I'd like to apologize for not finishing yesterday, as my computer was not cooperating. Anyways, he we go:

Consider the following circuit. Using superposition determine a) the magnitude and b) the direction of the current going through $\displaystyle R_1$ (is it away from node a?). Dtermine the c) magnitude and d) direction of the current through $\displaystyle R_2$ (is it away from node b?). e) Find $\displaystyle V_{ab}$.

a) First, I will find the equivalent resitance, and then short each of the voltage supplies and determine the current separately.

$\displaystyle R_{eq}=4\Omega+[(1/3\Omega)+(1/5\Omega)]^{-1}=5.875\Omega$

$\displaystyle I_{R_{1,1}}=(R_{eq})(R_{1})^{-1}(I_1)$

$\displaystyle I_{R_{1,1}}=(5.875\Omega)(4\Omega)^{-1}(4V/5.875\Omega)=1 A$ away from node a.

Due to the other voltage source, we have $\displaystyle I_{R_{1,2}}=(R_{eq})(R_{1})^{-1}(I_2)$

$\displaystyle I_{R_{1,2}}=(5.875\Omega)(4\Omega)^{-1}(6V/5.875\Omega)=1.5 A$ towards node a.

Therefore, the magnitude of the current going through $\displaystyle I_{R_1}=1.5A-1A=0.5A$. For b) it is directed towards a.

Next, this is where I get rather confused, and because of the ground location.. Let's say for a minute that the current goes through R1 and heads towards R2, what is stopping it from going through the ground rather than the resistor? Anyways, here is my attempt.

$\displaystyle R_{eq}=5.875\Omega$

$\displaystyle I_{R_{2,1}}=(R_{eq})(R_{2})^{-1}(I_1)$

$\displaystyle I_{R_{2,1}}=(5.875\Omega)(5\Omega)^{-1}(4V/5.875\Omega)=0.8 A$ away from node b.

Due to the other voltage source, we have $\displaystyle I_{R_{2,2}}=(R_{eq})(R_{2})^{-1}(I_2)$

$\displaystyle I_{R_{2,2}}=(5.875\Omega)(5\Omega)^{-1}(6V/5.875\Omega)=1.2 A$ away from b.

c) $\displaystyle 1.2 A+ 0.8A= 2 A $ d) away from b.

e) I find this one very confusing too. $\displaystyle V_{ab} = V_a-V_b=(4V+1.915V)-(1.277V+6V)=-1.36V$