Why do you think that it is impossible?I am confused with outer loop calculation whether it's possible
Notice that Vx is a voltage drop across 5 ohm's resistor. So Vx = I1 * 5Ω thus I1 = Vx/5ΩCan someone let me know how I1 = Vx/5 comes out?
Hello,View attachment 271549View attachment 271548View attachment 271550
Hello, I understand why I only need to find Rth in this case. But as you can see in second photo, first, I am confused with outer loop calculation whether it's possible and second, how come I1 = Vx/5. Can someone let me know how I1 = Vx/5 comes out?
The author of that example is doing you a disservice by playing fast and loose with the units. Also, they are making the analysis needlessly obscure by fixing Vt at a specific voltage and by not eliminating Vx as soon as possible.View attachment 271549View attachment 271548View attachment 271550
Hello, I understand why I only need to find Rth in this case. But as you can see in second photo, first, I am confused with outer loop calculation whether it's possible and second, how come I1 = Vx/5. Can someone let me know how I1 = Vx/5 comes out?
I doubt that that is something the very many people would notice.This problem can be solved in one's head by noticing that the dependent source is equivalent to a -20 ohm resistor.
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