Sam Nichols Posted October 11, 2010 Report Share Posted October 11, 2010 Ok I'm completely lost here. I can deal with series and I can deal with parallel but I'm trying to work out how you get the total resistance, total current and the voltage drops for each resistor on a parallel-series circuit.. Here's what I've got.. Basically I think the total R is 99Ω and the total I is 1.01A. Is this right? If not, how would I get the correct answer. Also, how on earth do I work out the V-drops? Many thanks! Hopefully someone on here can help.. Quote Link to comment Share on other sites More sharing options...
Will Arnold Posted October 11, 2010 Report Share Posted October 11, 2010 where did you get 99 from? i believe it's 130 ohms? been ages since i've done circuits! Quote Link to comment Share on other sites More sharing options...
Robwalker Posted October 11, 2010 Report Share Posted October 11, 2010 (edited) Adding resistors in parallel is 1/(1/r1 + 1/r2 + 1/r....) - so on and so forth. Adding resistors in series is as simple as r1 + r2. Resistors added in parrallel will always give a value lower than the lowest resistor when added together. Break the circuit down into 2 parts. Parallel = 1/(1/50 + 1/70) = 29.16 ohms (this is now theoretically in series with the other two resistances) Series 29.16ohms + 30ohms + 40ohms =99.17ohms. For future reference, when there are only two resistors in parrallel you can use the equation 'product over sum'. As so - Rt = R1 x R2 / R1 + R2 Edited October 11, 2010 by Robwalker Quote Link to comment Share on other sites More sharing options...
Ryan Trott Posted October 11, 2010 Report Share Posted October 11, 2010 (edited) . Edited May 7, 2011 by Ryan Trott Quote Link to comment Share on other sites More sharing options...
JT! Posted October 11, 2010 Report Share Posted October 11, 2010 v/r(equiv) = (v/r1)+(v/r2) 100/r(equiv) = 100/50 + 100/70 100/r(equiv) = 2 + 1.43 100/r(equiv) = 3.43 100 = 3.43r(equiv) 29.15 = the resistance over the parallel section. Then add 40 and 30 to get 99.15 total Resistance.The voltage drops over parallel resistors are the same no matter what the resistor values are. So replace the parallel section with the 29.15 (round it up to 30). Seing as though the total resistance adds up to 100 and the voltage is 100 the voltage drops over the 30ohm resistor is 30v, and 40v over the 40ohm resistor. Both the resistors on the parallel section both have a v drop of 30v Quote Link to comment Share on other sites More sharing options...
Robwalker Posted October 11, 2010 Report Share Posted October 11, 2010 out of interest, why do you ask? Quote Link to comment Share on other sites More sharing options...
Sam Nichols Posted October 11, 2010 Author Report Share Posted October 11, 2010 Thanks guys, I get it now! Todays lecturer was pretty crap so I struggled to grasp it! Rob, I'm training to be an aircraft engineer, doing some early modules at the moment (all really basic stuff). Nailed the maths and physics but I've never done electronics before.Thanks for the voltage drop explanation too, JT. Quote Link to comment Share on other sites More sharing options...
Robwalker Posted October 12, 2010 Report Share Posted October 12, 2010 ah sweet man, all the best with that. Any other problems hit me up, im doing an apprenticeship and covered all the basics last year, kept all my notes and most of my assignments which may be of some help at some point. Quote Link to comment Share on other sites More sharing options...
Sam Nichols Posted October 13, 2010 Author Report Share Posted October 13, 2010 A - licence? Quote Link to comment Share on other sites More sharing options...
Robwalker Posted October 13, 2010 Report Share Posted October 13, 2010 say wha? Quote Link to comment Share on other sites More sharing options...
Sam Nichols Posted October 13, 2010 Author Report Share Posted October 13, 2010 What apprenticeship are you doing? Quote Link to comment Share on other sites More sharing options...
Robwalker Posted October 13, 2010 Report Share Posted October 13, 2010 advanced apprenticeship with e.on, looking to specialise in control and instrumentation. Im half way through my national certificate, then onto my foundation degree. If i do well enough at that e.on pay for me to go and complete my full degree at uni. result. Quote Link to comment Share on other sites More sharing options...
ellingtj Posted October 18, 2010 Report Share Posted October 18, 2010 Thanks guys, I get it now! Todays lecturer was pretty crap so I struggled to grasp it! Rob, I'm training to be an aircraft engineer, doing some early modules at the moment (all really basic stuff). Nailed the maths and physics but I've never done electronics before. Thanks for the voltage drop explanation too, JT. Engineer or Technician? Are you designing or repairing? Electronics is about following some fundamental rules - learn the basic rules and you'll be able get yourself round almost any issue (voltage and current laws). I'm assuming all aircraft systems are DC so that simplifies the electrical calcultions (AC power factor, lead, lag etc.). Quote Link to comment Share on other sites More sharing options...
Sam Nichols Posted October 18, 2010 Author Report Share Posted October 18, 2010 I'm doing a combined course. A degree in aeronautical engineering (design) and a B1 aircraft maintenance licence. So all going well at the end of the course I can choose either. Quote Link to comment Share on other sites More sharing options...
ellingtj Posted October 18, 2010 Report Share Posted October 18, 2010 I'm doing a combined course. A degree in aeronautical engineering (design) and a B1 aircraft maintenance licence. So all going well at the end of the course I can choose either. Cool, having worked in engineering for years (R&D and design, though I do a totally different job now), it seemed to be, heavy electrical side anyway, the maintenance technicians that seemed to get the better pay working as contractors than the actual engineers. Not sure how it is in the aeronautical arena but worth looking into/considering when the time comes to decide. My first degree was in electrical and electronic engineering (also trained as an electriician whilst doing my PhD - C&G 2330) - any electrical questions feel free to PM me. Dr. Tim (BEng, PhD) Quote Link to comment Share on other sites More sharing options...
David Posted October 20, 2010 Report Share Posted October 20, 2010 (edited) I'm assuming all aircraft systems are DC so that simplifies the electrical calcultions (AC power factor, lead, lag etc.). That's not always the case although most do run from rectified AC. The airframe will supply a 115V (nominal) AC supply for use by all the on-board electronics. Using Ohms law and other calculations comes a lot easier with time and practice. Try not to get too bogged down with the maths though. On a day-to-day basis you won't be using much more than a few simple formula's. It is more important to understand how and why they are used. Dave (Electronics Engineer - Aerospace) Edited October 20, 2010 by David Quote Link to comment Share on other sites More sharing options...
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