Basically there are few different things here that appear lumped together in your mind and which may be better seen separately.

First you have the classical methods for solving circuits, such as the mesh analyses, node-voltage analyses, full equation system, reduced equation system and maybe some more I forgot.

First, the full equation system has no excuse to be used, so don't use it. The reduced equation system should only be used for very simple circuits where the preparation to use one of the two other systems would take more work than solving the circuit using reduced system. After some practice, you should be able to determine how much work you'll need to solve the circuit using each system (so try using all 3 systems on some problems, just so you can compare how much work each requires).

Next, you should have good theoretical knowledge of how each system works. The circuit may be such that some system cannot be easily used with it. You should also note the special cases where the mesh analyses and nodal analyses are especially useful. For example, in mesh analyses, if you have lots of constant current sources, you get lots of trivial equations for mesh currents. Same thing with nodal analyses and voltage sources. So basically, you should calculate how many equations you'll get with each and pick one with least equations, but do take into account special cases of current source and voltage sources when calculating the number of equations!

After that, you have the theorems which you use. I'd say that you should first see if any theorems can be applied to a circuit and only after applying them try to solve it using the systems. For example, you should use Norton/Thevenin when you have a big circuit where one for few elements change. For example, you have a circuit with a rheostat and you need to calculate say current through it on various settings. In this case, just replace the rest of the circuit with Thevenin's generator, since it doesn't change. In case of superposition theorem, it's useful in cases where you have sources that turn on and off and you should calculate their effects on some part of the circuit. Same thing goes for other theorems, like bisection (where you have symmetrical circuit, so you only need to solve one half of it) or compensation (which is useful when you have sources whose values change).

So for theorems, the general idea is to find use cases where each of them will actually allow the circuit to be simplified. So when you have a problem, ask yourself not "How am I going to solve this using method X?" but "Why am I going to solve this using method X?". This should work even on problems in textbooks where they are divided by areas. So as I said before, try solving one problem using several different methods. See which ones can be applied to the problem, which ones can't be applied to the problem, ask yourself why for each and then take a look and see which method is the most optimal (in sense that the least number of equations needs to be solved or that you get a significant number of simple equations) and try to understand why the most optimal method is the most optimal method. This way, you'll see when it's counterproductive to apply some theorem, when you gain nothing by applying some theorem and when applying a theorem actually helps. Same story goes for reduced system, nodal and mesh analyses too.

I'm not going to just give you the answer to your homework problem.

However, consider what it really means to be a voltage source. Let's say I have a 6V source that is good to 1 A. What voltage is it when nothing is connected? When I put a 60 Ω resistor accross it? A 30 Ω resistor accross it?

If I put the 6 V source and resistor inside a black box and brought out only the two leads, what difference could someone observe on the outside to distinguish the no load, 60 Ω and 30 Ω cases? Now consider this is the case you have. Your +6V and R are inside the box and the rest of the circuit on the outside.

## Best Answer

If I understand the diagram,

Iis the current through the 1 Ohm resistor at the far right of the schematic.What they're saying is, assume

Iis 1 A. Now you know, for example, the voltage at the next node left is 3 V. And you have a 1 A current through the 3-Ohm resistor. Since you have 2 A through the last two rungs of the ladder, you must have 4 V across the 2-Ohm resistor 2nd-from-the-right (see how much easier this would be if you justput designators in your schematic?). And so on, working back to the left.Now when you get back to the source, you will get some voltage other than 7 V. Say you get 21 V (I haven't calculated, I just made up a number).

Since the system is linear, you know that if you scale this voltage down to 7 V, the current

Iwill also scale down by factor of 3, so the actualIis 333 mA, not the 1 A you originally assumed.