(12C) Resistors in Parallel
10-25-2018, 10:18 AM (This post was last modified: 10-25-2018 10:26 AM by Gamo.)
Post: #1
 Gamo Senior Member Posts: 713 Joined: Dec 2016
(12C) Resistors in Parallel
Program that solve up to "Four Pair" of unequal value of "Resistors in Parallel"
and little add-on program that solve for one unknown out of two parallel resistors from your desire Total.

Procedure:

1st Program:
All resistors must be "unequal" in value.
Store each value from R0 to R7
[R/S]

2nd Program
Store R0 for known Total
Store R1 for one of the two known resistor.
To run first goto line 40
[R/S]

----------------------------------------------------

Example:

Program 1: FIX 4

What is the total with 4 resistors in parallel?
50, 65, 120 and 250

50 [STO] 0
65 [STO] 1
120 [STO] 2
250 [STO] 3
[R/S] 20.9565

------------------------------------------------------

Program 2: FIX 4

1st resistor is 50 and my desire total is 28.2608
What is the 2nd resistor value?

28.2608 [STO] 0
50 [STO] 1
[GTO] 40
[R/S] 64.9996 = 65

Program:
Code:
 01 RCL 0 02 1/x 03 RCL 1 04 1/x 05  + 06 RCL 2 07 X=0 08 GTO 37 09 1/x 10  + 11 RCL 3 12 X=0 13 GTO 37 14 1/x 15  + 16 RCL 4 17 X=0 18 GTO 37 19 1/x 20  + 21 RCL 5 22 X=0 23 GTO 37 24 1/x 25  + 26 RCL 6 27 X=0 28 GTO 37 29 1/x 30  + 31 RCL 7 32 X=0 33 GTO 37 34 1/x 35  + 36 GTO 38 37 X<>Y 38 1/x 39 GTO 00 -------------------------------------- 40 RCL 0 41 RCL 1 42  x 43 RCL 1 44 RCL 0 45  - 46  ÷

Gamo
10-25-2018, 12:50 PM
Post: #2
 Albert Chan Senior Member Posts: 1,983 Joined: Jul 2018
RE: (12C) Resistors in Parallel
Hi Gamo

Dieter had a better program, with more accuracy by doing R = R1 R2 / (R1 + R2)

Quote:What is the total with 4 resistors in parallel?
50, 65, 120 and 250

There is no need for more code. Just keep running each pair in parallel.

Parallel of 50, 65, R = 50*65/(50+65) = 28.261
"Add" 120, R = 120*28.261/(120+28.261) = 22.874
"Add" 250, R = 250*22.874/(250+22.874) = 20.957 (total of 4 resister in parallel)

Quote:1st resistor is 50 and my desire total is 28.2608
What is the 2nd resistor value?

Again, no need for more code. Remove a resister from total R, just negate the resister.

R1 = -50*28.2608/(-50+28.2608) = 65.000
10-25-2018, 02:57 PM
Post: #3
 Thomas Klemm Senior Member Posts: 1,767 Joined: Dec 2013
RE: (12C) Resistors in Parallel
Or then we can use the method described by Sam Levy in his article EE calculations, some not obvious:
Quote:Resistive Dividers
When calculating from known resistors, enter the lower resistor repeatedly.
• Key in the top resistor, and add, the result is the total resistance,
• then divide, the result is the fractional output voltage of the divider.
• Then multiply and subtract to see the output resistance from the divider tap.

This leaves us with a program that consists of just the 4 arithmetic operators:
Code:
+ ÷ × -

Example:

50 ENTER ENTER ENTER
65 R/S
28.2609

Kind regards
Thomas

This method uses:

$$R_1 - \frac{R_1^2}{R_1 + R_2} = \frac{R_1 R_2}{R_1 + R_2}$$
10-26-2018, 05:01 AM
Post: #4
 Gamo Senior Member Posts: 713 Joined: Dec 2016
RE: (12C) Resistors in Parallel
Thanks to Albert and Thomas

Many good varieties of ways to solve this type of problem.

Gamo
10-28-2018, 09:23 AM
Post: #5
 Dieter Senior Member Posts: 2,397 Joined: Dec 2013
RE: (12C) Resistors in Parallel
(10-25-2018 12:50 PM)Albert Chan Wrote:  Dieter had a better program, with more accuracy by doing R = R1 R2 / (R1 + R2)

Here is how you can do it on the 12C:

Code:
01 ENTER 02 ENTER 03 R↓ 04 R↓ 05 X<>Y 06 x 07 X<>Y 08 LstX 09 + 10 ÷

As already pointed out, this is all you need for both calculations:

Add up parallel resistors: enter resistor values, one ofter another, and press [R/S]
Simply think of [R/S] as an "add resistor" key.

Find the required resistor for a given total: enter resistors as before, finally enter the desired total and press [CHS] [R/S] [CHS]

Example 1:
Two, three, four or more resistors in parallel.

50 [ENTER] 65 [R/S] => 28,26
120 [R/S] => 22,87
250 [R/S] => 20,96

Example 2:
Find the missing resistor.
Assume you have 100 and 220 Ω, and a total of 50 Ω is required.

100 [ENTER] 220 [R/S] => 68,75
50 [CHS] [R/S] [CHS] => 183,33

Or for Gamo's example:
50 [ENTER] 28,2608 [CHS] [R/S] [CHS] => 65

Dieter
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