Ah, I see. This was a LOT easier than I was making it... haha.
I was caught up on the idea that the distance traveled would contract if he was traveling at 0.6c, and thought that if he didn't slow down, event B would happen behind him. I'm very intrigued by the effects of relativity but I'm...
Using the distance between the two events divided by the time (5 microseconds) as V:
V = 1000m/(5 E-6)s = 2E8 m/s
g = 1/√(1 - (2E8)2/c2) = 1.8
x' = 1.8(1000m - (2E8)(5E-6)) = 0
So V is the speed of the moving observer as seen by a hypothetical second person at rest. Here, V is just...
Homework Statement
Event A occurs at xA = 500m. Event B occurs 5 microseconds later at xB = 1500m. With what speed must an observer move in the positive x direction so that the events occur at the same point in space in the observer's frame?Homework Equations
Lorentz transformation...
Ah, that makes sense! Thank you!
Just to double check, I got for an answer:
i = Voeiwt (iwC + 1/R)
Does it matter that there's also an i in the exponent?
RIGHT that's what I meant.
Is that my answer (solving for i) or do I set that equal to I/C and solve for I?
I guess I'm getting hung up on the fact that there's an instantaneous current (i) and an average/overall current (I), and I don't know which to solve for.
Is the i in the voltage function V = Voeiwt the current I want? If so, I've been spending MUCH longer on this than I needed to... haha
In that case, it'd just be:
V = Voeiwt
dV/di = Vo(iwt)eiwt
I've gotten as far as:
i = C*dV/dt + V/R
Is it mathematically correct to say that V is common to both terms on the right, and therefore:
i = V(C*d/dt + 1/R) ?
If so, I would plug in the V value given in step (b):
i = Voeiwt(C*d/dt + 1/R)
How would I go about integrating this...
Homework Statement
I am given an RC circuit with an alternating current. The circuit contains a capacitor and a resistor in parallel. Part (a) says "Use KCL to find a differential equation for I in terms of V." Part (b) says "For an applied voltage V = Voexp(iwt), find the current I."...