Hi. your last problem has a bit of an issue. starting at 16:15 you say that "more resistance, less current" and that the 12R will get half as much current as the 6R because it's double the size. cool so far. Your very next calculation then incorrectly shows the opposite happening. Then when you go to solve that parallel resistor section in a different way, you end up saying that the 6R gets one third of an Amp and the 12R gets two thirds, which is also incorrect. It should read that the 6R gets 1/3A and the 12R gets 1/6A. the formula is I1 = (Rt/R1)*It and Rt for that section is 4ohms. so its ((4/6) *.5) = 1/3A and ((4/12)*.5)= 1/6A.
Here's two other's I have that touch on it- more lecture style, but the conceptual one will help out a lot with remembering how the laws work- ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-oH9w7jZt12c.htmlsi=7i8huLTXmtqGwh1m ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-tpheicgkW4E.htmlsi=vsmsMaerYWa0q615 Glad I could help! I'm going to try and make some more just straight practice videos soon!
Depends, what other information are you given? If you have current and resistance, use V=IR, if it’s a simple parallel, equal to battery voltage, anything more complex would need more steps
Depends on which type of circuit- in a parallel, typically you know its the same as the voltage of the battery (in a simple parallel), where in a series you typically solve for current first using the total resistance, then use V=IR, with the individual resistances to calculate the voltage drops.
Because the 12-ohm resistor is 2x the resistance of the 6-ohm resistance, it will get half as much current. Since the total current is 0.5A, if we divide it by 3 we can give the 6-ohm half the resistance (0.17 amps) and the 12-ohm twice that (0.33 amps). Basically, using the proportionality to get there. A better way would have been solving for the equivalent resistance (1/R= 1/12 + 1/6; R=4), then multiplying by the current (0.5 A* 4 ohms= 2 V) to get the voltage drop, then using V/R=I to get the individual currents. (2V/6= 0.33A and 2V/12= 0.17 A)
Maybe this video will be a good place to start? It's my attempt to explain circuit conceptually, so the math makes more sense. ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-oH9w7jZt12c.htmlsi=negipWxhGFTP_Ng7
