deathbyampere Posted February 7, 2005 Report Share Posted February 7, 2005 My apologies for a second post concerning dividers, but I didn't want to detract from the information being given to that other fellow. Also, please pardon my rather morbid profile; it seems to be true, but I will change it if anyone finds it offensive.Now, to the heart of the matter. I'm taking an electronics course and now I am at a loss. At present, the course requires that I read a simple schematic, calculate for the missing or needed value(s), build that circuit, and finally measure it. My calculations never match the measurements (if the circuit works at all). I couldn't find a good schematic for this, so pardon the use of random symbols:_____________________________| | || R1 || | ____________ |bat | | || R2 R5 R6| | | || |_______ | |gnd | | | | | | | | R3 R4 | | | | | | | | | | gnd gnd gnd gndWhere R1 = 30 ohmsR2 = 25 ohmsR3 = 10 ohmsR4 = 40 ohmsR5 = 20 ohmsR6 = 10 ohmsV(bat) = 9V To figure the total resistance, am I correct in doing the following:R3 x R4 / R3 + R4 = Re1...Re1 + R2 = Re2...Re2 x R5 / Re2 + R5 = Re3...Re3 + R1 = Re4Re4 x R6 / Re4 + R6 = Rtor Re1 = 8Re2 = 33Re3 = 14.2Re4 = 44.2Rt = 8.1____________Now, if that is right, the current should be appx. 1.125A, yes?____________Here's where I get more lost. As I try to figure the voltage drops, would I figure R1 and R6 separately (resulting in an untenable answer) or figuring that drop for R1-R6 as a parallel branch? If so, how do I find the voltage drops for R1 and R6 respectively, and on down the line?My apologies for this pedantic, long-winded, and ridiculous question. If anyone has articles that would be helpful, I'd love to see them. As it is, I'm probably going to pull my hair out and certainly going to electrocute myself at this rate.Sincerely. Quote Link to comment Share on other sites More sharing options...
trigger Posted February 7, 2005 Report Share Posted February 7, 2005 Just give you some hints:1. Try to make use of the ohm's law.2. Do not calculate the total current in only one equation.3. Try to divide the whole cirucit in parts and calculate their voltages and currents.4. Note that at R6, the voltage across it is the same as the battery voltage. Other parts will be more or less the same.5. Total current = current flow to R6 + current flow to R1 (and the branches)(try to name all node with voltages and currents so that will let you easily get the answer) Quote Link to comment Share on other sites More sharing options...
yousef_ob Posted March 18, 2005 Report Share Posted March 18, 2005 HELLO,YOU CAN USE THIS METHOD IN FINDING THE TOTAL RESISTANCE SEEN BY A VOLTAGE SOURCE.1. AT THE TERMINALS OF THE VOLTAGE SOURCE NAME (a,b), THEN WHEN U FIND ANY RESISTANCE NAME ©, THEN GO ON d,e AND SO ON.2. REDRAW THE CCT IN A NEAT FORM BY PUTTING THE LETTERS (a,b,c,d...) THEN SEE WHERE IS THE SERIES AND WHERE IS THE PARALLEL CONFIGURATIONS AND SIMPLFY THE CCT.3. DO ANY DELTA-STAR OR STAR-DELTA TRANSFORMATIONS.SAY THANX TO ME Quote Link to comment Share on other sites More sharing options...
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