Simple parallel circuit problem
08-25-2015, 12:14 AM (This post was last modified: 08-25-2015 12:15 AM by Gerson W. Barbosa.)
Post: #1
 Gerson W. Barbosa Senior Member Posts: 1,429 Joined: Dec 2013
Simple parallel circuit problem

R0 = 2048 Ω
R1 = R2 = R3 = ... = R512 = 163 Ω
Consider all components are ideal.
What should A, an 8-digit digital ammeter, read?
08-25-2015, 12:52 AM (This post was last modified: 08-25-2015 12:54 AM by Dave Frederickson.)
Post: #2
 Dave Frederickson Senior Member Posts: 2,114 Joined: Dec 2013
RE: Simple parallel circuit problem
(08-25-2015 12:14 AM)Gerson W. Barbosa Wrote:

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

A = 1/(163/512 // 2048) = 3.1415926 amps
08-25-2015, 01:00 AM
Post: #3
 Gerson W. Barbosa Senior Member Posts: 1,429 Joined: Dec 2013
RE: Simple parallel circuit problem
(08-25-2015 12:52 AM)Dave Frederickson Wrote:  A = 1/(163/512 // 2048) = 3.1415926 amps

That's correct!
08-25-2015, 02:21 AM (This post was last modified: 08-25-2015 02:25 AM by Steve Simpkin.)
Post: #4
 Steve Simpkin Senior Member Posts: 633 Joined: Dec 2013
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
08-25-2015, 03:20 AM (This post was last modified: 08-25-2015 03:46 AM by Gerson W. Barbosa.)
Post: #5
 Gerson W. Barbosa Senior Member Posts: 1,429 Joined: Dec 2013
RE: Simple parallel circuit problem
(08-25-2015 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 :-)
08-25-2015, 02:23 PM
Post: #6
 rprosperi Senior Member Posts: 5,069 Joined: Dec 2013
RE: Simple parallel circuit problem
(08-25-2015 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
08-25-2015, 02:59 PM
Post: #7
 Gerson W. Barbosa Senior Member Posts: 1,429 Joined: Dec 2013
RE: Simple parallel circuit problem
(08-25-2015 02:23 PM)rprosperi Wrote:
(08-25-2015 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...

Same as || on the WP 34S:

a//b = (a*b)/(a + b)

Gerson.
08-25-2015, 07:56 PM
Post: #8
 rprosperi Senior Member Posts: 5,069 Joined: Dec 2013
RE: Simple parallel circuit problem
(08-25-2015 02:59 PM)Gerson W. Barbosa Wrote:  Same as || on the WP 34S:

a//b = (a*b)/(a + b)

Gerson.

Thanks Gerson!

--Bob Prosperi
08-25-2015, 11:04 PM
Post: #9
 Brad Barton Member Posts: 189 Joined: Jan 2014
RE: Simple parallel circuit problem
(08-25-2015 02:59 PM)Gerson W. Barbosa Wrote:
(08-25-2015 02:23 PM)rprosperi Wrote:  What is the "//" symbol? Some of us aren't EE types...

Same as || on the WP 34S:

a//b = (a*b)/(a + b)

Gerson.

Haha, that makes me the dumb one. I thought it was a typo.
08-25-2015, 11:07 PM (This post was last modified: 08-25-2015 11:08 PM by Garth Wilson.)
Post: #10
 Garth Wilson Senior Member Posts: 479 Joined: Dec 2013
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 HP-41 programming):

Code:
100   \ First resistor value 1/X 330   \ Second resistor value 1/X + 910   \ Third resistor value 1/X + [...] \ Go as far as you want. 1/X   \ Turn the sum of the conductances into resistance.

Doing it in complex numbers, it works for admittance too, ie, where there may be capacitive or inductive reactances.

08-25-2015, 11:26 PM
Post: #11
 Dave Frederickson Senior Member Posts: 2,114 Joined: Dec 2013
RE: Simple parallel circuit problem
(08-25-2015 02:23 PM)rprosperi Wrote:
(08-25-2015 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...

// may be read as "in parallel with".
08-27-2015, 02:48 AM
Post: #12
 Gerson W. Barbosa Senior Member Posts: 1,429 Joined: Dec 2013
RE: Simple parallel circuit problem
(08-25-2015 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:

a // b = 1/ (1/a + 1/b)

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.
08-27-2015, 03:18 PM
Post: #13
 Gerson W. Barbosa Senior Member Posts: 1,429 Joined: Dec 2013
RE: Simple parallel circuit problem
Same theoretical reading using standard E24 resistors:

08-28-2015, 03:23 AM
Post: #14
 Garth Wilson Senior Member Posts: 479 Joined: Dec 2013
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 ENTER 15060 1/X 183 1/X + 24 1/X + 1 + 1/X    \ Not quite the most efficient, /      \ but just to show what's done.

08-28-2015, 04:42 AM (This post was last modified: 08-28-2015 07:31 PM by Dave Frederickson.)
Post: #15
 Dave Frederickson Senior Member Posts: 2,114 Joined: Dec 2013
RE: Simple parallel circuit problem

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.
08-28-2015, 05:31 AM (This post was last modified: 08-28-2015 05:35 AM by Gerson W. Barbosa.)
Post: #16
 Gerson W. Barbosa Senior Member Posts: 1,429 Joined: Dec 2013
RE: Simple parallel circuit problem
(08-28-2015 03:23 AM)Garth Wilson Wrote:  I don't know what you mean by "E24" resistors, but we have:
3 / (15.06K // 183 // 1 // 24) = Pi A
Code:
3 ENTER 15060 1/X 183 1/X + 24 1/X + 1 + 1/X    \ Not quite the most efficient, /      \ but just to show what's done.

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
08-28-2015, 05:52 AM
Post: #17
 Gerson W. Barbosa Senior Member Posts: 1,429 Joined: Dec 2013
RE: Simple parallel circuit problem
(08-28-2015 04:42 AM)Dave Frederickson Wrote:  Edit: I just realized I collect vintage TI calculators, too.

Only a TI-57 here, but I had a TI-51-III in 1982, replaced three months later by a TI-59 due to a faulty keyboard. The TI-59 worked flawlessly for about six months before eventually being stolen. The replacement HP-15C is still working to date :-)
08-28-2015, 01:47 PM (This post was last modified: 08-28-2015 01:53 PM by Dave Frederickson.)
Post: #18
 Dave Frederickson Senior Member Posts: 2,114 Joined: Dec 2013
RE: Simple parallel circuit problem
(08-28-2015 05:52 AM)Gerson W. Barbosa Wrote:
(08-28-2015 04:42 AM)Dave Frederickson Wrote:  Edit: I just realized I collect vintage TI calculators, too.

Only a TI-57 here, but I had a TI-51-III in 1982, replaced three months later by a TI-59 due to a faulty keyboard. The TI-59 worked flawlessly for about six months before eventually being stolen. The replacement HP-15C is still working to date :-)

I suspect others missed the joke, too. My calculators are the old Excel spreadsheet resistor calculators written by a guy at TI.

Dave
08-28-2015, 05:33 PM (This post was last modified: 08-28-2015 05:36 PM by Gerson W. Barbosa.)
Post: #19
 Gerson W. Barbosa Senior Member Posts: 1,429 Joined: Dec 2013
RE: Simple parallel circuit problem
(08-28-2015 01:47 PM)Dave Frederickson Wrote:
(08-28-2015 05:52 AM)Gerson W. Barbosa Wrote:  Only a TI-57 here, but I had a TI-51-III in 1982, replaced three months later by a TI-59 due to a faulty keyboard. The TI-59 worked flawlessly for about six months before eventually being stolen. The replacement HP-15C is still working to date :-)

I suspect others missed the joke, too. My calculators are the old Excel spreadsheet resistor calculators written by a guy at TI.

Quote:PARALLEL RESISTOR CALCULATOR, Version 1

By Bruce Carter, Texas Instruments Applications

This utility solves a common engineering problem - how to reach a target
resistance with two resistors in parallel. It assumes that you are free to choose
one of the two resistors that will be put in parallel - called the "seed" resistor.
The spreadsheet will then choose the closest resistor that, in parallel with
the seed resistor, will produce the target resistance. You have the choice of
using 1% resistors, or 5%.

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 WP-34S. Thanks, Walter!
11-04-2015, 01:57 AM (This post was last modified: 11-04-2015 02:16 AM by Vtile.)
Post: #20
 Vtile Senior Member Posts: 404 Joined: Oct 2015
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:
 << INV SWAP INV SWAP + INV >>

For imaginary circuit
Code:
         Z1               -+--####--+------------+   |              |                 |   #             #                #   #  Z4        #  Z3           #  Z2           #             #                #   |              |                 | -+----------+------------+

First enter two first parallel impedance values (in complex, s-domain, 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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