Recent content by Kavik

A very basic model of a transmission line is like that of a resistive divider. If we look at two examples I think it will be clear. First Example: V1 = 120 Volts. R_Line = 0.01 Ohms. R_Load = 2.00 Ohms. There's only one current to calculate since everything is in series: I =...

It is probably tinned copper. Tinned copper is a bit easier to work with when soldering so I could easily see it being included in any learning kit. It is also less prone to corrosion. Any differences in stats would be negligible for what you're doing since it's mostly copper. Try looking at...

I'm going to assume you meant 9kW instead of 9kV here. Are you asking if you should use something like a circuit breaker or fuse (overcurrent protection)? Here's what I would do if this was a factory power supply and not some independent generator: A 10kW supply to me means it can provide...

There's two main reasons I can think of that would affect your motor: The output voltage of a rectified signal is not a flat DC voltage, it is actually just a sin wave except the negative part of the wave is reversed to be positive. The average voltage of this wave is lower than your RMS...

Here is a table for wire gauge (AWG) and Ohms (resistance) per 1000 feet. The only columns you need to worry about are the first and third. You can see that 1000 feet of a 22 gauge wire has a resistance of 16.0 Ohms. I would say a good estimate for the length of your wire is 1 foot. This gives a...

Some diagrams of Op Amps leave out the voltage source (Vs+,Vs-). That is where the output current flows from.

Have you tried distributing j100 Ohm and -j100 Ohm and then combining like terms?

Yep! I was just making a few doodles on the picture to illustrate that, so I'm going to post it anyways, even though you already got it. :smile:

P = IV still applies. You know the current through the two resistors (0.4 A), and the equivalent resistance of the two series resistors is 4Ω. So to find the voltage from the top node to the bottom node, it's just: V = IR V = 0.4*4 V = 1.6 volts Then P = IV! Edit: And maybe I should be clear...

That's what I got! You can always do an extra check by calculating the power produced (by the two current sources) vs. power consumed (by the two resistors). These should always be the same.

Hmm, I may have given a more complicated method than necessary. If you were to use Mesh Current, you would draw the 2 current loops. When you have a current source in a loop, then you can just define that loop current as the value of the current source (i.e. I1 = 0.2 A, I2 = V1/4). I have...

Hi November, Are you familiar with Mesh Current analysis? I think you are getting ahead of yourself when you add the two resistors together. Try leaving them separate and applying Mesh Current analysis. That should give you two equations and two unknowns. Here is a short lesson on Mesh...

These are matrices, correct? Why not just try a couple examples of n, say n = 1. Then n = 2. Then n = 3. The pattern should be clear then and you could make a more general proof. Matrix multiplication is all you need, e.g.: [A]^2 = [A]*[A]

This will help you with #1: http://www.algebralab.org/lessons/lesson.aspx?file=Algebra_Exponents_e.xml What is the variable and the constant with regards to the limit in #2? "Limit as __ goes to __"

You have to simplify one branch at a time where resistors are clearly either in series or parallel. Starting at the left branch, it begins to simplify like so: [1] Recognize that the two 3Ω resistors are in series, add them: 3Ω + 3Ω = 6Ω. Replace that resistor in next step. [2] With the new...