# CIRCUIT ANALYSIS: Use superposition to find the current I

1. Jan 26, 2007

### VinnyCee

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

Use superposition to find I in the circuit below.

http://img249.imageshack.us/img249/5889/chapter4problem15hd1.jpg [Broken]

2. Relevant equations

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

3. 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 [Broken]

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 for$I_1$ and I get $I_1\,=\,\frac{3}{8}\,A$

Now, to the second source.

http://img263.imageshack.us/img263/5007/chapter4problem15part3jk5.jpg [Broken]

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:

$$I_2\,=\,\frac{V}{R}\,=\,\frac{\frac{15}{2}\,V}{3\Omega}\,=\,\frac{5}{2}\,A\,=\,2.5\,A$$

Now, the last source.

http://img412.imageshack.us/img412/3179/chapter4problem15part4ze2.jpg [Broken]

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 $I_3\,=\,-1\,A$.

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

EDIT: Thanks for the help chanvincent!!!

Last edited by a moderator: May 2, 2017
2. Jan 26, 2007

### chanvincent

For your second figure, I spoted a little mistake

The voltage accross 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 $$I_2 = 2.5$$V... which will yeild a correct answer.