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1. Homework Statement
Use superposition to find I in the circuit below.
http://img249.imageshack.us/img249/5889/chapter4problem15hd1.jpg [Broken]
2. Homework 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 [itex]I_1[/itex] through [itex]I_3[/itex].
http://img180.imageshack.us/img180/5125/chapter4problem15part2nx3.jpg [Broken]
Expressing the currents in terms of node voltages:
[tex]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}[/tex]
KCL @ V1) [tex]I_2\,+\,I_3\,+\,I_4\,=\,-2\,\longrightarrow\,7\,V_1\,-\,4\,V_2\,=\,-8[/tex]
KCL @ V2) [tex]I_3\,+\,2\,=\,I_1\,\longrightarrow\,3\,V_1\,-\,4\,V_2\,=\,-6[/tex]
Using those two equations in two variables, I get that [itex]V_2\,=\,\frac{9}{8}\,V[/itex].
Substitute back into equation for[itex]I_1[/itex] and I get [itex]I_1\,=\,\frac{3}{8}\,A[/itex]
Now, to the second source.
http://img263.imageshack.us/img263/5007/chapter4problem15part3jk5.jpg [Broken]
Express the currents:
[tex]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}[/tex]
KCL @ V2) [tex]I_1\,+\,I_3\,+\,I_4\,=\,0\,\,\longrightarrow\,\,6\,V_1\,+\,3\,V_2\,+\,4\,V_3\,=\,0[/tex]
KCL @ V3) [tex]I_3\,=\,I_2\,\,\longrightarrow\,\,3\,V_2\,-\,4\,V_3\,=\,0[/tex]
[tex]3\,V_2\,=\,4\,V_3[/tex]
Plug this into the equation for KCL @ [itex]V_2[/itex]:
[tex]6\,V_1\,+\,6\,V_2\,=\,0[/tex]
We know the voltage difference between [itex]V_1[/itex] and [itex]V_2[/itex]:
[tex]V_2\,-\,V_1\,=\,20\,V[/tex]
Solving: [tex]V_1\,=\,-10\,V\,\,&\,\,V_2\,=\,10\,V\,\,&\,\,V_3\,=\,\frac{15}{2}\,V[/tex]
Now plug the voltages into the [itex]I_2[/itex] v = i R equation:
[tex]I_2\,=\,\frac{V}{R}\,=\,\frac{\frac{15}{2}\,V}{3\Omega}\,=\,\frac{5}{2}\,A\,=\,2.5\,A[/tex]
Now, the last source.
http://img412.imageshack.us/img412/3179/chapter4problem15part4ze2.jpg [Broken]
Express the currents:
[tex]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}[/tex]
KCL @V1) [tex]I_1\,+\,I_2\,+\,I_4\,=\,0\,\longrightarrow\,7\,V_1\,-\,4\,V_2\,=\,-16[/tex]
KCL @ V2) [tex]I_2\,=\,I_3\,\,\longrightarrow\,3\,V_1\,-\,4\,V_2\,=\,0[/tex]
I get [itex]V_2\,=\,-3\,V[/itex] and I plug into the original current equation and I get that [itex]I_3\,=\,-1\,A[/itex].
Finally, add them up:
[tex]I\,=\,I_1\,+\,I_2\,+\,I_3\,=\,\frac{15}{8}\,=\,1.875\,A[/tex]
EDIT: Thanks for the help chanvincent!!!
Use superposition to find I in the circuit below.
http://img249.imageshack.us/img249/5889/chapter4problem15hd1.jpg [Broken]
2. Homework 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 [itex]I_1[/itex] through [itex]I_3[/itex].
http://img180.imageshack.us/img180/5125/chapter4problem15part2nx3.jpg [Broken]
Expressing the currents in terms of node voltages:
[tex]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}[/tex]
KCL @ V1) [tex]I_2\,+\,I_3\,+\,I_4\,=\,-2\,\longrightarrow\,7\,V_1\,-\,4\,V_2\,=\,-8[/tex]
KCL @ V2) [tex]I_3\,+\,2\,=\,I_1\,\longrightarrow\,3\,V_1\,-\,4\,V_2\,=\,-6[/tex]
Using those two equations in two variables, I get that [itex]V_2\,=\,\frac{9}{8}\,V[/itex].
Substitute back into equation for[itex]I_1[/itex] and I get [itex]I_1\,=\,\frac{3}{8}\,A[/itex]
Now, to the second source.
http://img263.imageshack.us/img263/5007/chapter4problem15part3jk5.jpg [Broken]
Express the currents:
[tex]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}[/tex]
KCL @ V2) [tex]I_1\,+\,I_3\,+\,I_4\,=\,0\,\,\longrightarrow\,\,6\,V_1\,+\,3\,V_2\,+\,4\,V_3\,=\,0[/tex]
KCL @ V3) [tex]I_3\,=\,I_2\,\,\longrightarrow\,\,3\,V_2\,-\,4\,V_3\,=\,0[/tex]
[tex]3\,V_2\,=\,4\,V_3[/tex]
Plug this into the equation for KCL @ [itex]V_2[/itex]:
[tex]6\,V_1\,+\,6\,V_2\,=\,0[/tex]
We know the voltage difference between [itex]V_1[/itex] and [itex]V_2[/itex]:
[tex]V_2\,-\,V_1\,=\,20\,V[/tex]
Solving: [tex]V_1\,=\,-10\,V\,\,&\,\,V_2\,=\,10\,V\,\,&\,\,V_3\,=\,\frac{15}{2}\,V[/tex]
Now plug the voltages into the [itex]I_2[/itex] v = i R equation:
[tex]I_2\,=\,\frac{V}{R}\,=\,\frac{\frac{15}{2}\,V}{3\Omega}\,=\,\frac{5}{2}\,A\,=\,2.5\,A[/tex]
Now, the last source.
http://img412.imageshack.us/img412/3179/chapter4problem15part4ze2.jpg [Broken]
Express the currents:
[tex]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}[/tex]
KCL @V1) [tex]I_1\,+\,I_2\,+\,I_4\,=\,0\,\longrightarrow\,7\,V_1\,-\,4\,V_2\,=\,-16[/tex]
KCL @ V2) [tex]I_2\,=\,I_3\,\,\longrightarrow\,3\,V_1\,-\,4\,V_2\,=\,0[/tex]
I get [itex]V_2\,=\,-3\,V[/itex] and I plug into the original current equation and I get that [itex]I_3\,=\,-1\,A[/itex].
Finally, add them up:
[tex]I\,=\,I_1\,+\,I_2\,+\,I_3\,=\,\frac{15}{8}\,=\,1.875\,A[/tex]
EDIT: Thanks for the help chanvincent!!!
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