Simple parallel circuit problem

Simple parallel circuit problem - Printable Version

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Simple parallel circuit problem - Gerson W. Barbosa - 08-25-2015 12:14 AM

[Image: circuit_zpsfvs5qrsy.png]

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

RE: Simple parallel circuit problem - Dave Frederickson - 08-25-2015 12:52 AM

(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

RE: Simple parallel circuit problem - Gerson W. Barbosa - 08-25-2015 01:00 AM

(08-25-2015 12:52 AM)Dave Frederickson Wrote: A = 1/(163/512 // 2048) = 3.1415926 amps

That's correct!

RE: Simple parallel circuit problem - Steve Simpkin - 08-25-2015 02:21 AM

That was fun! Well not as much fun as typing "58008" on an LED calculator and reading it upside down, but fun neverless Smile

RE: Simple parallel circuit problem - Gerson W. Barbosa - 08-25-2015 03:20 AM

(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 :-)

RE: Simple parallel circuit problem - rprosperi - 08-25-2015 02:23 PM

(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...

RE: Simple parallel circuit problem - Gerson W. Barbosa - 08-25-2015 02:59 PM

(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.

RE: Simple parallel circuit problem - rprosperi - 08-25-2015 07:56 PM

(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!

RE: Simple parallel circuit problem - Brad Barton - 08-25-2015 11:04 PM

(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.

RE: Simple parallel circuit problem - Garth Wilson - 08-25-2015 11:07 PM

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.

RE: Simple parallel circuit problem - Dave Frederickson - 08-25-2015 11:26 PM

(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".

RE: Simple parallel circuit problem - Gerson W. Barbosa - 08-27-2015 02:48 AM

(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.

RE: Simple parallel circuit problem - Gerson W. Barbosa - 08-27-2015 03:18 PM

Same theoretical reading using standard E24 resistors:

[Image: circuit2_zpsjjxopivv.png]

RE: Simple parallel circuit problem - Garth Wilson - 08-28-2015 03:23 AM

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.

RE: Simple parallel circuit problem - Dave Frederickson - 08-28-2015 04:42 AM

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. Confused

RE: Simple parallel circuit problem - Gerson W. Barbosa - 08-28-2015 05:31 AM

(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=Egyptian+fraction+2718281828%2F1000000000

http://www.wolframalpha.com/input/?i=Egyptian+fraction+3141592654%2F1000000000

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/resistor_values.html

RE: Simple parallel circuit problem - Gerson W. Barbosa - 08-28-2015 05:52 AM

(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 :-)

RE: Simple parallel circuit problem - Dave Frederickson - 08-28-2015 01:47 PM

(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. Smile

Dave

RE: Simple parallel circuit problem - Gerson W. Barbosa - 08-28-2015 05:33 PM

(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!

RE: Simple parallel circuit problem - Vtile - 11-04-2015 01:57 AM

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