**1. The problem statement, all variables and given/known data**

Screenshots have everything.

**2. Relevant equations**

##V = iR##

Voltage/current division

Mesh analysis

Adding resistors in parallel/series

Kirchhoff’s Laws

**3. The attempt at a solution**

**First problem**

I started with turning off the current source. You are left with a simple series circuit with a dependent source because the part with the current source becomes an open circuit. I used Kirchhoff’s voltage rule around the loop and got: $$-12 + i_{1} + 3i_{1} + 2i_{1} = 0$$

Which yields ##i_{1} = .5##

I then turned off the voltage source and activated the current source. I chose to use mesh analysis because it seemed to be the quickest way. The current you are left with is a resistor and a dependent source, another resistor, and your current source in parallel. The mesh current on the left is ##i_{x}## and on the right is ##i_{y}##. Using KVL for the supermesh around the perimeter: $$i_{x} +3i_{y}+2i_{y} = 0$$

and also

$$i_{y} – i_{x} = 6$$

Which solves ##i_{x}## as -5 amperes. So, adding the first current and the second current gives -4.5 Amperes?

**Second question**

Again, I turned off the current source first. This leaves you with 2 resistors in series (parallel with the V source) and another 30 Ohm resistor in parallel. Since they all shared one node, the voltage should stay the same, right?

If that’s true, you can use Ohm’s law to find the current over the strand on the right will be $$\frac {200}{100} = 2$$

This next part is where I have a bit of a trouble. Turn off voltage source, turn on current source. If you look at the second picture, and replace the voltage source with a short circuit, does this mean that the 30 Ohm resistor can be ignored, because the current will always go through the short circuit instead of the path with the resistor?

If this is true, you can just use current division, like so $$3 \frac {60}{40+60} = 1.8$$

And you know this will be in the opposite direction from the previous current value, so ##2 – 1.8 = .2## so the current ##I_{0}## will be .2 Amperes. Is this correct? Thanks!

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