# Circuit Analysis

• Mar 1st 2011, 12:16 PM
quantoembryo
Circuit Analysis
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 $R_1$ (is it away from node a?). Dtermine the c) magnitude and d) direction of the current through $R_2$ (is it away from node b?). e) Find $V_{ab}$.

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

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

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

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

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

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

Therefore, the magnitude of the current going through $I_{R_1}=1.5A-1A=0.5A$. 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.

$R_{eq}=5.875\Omega$

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

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

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

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

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

e) I find this one very confusing too. $V_{ab} = V_a-V_b=(4V+1.915V)-(1.277V+6V)=-1.36V$
• Mar 2nd 2011, 06:35 AM
quantoembryo
There we go.

Moderator edit: Due to computer issues the OP was unable to post the complete question until now. The date stamp can't be changed to accomadate this so it is reasonable on this occassion that the OP 'bump' this thread.
• Mar 4th 2011, 06:48 AM
Ackbeet
Let's call the resistor in the middle $R_{3}.$ I would definitely go with KVL analysis (mesh current analysis) on this one, since you have three nodes but only two meshes.

First Case: Short out the $V_{1}=4\,\text{V}$ source. Call the left mesh current $i_{1},$ going in the clockwise direction. Call the right mesh $i_{2},$ again going in the clockwise direction.

Mesh 1: $i_{1}R_{1}+V_{2}+(i_{1}-i_{2})R_{3}=0,$ and

Mesh 2: $i_{2}R_{2}+(i_{2}-i_{1})R_{3}-V_{2}=0.$

The system becomes

$7i_{1}-3i_{2}=-6$

$-3i_{1}+8i_{2}=6,$ with solution $i_{1}=-0.638,\;i_{2}=0.51.$

Similar analysis with $V_{2}$ shorted out yields, with the same mesh current definitions, $i_{1}=0.681,\; i_{2}=0.255.$

So I get $i_{1}=0.043,\; i_{2}=0.765.$ The $i_{1}$ current is directed away from a, and the $i_{2}$ current is directed away from b.

I'll need to refresh my memory on the passive sign convention in order to finish the problem.

Does this help?
• Mar 7th 2011, 04:47 AM
quantoembryo
THanks
• Mar 7th 2011, 04:57 AM
Ackbeet
You're welcome!