Asescu, you're very welcome. I'm super glad the videos are helping even if it's a little bit. Your words are very encouraging so thank you so much. I try to do these review videos from a student perspective so let me know of any suggestions to improve in any way possible. Thank you :)
When you were doing loop one, you went from the voltage source to the .25 ohm resistor. You assigned its value a negative because it was a voltage drop. That makes sense. But when you were doing loop two, you went from the voltage source to the .4 ohm resistor (after changing the polarity). Why did you assigned its value as a positive when you made the value for the .25 ohm resistor negative?
John, it's because assigning a negative sign would affect the values inside the bracket so I choose to put positive sign so I can it won't affect my analysis of voltage drop/rise. The voltage drop only applies with the current I2 (hence negative sign) and not I1. With the .25 ohm, only current I1 passes thru that resistor as a voltage drop. You get it? Let me know...
Sir, Please teach me the loop current method if you got to know crack this method by now since i see that you posted this comment a year ago. Thank you.
I am clearly confused now with this loop current method and assigning the polarity.Could you please upload a separate video for this analysis alone would be better for clear understanding. Thank you.
In the loop 2 across .4 ohm resistor why did u change the +/- sign. it could be same as it was for loop one. you did that way for a problem in DC ckt example problem 2 in other video. I am confused.
Asescu, you're very welcome. I'm super glad the videos are helping even if it's a little bit. Your words are very encouraging so thank you so much. I try to do these review videos from a student perspective so let me know of any suggestions to improve in any way possible. Thank you :)
You could have used superposition to find the answers more easily.
When you were doing loop one, you went from the voltage source to the .25 ohm resistor. You assigned its value a negative because it was a voltage drop. That makes sense.
But when you were doing loop two, you went from the voltage source to the .4 ohm resistor (after changing the polarity). Why did you assigned its value as a positive when you made the value for the .25 ohm resistor negative?
John, it's because assigning a negative sign would affect the values inside the bracket so I choose to put positive sign so I can it won't affect my analysis of voltage drop/rise. The voltage drop only applies with the current I2 (hence negative sign) and not I1. With the .25 ohm, only current I1 passes thru that resistor as a voltage drop. You get it? Let me know...
Sir, Please teach me the loop current method if you got to know crack this method by now since i see that you posted this comment a year ago. Thank you.
I am clearly confused now with this loop current method and assigning the polarity.Could you please upload a separate video for this analysis alone would be better for clear understanding. Thank you.
Idk why but Indian accents give me ASMR so I get that while I learn
In the loop 2 across .4 ohm resistor why did u change the +/- sign. it could be same as it was for loop one. you did that way for a problem in DC ckt example problem 2 in other video. I am confused.