Recent content by stiwi_xyz

Yup! Everything worked out! Much thanks man.

oh that's sneaky. So (1/((1/6)+(1/3)) and my overall resistance for the triangle is 2k Ohms now I add that to the 3k Ohms in the rest of the series and I end up with 5k Ohms?

So something similar to what we did in the first example? In this case, how would I get the overall resistance of the triangle portion?

So the Thevenin voltage is 0.003333*3, and we end up with 0.009. Now to calculate the resistance, is it possible to do it in one operation knowing the Thevenin voltage and overall current?

Yea so that's what I think I meant before when I said that the currents added up to 0, I just forgot that u'd have to also multiply them by the resistances. So now that we have our current of 3.333 amps, what can we do with that to actually find the overall resistance and the Thevenin voltage?

so the current would be 10amps? Using Ohm's Law I can find the current by dividing 30 by 3.

So if the triangle is basically a loop, doesn't that mean that the current through it should add up to 0

I mean I know what it is, but I haven't ever run into it. I'm pretty new to electrical engineering cause I've mostly just stuck to programming until recently.

All right so I got to do (1/((1/4)+(1/4)) and that gets me 2𝑘Ω for part a.

So I end up with an equivalent 4𝑘Ω resister on both sides and I can just do (1/4𝑘Ω) + (1/4𝑘Ω) and I get an equivalent 0.5𝑘Ω resistor?

I'm going to say 0, probably because its going both ways and if the resistance is equal on both sides, then the voltages should cancel out

We can definitely simplify by using the parallel and series resistor rules but in this question specifically, I'm not sure of which ones are arranged in a parallel or series orientation.

Consider the resistor network shown in Figure 1-1, where 𝑅1=2𝑘Ω and 𝑅2=5𝑘Ω. (a) Calculate the numerical value of the single equivalent resistance, in 𝑘Ω, as seen from the terminals 𝐴−𝐴′. Express your answer to two decimal places.Next, consider the circuit shown in Figure 1-2, where 𝑉=30𝑉 and...