 |
 | Home | Contact Us | FAQ | Search & Site Map | Link to Us |
 | Sign In | Join | Other 45 Sites in Network |
| HomeDiscussion GroupsElectronicsBasicsRepairDesignCADComponentsEquipmentElectrical Engineering ElectronicsKB.com Contact UsLink To UsSearch & Site Map |
Electronics Forum / Basics / March 2006Newbie: Series/Parallel DC circuit current question
Hi, I'm trying to figure out the current through each parallel branch of a series/parallel circuit. I can't draw the circuit so I'll describe it. We have a 100v battery, the current leaves the neg terminal of the battery travels along a single path hits R1 5ohms, then hits junction A where it splits into 3 parallel branches containing one resistor of 30 ohms each, R2, R3, R4 then combines at junction B travels along a single path where it hits R5 15 ohms, continues along a single path where it hits R6 20 ohms then back to the pos side of the battery. Calculation of Rt: (R1) 5 ohms + Req(R2, R3, R4): 10 ohms ( 3 resistors of equal value 30/3 = 10) + (R5) 15 ohms + (R6) 20 ohms = 50 ohms. (5+10+15+20 = 50) Calculation of It through mains: 100v/50ohms = 2 Amps Now, the voltage drop of R1 is V=R*I or 5 ohms * 2 amps = 10 volts so R1 drops 10 volts. So, now at junction A we have 90v & still 2 amps right ? What is the current through each of the 30 ohm parallel branches ? Kirchoff's current law says whatever value of current enters a junction, the same value must leave the junction. So if 2 amps enters the junction at A, 2 amps must leave at junction B. But when I calculate the actual current through each branch using Ohms Law I get 3 amps each. 90v/30ohms of R2 = 3 amps across R2. Same calculation for the other 2 branches so 3 amps across R3 & 3 amps across R4. How can there be 2 amps entering, 2 amps leaving but 3 amps running along each branch in-between ? If I sum the 3 branches of 3 amps I'd get 9 amps. What am I doing wrong ? TIA J > 50 ohms. Good > Calculation of It through mains: 100v/50ohms = 2 Amps Good > R1 drops 10 volts. Good > right ? What is the current through each of the 30 ohm parallel > branches ? 0.6667 amp each > when I calculate the actual current through each branch using Ohms Law > I get 3 amps each. 90v/30ohms of R2 = 3 amps across R2. Same Not 90V. Read picture. Picture: http://img504.imageshack.us/img504/8222/drop117sz.jpg >> Picture: http://img504.imageshack.us/img504/8222/drop117sz.jpg The pic doesn't show how they come up with using 2/3. I can say 2 amps enters the junction, splits 3 equal ways so we have .666 amps across each branch but what about Ohms law ? If 90 volts is being applied at junction A, then each 30 ohm resistor sees the 90 volts and again 90v/30ohms is 3 amps ? If I can't use Ohms law there, when will I know to use it ? >>> Picture: http://img504.imageshack.us/img504/8222/drop117sz.jpg > [quoted text clipped - 4 lines] > 90v/30ohms is 3 amps ? If I can't use Ohms law there, when will I know > to use it ? Not 90V. The drop across the junction is 20V. >> Not 90V. The drop across the junction is 20V. can you tell me how you get the 20v ? you forgot about the drop from the R5 and R6. The parallel resistors don't drop the full 90v only 20v. Also you might want to look at the current divider principle, it explains why you have 2/3A for each resistor. >>> Not 90V. The drop across the junction is 20V. > > can you tell me how you get the 20v ? The resistance seen across the junction is 10 ohms. The voltage drop is therefore 2 amp * 10 ohms = 20 volts. >>The resistance seen across the junction is 10 ohms. The voltage drop is therefore 2 amp * 10 ohms = 20 volts. So there is no 90v anywhere ? The way I think of it now is, 100v source, hits R1 which drops 10v so after R1 the EMF is 90v, then it hits junction A of the parallel branch which "consumes" 20v of the 90v applied so the EMF leaving junction B is 70v and so on down the line till all EMF is consumed. First of all, is this correct ? and if so, we don't use the applied voltage but rather the consumed voltage through that part of the circuit ? >>>The resistance seen across the junction is 10 ohms. The voltage drop is therefore 2 amp * 10 ohms = 20 volts. > [quoted text clipped - 5 lines] > so, we don't use the applied voltage but rather the consumed voltage > through that part of the circuit ? Volts is not something that takes place at a point, as far as network analysis is concerned. Each component in the network has some voltage across it. The total of all the voltages around any loop has to add up to zero. So the supplied voltage has to equal the sum of the consumed voltages. 100 volts is applied across the whole network. Parts of that 100 volts appear across each of the components inside the network. Each component responds to the voltage difference between its ends. So the first resistor you mention has 100 volts on one side, and 90 on the other, so it sees the difference of these two, or 10 volts. The 3 parallel resistors each see the difference of the 90 volts on one side and 70 volts on the other side, or 20 volts difference., etc. >> Each component in the network has some voltage across it. I got it. I have to use the value of the voltage across that component or section not what's left over from other parts. That's where I went wrong. I have to use the 20v being used between junctions A & B or that "section" of the circuit then use Ohms law to find the value of current thru each branch. To make sure I have it, let's redo this circuit without R5 & R6. So if R5 & R6 didn't exist, we'd have just the 5ohm value of R1 then the 10ohm value of the parallel section of R2, R3, R4 so then Rt would be 15 ohms. And current would be 100v/15ohms = 6.666 amps. The voltage drop of R1 = 6.666 amps * 5 ohms = 33.33 volts "consumed" by R1. And the voltage drop of the parallel section of R2, R3 & R4 would be 6.666 amps * 10 ohms = 66.66 volts dropped or consumed by the parallel section. So since 6.666 amps flows into the parallel section at junction A, it must flow out at junction B. What is the current thru R2 you ask ? :) IR2 = 66.66volts / 30 ohms = 2.222. Multiply that by 3 since each branch is of equal resistive value and you have 6.666 amps total. Thanks I can stop banging my head against the wall...at least until I come to the AC section ! j >>> Not 90V. The drop across the junction is 20V. > > can you tell me how you get the 20v ? 2 amps through 10 ohms, which equates to 1/3 of 2 amps for each. And, of course, Ohm's law applies to R5 and R6, too. Look at the picture again: http://img504.imageshack.us/img504/8222/drop117sz.jpg Cheers! Rich >I can't draw the circuit so I'll describe it. > hdjim69 This is what most folks use: http://translate.google.com/translate?u=http://www.tech-chat.de/aacircuit.html+& langpair=de%7Cen (Word-for-word language translation is always entertaining.) >Hi, >I'm trying to figure out the current through each parallel branch of [quoted text clipped - 6 lines] >single path where it hits R6 20 ohms then back to the pos side of the >battery. R2 +--[30R]--+ A | | B R1 \ | R3 | / R5 R6 +--------[5R]-------+--[30R]--+----[15R]---[20R]---+ | | | | |-100V | R4 | | [BATTERY] +--[30R]--+ | | | | | +--------------------------------------------------+ >Calculation of Rt: (R1) 5 ohms + Req(R2, R3, R4): 10 ohms ( 3 resistors >of equal value 30/3 = 10) + (R5) 15 ohms + (R6) 20 ohms = 50 ohms. [quoted text clipped - 5 lines] >R1 drops 10 volts. So, now at junction A we have 90v & still 2 amps >right ? --- Right --- > What is the current through each of the 30 ohm parallel >branches ? Kirchoff's current law says whatever value of current >enters a junction, the same value must leave the junction. --- Forget about Kirchoff for a moment and think about how that current has to be shared in each of the branches. If you have 2 amperes being shared by three equally valued resistors in parallel, then each one must have one third of the total current going through it. --- >So if 2 >amps enters the junction at A, 2 amps must leave at junction B. But >when I calculate the actual current through each branch using Ohms Law >I get 3 amps each. 90v/30ohms of R2 = 3 amps across R2. --- That's wrong. If you have 2 amps in the circuit and the total resistance of R2, R3, and R4 is 10 ohms, then the voltage dropped across all of the resistors is E = IR = 2A * 10R = 20V, _not_ 90V. 90V is what's left for the rest of the circuit (R5 and R6) of which R1||R2||R3 only takes 20V. The current in any of the resistors is equal to the voltage dropped across it divided by its resistance: E 20V I = --- = ----- = 0.666... amperes R 30R >Same calculation for the other 2 branches so 3 amps across R3 & 3 amps >across R4. --- Same error! ;) --- >How can there be 2 amps entering, 2 amps leaving but 3 amps >running along each branch in-between ? --- There can't be. --- > If I sum the 3 branches of 3 >amps I'd get 9 amps. > >What am I doing wrong ? --- You forgot about R5 and R6.  SignatureJohn Fields Professional Circuit Designer |
Reply to this Message |
 Sign In
 Join
 My Latest Posts My Monitored Threads My Blog My Photo Gallery My Profile My Homepage Start New Thread Enable EMail Alerts Rate this Thread
|
©2009 Advenet LLC Privacy Policy - Terms of Use
This website includes both content owned or controlled by Advenet as well as content owned or controlled by third parties.