This was the only video that was actually quite good about this topic. But, as a suggestion, I would say it could be better if you could use drawings when explaining the signs instead of just telling them verbally. It seems complex for many as to why the potential differences are taken positive or negative in the loop, so a bit more clarification on that would be extremely helpful, Sir. Regardless, thanks a lot.
Thanks for the video, it was really helpful! At 1:08, did you mean "the battery subtracts voltage and resistors add voltage"? Since V=IR I guess it's the same thing but I was wondering if the focus here should be on voltage instead. Thank you!
why would the 18 volts in loop 3 not be positive? Would it not be 3V - I3(400) - (-18V) since the current going in the battery is in opposite direction?
you said something wrong at 9:06 it should be I2 time R2 but you said I1 time R1. also for the same part, for V2, you said R3 times the second current where it should have been the second resistor. Beside that, this has been one of the most helpful video I have seen on Kirchhoff's law.
Hi, sir! Thanks for the great content. I am wondering how does two junctions being both part of a same junction (because they yield the same equation) matter when we solve the problem? Thank you
Can you comment on what determines our choice of loops should be in a given circuit generally. In particular, should no loop be the combination of any of the other loops chosen. For example, would choosing the outer loop also here at 5:10 just yield the same information as the two inner loops. Also, can you give any intuition whether Kirchoff rules always lead to a solution: ie that we always get as many independent equations as variables. Thanks.
Find I3 first. It turns out to be negative (I3= -0.0375). Then find I2 = 0.015.So when plugging that back into the JA equation JA: I1 -I2-I3=0 becomes I1-0.015-(-0.0375)=0, I1= -0.0375+0.015 I1= -0.0225
