thanks , but can you explain it a little bit as i dont know much about what you did after the Vab=Rth+Vth?
That's the thevenin equivalent, then just make a voltage divisor to know the voltaje over the resistor
Well, Vab=Iab*Rth+Vth is the result of the thevenin theorem, the next thing I did was KCL on the node A, so entering currents are equal to the currents leaving, the most important thing is to let Vab in terms of Iab, the coefficient of Iab is Rth (thevenin resistance) and the other term is Vth (thevenin voltage)thanks , but can you explain it a little bit as i dont know much about what you did after the Vab=Rth+Vth?
i disappeared because the solutions given here are too abstract for me , no one solved it step by step in order for me to understand it ....btw electronics is not my major , i am just new to itHi,
Yeah, almost like a trick question
Another case of the disappearing, low posts count OP ?
well , according to what thevinin's theorem says on this site http://www.allaboutcircuits.com/textbook/direct-current/chpt-10/thevenins-theorem/ , its almost impossible for me to apply that same rules to this questionOh, and if the TS would look at the problem carefully, they'll see that there are only two sources that matter at all. Everything else is just a distraction. One node equation yields the answer and is simple enough that it can be done, with some thought, by inspection. Hint: -15 V < v < -10 V.
okay much better nowThere's nothing abstract at the Thevenin theorem that I showed to you
I gave you almost the 90% of the exercise, you only have to do this at the end