This wasn't an assigned problem, but I noticed it and wanted to try it but my currents from the two methods aren't equal: http://img8.imageshack.us/img8/5228/4444tnc.jpg http://img8.imageshack.us/img8/1492/4441.jpg
In the first image, not quite halfway down, on the right, you have: 6/9 = 1/3 A You've got to watch for these silly arithmetic mistakes; they lose you points on the final. If you fix that error and redo your calculations the rest of the way, you'll get i=1/2 amp. Now, for the nodal method in the second image. You got 1/2 amp, which is correct, but it was a coincidence that allowed you to get that answer. Your first nodal equation is wrong because you didn't take into account the leftmost 6 ohm resistor. Remember in an earlier post, I told you that when you're using the nodal method, and you have voltage sources connected to nodes, you have to do something special. You have to use the supernode method. You need a third equation for the node to the left of the 3 volt source. The coincidence that allowed you to get the right answer by accident is this: Remove the 3 volt source and the 6 ohm resistor to its left. Solve the resulting network. You'll find that the voltage at the node you've called e1 is 3 volts. This means that when you add back in the 3 volt source and 6 ohm resistor, they have absolutely no effect on the circuit; the polarity of the 3 volt source is such that it opposes the 3 volts at e1 and no current flows in the 6 ohm resistor or the 3 volt source. That's why you got the correct answer even though you didn't take into account the 6 ohm resistor to the far left. To see that your equations won't work right, change the 3 ohm resistor that goes to ground from node e2 to 1 ohm and solve the network. The correct voltage at node e1 is then 12/7 volt, but your equations won't give that value.
Oh, I remember you mentioning the supernode part, but I didn't catch that it applied to this situation... which is actually important for me. I thought I did take the 6 ohm resistor into account in the first term of my equation: (e1-3)/6 which I'm sure is wrong but was an attempt to take it into account nonetheless This was from an earlier test where of course she didn't teach us any of this stuff, so it's apparent in my work there that I was very unfamiliar with the node method at the time: (the circuit drawing under the actual problem is just me writing the answer she gave us after handing them back using superposition) http://img266.imageshack.us/img266/4687/777t.jpg Is what she did the way to use node method for that problem? Could you show me how to use the supernode method starting with the original problem of this thread, I'm looking at putting another equation for the node at the top-left corner of the circuit between the 6 ohm resistor and the voltage source, but I'm confused
Whenever you're using the nodal method, and there is a voltage source connected to a node, or between nodes, you may have to do something special, depending on just how complicated the circuit is. Looking back at what you did, I see that you did take the leftmost 6 ohm resistor into account; the fact that there are two 6 ohm resistors confused me. In this case, where you have a voltage source in series with a single resistor to ground, you can do it the way you did, and you don't need to use the supernode method. As long as you have this topology, without any more branches connected to the junction between the voltage source and the 6 ohm resistor to ground, you can use that technique. That's probably what you'll have on your exam. The way she did it is fine, and you did it her way. If you want to know about the supernode method, search the web, although it might not be a valuable use of your time right now. Here's a good start: www.ece.uvic.ca/~sneville/Teaching/Elec250/LectureNotes/Chapter 4.pdf
Ya you're right, she said no suprises on exam... it will be whatever we've had so far. I'll be sure to learn about supernodes after the final. But to make sure - What I did in my written example above with my equation and subsequent steps for the nodal analysis was correct and done the right way and not just correct by coicidence? And aside from my silly algebraic mistake in superposition I did that correct also?
Yes, it it's fine. I was confused by the two 6 ohm resistors in the circuit. But, interestingly enough, the coincidence I mentioned is quite true. If you leave out the (e1-3)/6 term from your first equation, you still get the same result! So the 3 volt source and its associated 6 ohm resistor do nothing! Since e1 = 3 volts, the (e1-3)/6 term evaluates to zero. This fact also added to my mistaken impression that you hadn't taken the leftmost 6 ohm resistor into account, and that coincidentally it didn't matter. Your superposition work looks ok, too.