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Sorry for the bad title, but I'm trying to get the Thevenin equivalent circuit and then use that circuit to calculate a current i. The problem I'm having is I'm not sure what to get the open circuit voltage with the terminals in this arrangement. Here's my circuit and the work I've done: enter image description here

So I know these equations: $$V_{oc}=V_{th}\;\;\;,\;\;I_{sc}=I_N\;\;\;,\;\;R_{th}=R_{N}={V_{oc}\over I_{sc}}$$ So to start, I will find $$I_N$$ Here's the circuit for it: enter image description here

Using the supermesh: $$-20+15i_1+5i_1=0\;\;\;\;\;\;\;\;I_{sc}=i_1=1A$$

Now I get stuck. I've tried finding R Thevenin and R open circuit but with both of those, I come to equations that have more unknowns than equations and thus can't solve them. What could I do? If I do the open circuit way should I omit the 6 ohm resistor and use KCL?

(just noticed that there is a schematic editor, but anyway, that current source is controlled, sorry for the confusion)

I'm not sure where to go from here and would really appreciate some help. Thank you.

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  • \$\begingroup\$ The easiest path is to use superposition: set the 20-V source to 0 V (replace it by a short circuit) and calculate \$V_{ab}\$ with the second source alive. Then set the second source to 0 V or 0 A (I'm not sure if this is a current source labeled \$V_x\$) and determine \$V_{ab}\$ in this condition with the 20-V source alive. The complete answer is the sum of both results. \$\endgroup\$ – Verbal Kint Oct 5 '18 at 5:34
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From the circuit drawing, we can see that the "load" is the \$6\Omega\$ resistor. So

If I do the open circuit way should I omit the 6 ohm resistor and use KCL?

To which the answer is yes. It may not be the easiest way, but it is the most general.

If you have equations where you have more unknowns than equations, then you're probably missing something. Calling the node between the \$15\Omega\$ and \$5\Omega\$ resistor \$v_1\$, the KCL equations in each node would be:

$$\frac{20V-v_1}{15\Omega}=\frac{v_1-v_a}{5\Omega}$$

$$\frac{v_1-v_a}{5\Omega} + 1S\cdot (v_1-v_a) = \frac{v_a}{10\Omega}$$

You also may need to look to your supermesh method. I believe you need to add a contribution in there from your current source. Or

$$V_x = i_1\cdot 5\Omega = I_{sc} - i_1 \Rightarrow I_{sc} = i_1 + i_1\cdot 5\Omega$$

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