CIRCUIT ANALYSIS: Use superposition to find the current I

[VinnyCee]

Homework Statement

Use superposition to find I in the circuit below.

http://img249.imageshack.us/img249/5889/chapter4problem15hd1.jpg

Homework Equations

KCL, KVL, Super-position principle, V = i R

The Attempt at a Solution

I turn only one source on at a time and solve for I_1 through I_3.

http://img180.imageshack.us/img180/5125/chapter4problem15part2nx3.jpg

Expressing the currents in terms of node voltages:

I_1\,=\,\frac{V_2}{3}\,\,&\,\,I_2\,=\,\frac{V_1}{2}\,\,&\,\,I_3\,=\,\frac{V_1\,-\,V_2}{1}\,\,&\,\,I_4\,=\,\frac{V_1}{4}

KCL @ V1) I_2\,+\,I_3\,+\,I_4\,=\,-2\,\longrightarrow\,7\,V_1\,-\,4\,V_2\,=\,-8

KCL @ V2) I_3\,+\,2\,=\,I_1\,\longrightarrow\,3\,V_1\,-\,4\,V_2\,=\,-6

Using those two equations in two variables, I get that V_2\,=\,\frac{9}{8}\,V.

Substitute back into equation forI_1 and I get I_1\,=\,\frac{3}{8}\,ANow, to the second source.

http://img263.imageshack.us/img263/5007/chapter4problem15part3jk5.jpg

Express the currents:

I_1\,=\,\frac{V_1}{2}\,=\,-10\,A\,\,&\,\,I_2\,=\,\frac{V_3}{3}\,\,&\,\,I_3\,=\,\frac{V_2\,-\,V_3}{1}\,\,&\,\,I_4\,=\,\frac{V_2}{4}

KCL @ V2) I_1\,+\,I_3\,+\,I_4\,=\,0\,\,\longrightarrow\,\,6\,V_1\,+\,3\,V_2\,+\,4\,V_3\,=\,0

KCL @ V3) I_3\,=\,I_2\,\,\longrightarrow\,\,3\,V_2\,-\,4\,V_3\,=\,0

3\,V_2\,=\,4\,V_3

Plug this into the equation for KCL @ V_2:

6\,V_1\,+\,6\,V_2\,=\,0

We know the voltage difference between V_1 and V_2:

V_2\,-\,V_1\,=\,20\,V

Solving: V_1\,=\,-10\,V\,\,&\,\,V_2\,=\,10\,V\,\,&\,\,V_3\,=\,\frac{15}{2}\,V

Now plug the voltages into the I_2 v = i R equation:

I_2\,=\,\frac{V}{R}\,=\,\frac{\frac{15}{2}\,V}{3\Omega}\,=\,\frac{5}{2}\,A\,=\,2.5\,ANow, the last source.

http://img412.imageshack.us/img412/3179/chapter4problem15part4ze2.jpg

Express the currents:

I_1\,=\,\frac{V_1}{2}\,\,&\,\,I_2\,=\,\frac{V_1\,-\,V_2}{1}\,\,&\,\,I_3\,=\,\frac{V_2}{3}\,\,&\,\,\frac{V_1\,+\,16}{4}

KCL @V1) I_1\,+\,I_2\,+\,I_4\,=\,0\,\longrightarrow\,7\,V_1\,-\,4\,V_2\,=\,-16

KCL @ V2) I_2\,=\,I_3\,\,\longrightarrow\,3\,V_1\,-\,4\,V_2\,=\,0

I get V_2\,=\,-3\,V and I plug into the original current equation and I get that I_3\,=\,-1\,A.

Finally, add them up:

I\,=\,I_1\,+\,I_2\,+\,I_3\,=\,\frac{15}{8}\,=\,1.875\,A

EDIT: Thanks for the help chanvincent!

 

[chanvincent]

For your second figure, I spoted a little mistake

The voltage across the 2\Omega resister is V_1, not -20.

-20 is the voltage of V_2 - V_1

If you correct this little error, you will get I_2 = 2.5V... which will yeild a correct answer.