At 31:49, I believe you have overlooked the fact that R_3 is connected to the node that you've performed the KCL. There is a current entering that node, i_3 = V_1 / R_3 , meaning that the final Nortons current will be i_N = i_1 - V_1 (1 / R_2 + 1/ R_3)
@azizgundogdu9916
2 жыл бұрын
I wish these video series were released 3 years ago :)
@pubgmobile8202
2 жыл бұрын
Do you complete all these lectures?..
@azizgundogdu9916
2 жыл бұрын
@@pubgmobile8202 learned it by myself 3-4 years ago
@pubgmobile8202
2 жыл бұрын
@@azizgundogdu9916 I just started....
@selimyener2103
2 жыл бұрын
At 50:19, to find In, don’t we need to consider that kV1 is 0. Because we said that the most right side of circuit is short. It means, all current coming from V1 node through Rf should be equal to In. Please, explain it for me, thank you :)
@selimyener2103
2 жыл бұрын
At 50:19, to find In, don’t we need to consider that kV1 is 0. Because we said that the most right side of circuit is short. It means, all current coming from V1 node through Rf should be equal to In. Please, explain it for me, thank you :)
@moslemasgari7388
2 жыл бұрын
You know kV1 depends on the voltage across the resistance r_pi. It means that we don't consider that KV1 is 0 unless the voltage across r_pi becomes 0. so in this situation, kV1 has its own current that has been generated from R_F.
@betulozkan171
6 ай бұрын
28.55, why IN is going the way it goes? I cant figure it out
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