, it is bounded by pi/2, so does that mean it converges to pi/2, because the limit of f(x) = L, and therefore the limit of the sequence is pi/2?