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My class had come to discuss the Thevenin's theorem and we stumbled upon this particular problem which quite confuses me in a way that what I found is different than what my lecturer taught us. So, the problem goes like this enter image description here

schematic

simulate this circuit – Schematic created using CircuitLab

Apparently, we can make use of the leftmost loop there and get

$$2k\cdot i(t) + 3v(t) = 5 V$$

which is equal to

$$i(t) = \frac{5 - 3v(t)}{2000} ...(1)$$

I can understand that. I also understand that looking from nodes A-B's point of view, we have

$$v(t) = V_{Th} ...(2)$$

and therefore, making use of the rightmost loop before nodes A-B, we have

$$V_{Th} = 20i(t) \cdot 25 ...(3)$$

Inserting (1) to (3) then gives us

$$V_{Th} = v(t) = \frac{5}{7} \approx 0.714 A$$

which I understand too. However, my lecturer somehow put a negative sign on (3) and thus the original equation would turn out like this

$$V_{Th} = -20i(t) \cdot 25 ...(4)$$

These two equations would, of course, lead to totally different answers. My lecturer's method would lead to \$E_{Th} = -5 V$. The thing is, I couldn't figure out why it is necessary to put a negative sign there. So, in short, my question is, why is there a negative sign? Any answers would be much appreciated.

To add some context, the complete solution (according to my lecturer) to the problem goes like this. We short A and B and assume that there is a short-circuited current that goes through both nodes A and B. Let's call it \$i_{sc}(t)\$. And thus $$i_{sc}(t) = -20 \cdot i(t)...(5)$$ (notice that there is still the negative sign) and since we shorted nodes A and B, there will be no voltage between the two, and so we can write it down as \$V_{25 \Omega} = v(t) = 0\$.

As \$v(t)\$ is equal to 0, we can infer that the dependent voltage source on the leftmost loop becomes irrelevant to the equation of that loop, and so we get $$i(t)=\frac{5}{2000}=2.5 mA...(6)$$ Inserting this to (5) will give us $$i_{sc}(t)=-20\cdot 2.5 = -50 mA...(6)$$ This will then lead us to the value of the last unknown $$R_{Th} = \frac{E_{Th}}{i_{sc}(t)}=\frac{-5V}{-50mA}$$

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  • \$\begingroup\$ Post a legible schematic. \$\endgroup\$
    – Andy aka
    Apr 27 '20 at 12:02
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    \$\begingroup\$ That is still illegible - take a look at what you have reposted - take a good luck. \$\endgroup\$
    – Andy aka
    Apr 27 '20 at 12:44
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    \$\begingroup\$ If you publish a larger sketch, we can give it a look otherwise, you'll have to supply the magnifier. \$\endgroup\$ Apr 27 '20 at 15:44
  • \$\begingroup\$ Thanks to Ariser who has added a schematic on my original post. \$\endgroup\$ Apr 27 '20 at 23:23
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I’m using an iPhone to answer so pardon the math. The reason Vth=-20*i(t)*25 and not Vth= 20*i(t)*25 where * is just multiplication is due to the simple rule of passive sign convention. When current flows through the positive side of terminal of passive component it establishes positive voltage and when current flows into negative terminal it produces negative voltage.

In your circuit current from the voltage controlled current source enters into the negative side of 25ohm. Producing a voltage of V(t)=-20*i(t)*25.

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  • \$\begingroup\$ Oh! What a trivial miss... Thanks btw. \$\endgroup\$ Apr 28 '20 at 7:28

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