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What is the Thevenin voltage and resistance at terminals a,b? Specifically, how do you go about finding these? I tried this problem and I get a Thevenin voltage of 57.6 V and a Thevenin resistance of 6 Ohms.

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    \$\begingroup\$ Show how you got to that result. We can help in solving problems, but we don't make your homeworks for you. \$\endgroup\$
    – DavideM
    Dec 30, 2016 at 20:06
  • \$\begingroup\$ I can explain what I did to you - I did some source transformations to find the Thevenin equivalent circuit instead of doing an open circuit voltage and short circuit current method. First, I converted the 72 V source into a current source (14.4 A) and put the 5 Ohm in parallel. Then I combined the 5 and 20 Ohm resistors (which gave me 4 Ohms). After that I did a second source transformation which gave me a voltage source of 57.6 Ohms. I then combined the 12 Ohm with the 4 Ohm and 8 Ohm resistors. I finally ended up with 57.6 V and 6 Ohms. \$\endgroup\$
    – ragzputin
    Dec 30, 2016 at 20:19
  • \$\begingroup\$ As a general rule, always post in your question (before people ask you) all the math and the reasoning behind it, so that people can look at them and find the mistakes you've made, if any. \$\endgroup\$
    – DavideM
    Dec 30, 2016 at 22:13
  • \$\begingroup\$ Okay got it. I tried to post some diagrams about what I did but it has been hard to find a good schematic/diagram editor. I will post the details soon. \$\endgroup\$
    – ragzputin
    Dec 30, 2016 at 22:51

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The equivalent resistance, short-circuiting the voltage source is: $$R_{eq}=(5||20+8)||12=(5(1||4)+8)||12=(5\frac{4}{5}+8)||12=12||12=12(1||1)=6\Omega$$ where \$||\$ is the parallel resistance operator: $$r=\frac{r_1r_2}{r_1+r_2}$$

The equivalent voltage source is obtained by calculating the open-circuit voltage. There are two loops, one for the power source (\$i_1\$) and other for the \$12\Omega\$ resistor (\$i_2\$). Hence the equations are: $$ 5(i_1-i_2)+20i_1=72\\ 12i_2+8i_2+5(i_2-i_1)=0 $$ In matrix notation: $$ \left[ {\begin{array}{cc} 25 & -5 \\ -5 & 25 \end{array} } \right] \left[ {\begin{array}{cc} i_1 \\ i_2 \end{array} } \right]= \left[ {\begin{array}{cc} 72 \\ 0 \end{array} } \right] $$ Hence: $$ \left[ {\begin{array}{cc} i_1 \\ i_2 \end{array} } \right]= \left[ {\begin{array}{cc} 3 \\ 3/5 \end{array} } \right] $$ And thus, the voltage is: $$v_{eq}=20i_1=60V$$

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