Recent content by johnsmith7565

I’m going to do the first part of your question first. Here is the node voltage equation: (v-40)/500 + v/2000 + v/6000 = 0 Using a calculator I get v=30 V. By ohms law: i2(0-) = -30/2000 = -0.0015 A. That’s right. I1(0-) = 30/6000 = 0.005 A which is right. Why do I get the correct...

I shorted the inductor and performed mesh analysis. The solutions to the linear system were done using a calculator. The book says that the value for i2(0-) should be 15 mA but I'm getting -2mA. What am I doing wrong? I'm completely confused. Maybe mesh isn't the most efficient way to find I2...

Thanks a million! I got the correct answer.

I am not sure where I made the mistake. If someone could point that out that would be much appreciated!

Yes, the answer is in the back of the textbook. I don’t really need to use nodal analysis since I’m just doing this for my own pleasure. I’m just trying to learn the basics of circuits so I just care about arriving at the correct answer, not really the method. Initially, I thought nodal analysis...

24 V for v0.

I taught myself the loop current method today, and I've just solved this circuit using the loop-current method. IMO mush easier and straightforward with the circuit using the loop-current method.

Sorry for not responding earlier. Trying to make the other node a ground node went no where. To be honest I am a little bit confused... why would I plug in infinity for the 40 ohm resistor? How do I even do that? This problem seems very difficult for a basic text on node voltages! It seems...

I'm in school currently, but when I have some down time I'll try setting the node labeled v2 as the ground. I think this might help out because the 60 V source would be in-between the current ground node and v1, which would make a supernode and give a constraint equation for idelta in terms of...

For v1: -i +(v1-v0)/10 + (v1 - v2)/30 For v2: I don’t know the current going through the dependent voltage source but I do know that current + (v2-v0)/20 + (v2-v1)/30.

What I’m stuck on is finding the constraint equation for i on the top question . I don’t know how to find it when it’s through a voltage source and not over any resistors. (I can’t use ohm‘s law) After I find i the problem should become easy to solve. I know that v1 = 10 and v2 = 20i . The KCL...

I still don’t understand. So the 40 ohm resistor and the 80 ohm one are in series so I add them, and now my equation is 4=(20)(120/R+120) and R is still too large.

The resistors are all in series , with exception of R, so I added them together and then used the current divider equation to solve for R, and I got 720 ohms. The textbook says R should be 30 ohms. I’m completely lost.