Recent content by Kruum

A much simpler way, for me at least, is to use the delta-Y conversion. That way you can use the normal series and parallel connections, without having to resort to KCL and KVL.

I get the same V=-11.66V as well. Is this an online assignment, or how do you know it's not the right answer? If the problem was in degrees, I would assume the angular frequency would have the degree symbol added. After all, rad/s is the more commonly used unit.

I'm not quite following you here. What does the last part, p-p = 120, mean? When I graphically checked the answer, I got the same as plugging in the values to the given signal.

Thanks for the replies, guys! I forgot the one n in my first post. I found out my error was in the easy stuff after the integration, I'd done a mistake in adding fractions. :redface:

Welcome to PF, newbie991! The 60º only sifts the angle, well 60 degrees. You've probably done trigonometry, where sin(∏+∏)=sin(2∏). You just got to be careful, though. Here the voltage is in the form v=60sin(ωt+60º).

Check out the last two lines again. You've got a mistake there...

Homework Statement A particle is in a box of width L. Calculate the probability to find the particle in the region [L/4, 3L/4] when the particle is a) in the ground state b) in the first excited state. Homework Equations (2/L)sin(n*π*x/L)^2 dx is the probability in [x, x+dx] The...

I'm sorry if I'm doing something wrong here, but my question is close to this topic and I don't want to start a new one. Anyways, I'm reading a book that says the following: Source: Michio Kaku, Physics of the Impossible, 1st edition, Anchor Books (2008). I know the book has very...

Welcome to PF, m.farrukh! You might want to check out the http://www.allaboutcircuits.com/worksheets/index.html" at allaboutcircuits.com.

Can you post the whole problem? I'm quite sure you have left something important out. One error I can see right now is, you haven't laplace transformed the pulse signal.

I'm getting a bit confused. If you know how to determine the directions of the currents, are you still saying the current is .8A+.14A=.94A?

Welcome to PF, stimpyholder! You could describe solving this one backwards. You just need work your way through the circuit resistor pair at a time. Star with the connection of the two 20k\Omega resistors.

Here's a tip for you how to determine the direction of the current. Voltage sources produce current going from their + -terminal to - -terminal and current sources to the direction in which their arrow point. I'm also curious, how you got .114A as the other current. .8A-.114A=.686A Maybe a typo?

If there's a short circuit in the circuit, all the current will go through it. No current through R2 or R3, no voltage drop.

If you are familiar with Laplace transformation, this problem can simply be solved by making a laplace transformation of the circuit and calculating the necessary value.