Simple parallel circuit problem

08252015, 12:14 AM
(This post was last modified: 08252015 12:15 AM by Gerson W. Barbosa.)
Post: #1




Simple parallel circuit problem
R0 = 2048 Ω R1 = R2 = R3 = ... = R512 = 163 Ω Consider all components are ideal. What should A, an 8digit digital ammeter, read? 

08252015, 12:52 AM
(This post was last modified: 08252015 12:54 AM by Dave Frederickson.)
Post: #2




RE: Simple parallel circuit problem  
08252015, 01:00 AM
Post: #3




RE: Simple parallel circuit problem  
08252015, 02:21 AM
(This post was last modified: 08252015 02:25 AM by Steve Simpkin.)
Post: #4




RE: Simple parallel circuit problem
That was fun! Well not as much fun as typing "58008" on an LED calculator and reading it upside down, but fun neverless


08252015, 03:20 AM
(This post was last modified: 08252015 03:46 AM by Gerson W. Barbosa.)
Post: #5




RE: Simple parallel circuit problem
(08252015 02:21 AM)Steve Simpkin Wrote: That was fun! Well not as much fun as typing "58008" on an LED calculator and reading it upside down, but fun neverless Starting with 163*π ~ 512, I came up with \[\frac{163+2^{20}}{163\times 2^{11}}\] The rational approximation is interesting perhaps, as it involves two instances of the last Heegner number and two powers of two, but it is only good to eight places. So I thought the parallel circuit would be a little more interesting :) 

08252015, 02:23 PM
Post: #6




RE: Simple parallel circuit problem
(08252015 12:52 AM)Dave Frederickson Wrote: A = 1/(163/512 // 2048) = 3.1415926 amps What is the "//" symbol? Some of us aren't EE types... Bob Prosperi 

08252015, 02:59 PM
Post: #7




RE: Simple parallel circuit problem  
08252015, 07:56 PM
Post: #8




RE: Simple parallel circuit problem
(08252015 02:59 PM)Gerson W. Barbosa Wrote: Same as  on the WP 34S: Thanks Gerson! Bob Prosperi 

08252015, 11:04 PM
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RE: Simple parallel circuit problem  
08252015, 11:07 PM
(This post was last modified: 08252015 11:08 PM by Garth Wilson.)
Post: #10




RE: Simple parallel circuit problem
Often overlooked, and I don't know why since this is what's happening: Add the conductances, then invert again to get resistance:
a // b = 1/ (1/a + 1/b) To do several at once do for example (in HP41 programming): Code: 100 \ First resistor value Doing it in complex numbers, it works for admittance too, ie, where there may be capacitive or inductive reactances. http://WilsonMinesCo.com (Lots of HP41 links at the bottom of the links page, http://wilsonminesco.com/links.html ) 

08252015, 11:26 PM
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RE: Simple parallel circuit problem  
08272015, 02:48 AM
Post: #12




RE: Simple parallel circuit problem
(08252015 11:07 PM)Garth Wilson Wrote: Often overlooked, and I don't know why since this is what's happening: Add the conductances, then invert again to get resistance: I think the alternative form a // b = a*b/(a + b) was an easier option long ago, when these calculations were done by hand. Dividing 1 by a large number (more than two significant digits) was a cumbersome calculation to do manually, I remember. For this particular problem, i = 1/2048 + 512/163 might be better. Gerson. 

08272015, 03:18 PM
Post: #13




RE: Simple parallel circuit problem
Same theoretical reading using standard E24 resistors:


08282015, 03:23 AM
Post: #14




RE: Simple parallel circuit problem
I don't know what you mean by "E24" resistors, but we have:
3 / (15.06K // 183 // 1 // 24) = Pi A Code: 3 http://WilsonMinesCo.com (Lots of HP41 links at the bottom of the links page, http://wilsonminesco.com/links.html ) 

08282015, 04:42 AM
(This post was last modified: 08282015 07:31 PM by Dave Frederickson.)
Post: #15




RE: Simple parallel circuit problem
For your amusement.
Parallel Resistor Calculator in Excel Voltage Divider Calculator in Excel The calculators use the E24 and E96 Standard EIA Decade Resistor Values. Courtesy of Texas Instruments Edit: I just realized I collect vintage TI calculators, too. 

08282015, 05:31 AM
(This post was last modified: 08282015 05:35 AM by Gerson W. Barbosa.)
Post: #16




RE: Simple parallel circuit problem
(08282015 03:23 AM)Garth Wilson Wrote: I don't know what you mean by "E24" resistors, but we have: Likewise, 2 ENTER 20000 1/x 110 1/x + 10 1/x + 4 1/x + 1 + 1/x / = e A. Parallel resistors values courtesy of Wolfram Alpha: http://www.wolframalpha.com/input/?i=Egy...1000000000 http://www.wolframalpha.com/input/?i=Egy...1000000000 I should have written "E24 series resistors", but since this involves resistors in parallel I thought this might be a bit confusing. Those E24 resistors are easily available, but I guess no one would want to build a physical circuit. Anyway, 5% tolerance would be too much for this application :) http://www.logwell.com/tech/components/r...alues.html 

08282015, 05:52 AM
Post: #17




RE: Simple parallel circuit problem
(08282015 04:42 AM)Dave Frederickson Wrote: Edit: I just realized I collect vintage TI calculators, too. Only a TI57 here, but I had a TI51III in 1982, replaced three months later by a TI59 due to a faulty keyboard. The TI59 worked flawlessly for about six months before eventually being stolen. The replacement HP15C is still working to date :) 

08282015, 01:47 PM
(This post was last modified: 08282015 01:53 PM by Dave Frederickson.)
Post: #18




RE: Simple parallel circuit problem
(08282015 05:52 AM)Gerson W. Barbosa Wrote:(08282015 04:42 AM)Dave Frederickson Wrote: Edit: I just realized I collect vintage TI calculators, too. I suspect others missed the joke, too. My calculators are the old Excel spreadsheet resistor calculators written by a guy at TI. Dave 

08282015, 05:33 PM
(This post was last modified: 08282015 05:36 PM by Gerson W. Barbosa.)
Post: #19




RE: Simple parallel circuit problem
(08282015 01:47 PM)Dave Frederickson Wrote:(08282015 05:52 AM)Gerson W. Barbosa Wrote: Only a TI57 here, but I had a TI51III in 1982, replaced three months later by a TI59 due to a faulty keyboard. The TI59 worked flawlessly for about six months before eventually being stolen. The replacement HP15C is still working to date :) Quote:PARALLEL RESISTOR CALCULATOR, Version 1 I did follow the links and noticed the TI reference. I even wondered about the connection to TI calculators. Sarcasm sign needed here sometimes :) I usually solve the first problem on the WP 34S by following this procedure: 1000 ENTER 50 +/ g  > 52.6315789474 Then I manually select one from the tables (or from my spare parts box), closest positive value, of course :) 1000 ENTER 52.3 g  > ~ 49.7 ohms. Gerson. P.S.: I love the  function on the WP34S. Thanks, Walter! 

11042015, 01:57 AM
(This post was last modified: 11042015 02:16 AM by Vtile.)
Post: #20




RE: Simple parallel circuit problem
Here is simple program (I just typed) in UserRPL to make it easy with 28..50g. The original idea is from L.R.Linares (He have done really nice series of HP49/HP50g instruction videos to youtube), but I'm not sure if this is 1:1 his solution for this job or were it even Linares original idea at first place.
I named it Parallel Impedance, Pimp for short. Code:
For imaginary circuit Code:
First enter two first parallel impedance values (in complex, sdomain, resistance form) in input line like: Z2 [SPaCe] Z3 Pimp => 1/(1/Z2)+(1/Z3) Z1 + => (1/(1/Z2)+(1/Z3))+Z1 Z4 Pimp => 1/(1/((1/(1/Z2)+(1/Z3))+Z1))+(1/Z4) < Parentheses look out.. That is it, now only should one use >num etc. to get decimal form. Middle of night, LV 

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