Guest Posted May 5 Report Posted May 5 Okay, so I have a circuit. I've previously solved said circuit but I have forgotten how! So I have the overall current of this circuit. 1.1 Amps, but I forgot how to solve parallel circuits specifically 10,15,20 Ohms here, I HAVE THE ANSWER BUT IDK HOW THE FUCK I GOT IT 10 --> .5 Amps 15---> .33 Amps 20---> .25 Amps Same for 30 & 35 30--> .59 Amps 35 --> .507 Amps If anyone knows how I got to these numbers I will greatly appreciate it. Dont fucking delete this milo this has a faster response rate than chegg tbh Quote
Guest Posted May 5 Report Posted May 5 20 minutes ago, HydraPOX said: https://en.wikipedia.org/wiki/Current_divider# this man solved it he gave the equation to do so in steam Everyone a round of applause for @ HydraPOX Quote
Doc 963 Posted May 5 Report Posted May 5 Consider the following for parallel and series circuits: Series: Total V = V1 + V2 + V3 (aka the total voltage is the sum of each part) Total A = A1 = A2 + A3 (aka the total current remains the same between each part) Total R = R1 + R2 + R3 (same as voltage) Parallel: Total V = V1 = V2 = V3 Total A = A1 + A2 + A3 1 / Total R = 1/R1 + 1/R2 ... Extra Formulas: P=IV (aka Power = Current * Voltage) V=IR (aka Voltage = Current * Resistance) For this specific circuit the parallel parts can be combined into one big part allowing the whole thing to be considered as a series circuit. Going clockwise from the power supply. 1) R = 5 2) Combined Parallel Parts - Using resistance rules for parallel circuits we can determine that the total resistance is 1 / R = 1 / 10 + 1 / 15 + 1 / 20 thus R = 4.615 3) R = 25 4) Combine Parallel Parts - Same as 2 - 1 / R = 1 / 30 + 1/35 thus R = 16.154 5) R = 40 Because we can combine the parallel parts into one big one and consider the whole circuit as a series we can determine the total resistance for the whole: R = 90.769 .... 1 minute ago, Moonini said: this man solved it he gave the equation to do so in steam Everyone a round of applause for @ HydraPOX Welp I got too distracted with my league game so someone beat me to it lol 1 Quote
HydraPOX 107 Posted May 5 Report Posted May 5 2 minutes ago, Moonini said: Everyone a round of applause for @ HydraPOX Happy to be of service Quote
Tman15tmb 1940 Posted May 5 Report Posted May 5 Can someone point me to the retard area. I seemed to have stumbled into a room full of smart people. 1 1 Quote
Event Admin Mighty 1556 Posted May 5 Event Admin Report Posted May 5 6 hours ago, Doc said: Consider the following for parallel and series circuits: Series: Total V = V1 + V2 + V3 (aka the total voltage is the sum of each part) Total A = A1 = A2 + A3 (aka the total current remains the same between each part) Total R = R1 + R2 + R3 (same as voltage) Parallel: Total V = V1 = V2 = V3 Total A = A1 + A2 + A3 1 / Total R = 1/R1 + 1/R2 ... Extra Formulas: P=IV (aka Power = Current * Voltage) V=IR (aka Voltage = Current * Resistance) For this specific circuit the parallel parts can be combined into one big part allowing the whole thing to be considered as a series circuit. Going clockwise from the power supply. 1) R = 5 2) Combined Parallel Parts - Using resistance rules for parallel circuits we can determine that the total resistance is 1 / R = 1 / 10 + 1 / 15 + 1 / 20 thus R = 4.615 3) R = 25 4) Combine Parallel Parts - Same as 2 - 1 / R = 1 / 30 + 1/35 thus R = 16.154 5) R = 40 Because we can combine the parallel parts into one big one and consider the whole circuit as a series we can determine the total resistance for the whole: R = 90.769 .... Welp I got too distracted with my league game so someone beat me to it lol 1 Quote
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