Recent content by bnosam

Thank you very much for you assistance!

Then I have the source, diode and two resistors in series in the circuit? So I can add the two resistors to make a 5k ohm resistor in series with the source and diode now?

So I would remove the voltage source and then with that voltage source set to short circuit, the two 5K resistors are in parallel and the Rth = 2.5k Ohms. Is that correct?

Ohh ok. I'm not used to drawing circuits in the manner that I presented it in, that's just the form my book uses. I struggle a lot with circuits in general. It makes sense that it is a voltage divider, though. So the voltage divider where R1 = R2 the output voltage is half of the input voltage...

Then everything on the left would simplify to the 10V source and just a single 10K resistor, no? The reason I want to simplify it is to apply the real diode model's properties to find current the compute output voltage of the diode using iterative analysis.

Are you meaning like this?

I'm assuming the section near the battery. I need to reduce the whole thing to a equivalent circuit with just a battery, resistor and diode. So whichever would get me to that the easiest would be best, so I can apply the real properties on a simplified model of the diode's circuit. So if I...

Homework Statement I have a circuit with a diode in it. I have to find the thevenin equivalent in order to reduce the circuit to a DC source, single resistor and single diode. How do I account for a diode when trying to find Rth and Vth? Do I just remove it and have it act as a short circuit or...

Homework Statement I have to do nodal analysis and I was wondering if there is any transformations or simplifications I can do to the circuit. In this class we are supposed to be able to look at the circuit and input what we need into the matrix based on just looking at it. I am not sure that I...

Whoops, should be in ksi.

So each little cube is (4*2) = 8 in^2 so τ = V/A τ = (4 kips)/ (8) = 1/2 kips Dividing by two for double sheet 1/4 kips ?

Sorry, I'm just trying to understand why...if I don't understand why then I can't actually continue with problems in the future... σ = P/A σ = 2 ksi A = (2 * 1) = 2 in^2 2 ksi = P/(2 * 1) P = 4 kips

Ohh I get it,I think because of double shear on this, the two edge pieces on the jaw will half the force on each, but the middle piece will have the full force so we use that for this stress? I hope that's what you meant.

But if we want the area of the part of the wood in contact with the part pulling, how do I assume from the diagram those are 1" because the middle is 1 inch"? The axial force is acting on the two parts in contact, right?

Plane a-a looks to be 4 x 2 but the plane we're looking for the area doesn't appear to have a measurement to me? it would be 2 * something for one contact area. Unless we can just assume it's 1 inch also because the middle measurement is 1 inch and we assume it's equal measurement for those...