Recent content by Loopas

Got it, thanks guys!

Do I add a constant to both sides of the equation? e^{-5u} + C_{u} = \frac{-5e^{7t}}{7} + C_{t}

Ok so I can rewrite as: \frac{du}{dt} = e^{5u} * e^{7t} e^{-5u} du = e^{7t} dt After finding the antiderivatives I cross-multiplied: 5e^{7t} = -7e^{-5u} \frac{-7}{5} = \frac{e^{7t}}{e^{-5u}} \frac{-7}{5} = e^{7t + 5u} This is now where I'm running into a problem. When I try...

Homework Statement \frac{du}{dt} = e^{5u + 7t} Solve the separable differential equation for u: Use the following initial condition: u(0) = 6. The Attempt at a Solution I tried to take the natural log of each side but now I'm stuck. How can I separate the equation when both the u...

Hi everyone, For my physics lab I have to solve for inductance, given the following equation and values for C, R, and f: 2*pi*f = \sqrt{\frac{1}{LC}-\frac{R^{2}}{4L^{2}}} I'm just not really sure how to solve for L. I started by squaring both sides but it just seems like I can't solve...

Is the individual impedance of the capacitor equal to XC? And for the inductor XL and resistor R? I found the RMS voltage using I = Vsource/Z. After that I found the voltage across each component by multiplying the individual impedances by the RMS voltage.

Well I have \frac{V_{out}}{V_{in}} = \frac{XC}{\sqrt{R^{2}+(XL-XC)^{2}}} How can I apply this to finding those functions that are part of the phase shift formula?

Homework Statement An LRC series circuit with R = 200Ω , L = 31mH , and C = 1.8μF is powered by an ac voltage source of peak voltage V0 = 540V and frequency f = 550Hz. What are the peak voltages across the inductor, capacitor, and resistor and their phase angles relative to the source...

PS - I'm using n = 30000 turns/m since n is originally given in turns/cm

Homework Statement A coil with 150 turns, a radius of 5.0 cm, and a resistance of 12 Ω surrounds a solenoid with 300 turns/cm and a radius of 4.3cm; see the figure. The current in the solenoid changes at a constant rate from 0 to 1.8 A in 0.12 s. Homework Equations B from solenoid =...

:approve: Finally found it! emf = -(5.3*10^-4)(cos(2500t)) Thanks for the help!

Actually I messed up before, this should be correct I think \frac{μ*I(t)*a}{2*pi}(ln(a+b)-ln(a))

I just want to make sure I'm doing this right -- I found the integral to be \frac{μI(t)a}{2pi}(ln(2pir)) Integrating from a to a+b it would be -- \frac{μI(t)a}{2pi}(ln(2pia+2pib)-ln(2pia))

Ok, so \int\frac{μ*I(t)}{2*pi*r}*adr And this will be integrated from b to b+a?

Well let's put it this way... I'm supposed to have a clue about integrals, but it's something that still eludes my full understanding, especially when integrating things like Faraday's Law, Ampere's Law, all of that good stuff. So I'm guessing that the first step in this problem should be to...