Recent content by GreenLantern674

Okay, I found the formula IE=RI^2+IL(dI/dt) and I used that to solve (b), which turned out to be 19 W, but that formula didn't work when I tried it on (c). I think what I have to do is find the total power in the system and subtract my answer from (b) from that. But I'm not sure. Any suggestions?

An RL circuit in which L = 9.00 H and R = 5.00 is connected to a 24.0 V battery at t = 0. (a) What energy is stored in the inductor when the current is 0.500 A? (b) At what rate is energy being stored in the inductor when I = 1.00 A? (c) What power is being delivered to the circuit by the...

Never mind. I got it.

Okay, so what you said got me thinking, and I remembered that the time constant equals 1/t and that equals L/R so I set 1/1.4 = L/0.3 but that didn't work. Am I on the right track? P.S. is inductance negative?

[SOLVED] Inductance RL Circuit Calculate the inductance in an RL circuit in which R = 0.300 and the current increases to one fourth its final value in 1.40 s. I tried doing this with V=IR, (I know, the easy way didn't work). I also tried I= V/R(1-e^(Rt/L)) but I don't know what V would...

Help please?

Also, I know that the direction of the magnetic field is to the left of the page.

Can anyone help me with this problem? In Figure P22.31, the current in the long, straight wire is I1 = 8.00 A and the wire lies in the plane of the rectangular loop, which carries 10.0 A. The dimensions are c = 0.100 m, a = 0.150 m, and = 0.350 m. Find the magnitude and direction of the net...

Yes, but the electric field in the cavity isn't zero. That's what initially confused me.

A sphere of radius 2a is made of nonconducting material that has a uniform volume charge density . (Assume that the material does not affect the electric field.) A spherical cavity of radius a is now removed from the sphere, as shown in Figure P19.62. Show that the electric field within the...

Also, the problem defines L as the horizontal aspect of the triangle. Can I use the small angle theorum to assume that the hypotenuse is the same as the horizontal component?

I know that max acceleration equals Aw^2 and we've used the formula a=Aw^2 cos(wt+phi)

I've never seen that kind of equation before, except in the equation A=[F/m]/(sqrt(w2 - (k/m)2) , which turns into something kinda similar when you solve for w. Also, what is the "d" in md^y/dt^2 = -2Ty/L?

A ball of mass m is connected to two rubber bands of length, L, each under tension T, as in Figure P12.49. The ball is displaced by a small distance y perpendicular to the length of the rubber bands. (a) Assuming that the tension does not change, show that the restoring force is -2yT/L ...

Never mind. It turns out I was using the wrong formula.