# Find the Thevenin's equivalent of a circuit between nodes A and B

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

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}$$

• Post a legible schematic. Apr 27 '20 at 12:02
• That is still illegible - take a look at what you have reposted - take a good luck. Apr 27 '20 at 12:44
• If you publish a larger sketch, we can give it a look otherwise, you'll have to supply the magnifier. Apr 27 '20 at 15:44
• Thanks to Ariser who has added a schematic on my original post. Apr 27 '20 at 23:23