(12C) Resistors in Parallel

10252018, 10:18 AM
(This post was last modified: 10252018 10:26 AM by Gamo.)
Post: #1




(12C) Resistors in Parallel
Program that solve up to "Four Pair" of unequal value of "Resistors in Parallel"
and little addon 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:
Gamo 

10252018, 12:50 PM
Post: #2




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? 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 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 

10252018, 02:57 PM
Post: #3




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 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: \(R_1  \frac{R_1^2}{R_1 + R_2} = \frac{R_1 R_2}{R_1 + R_2}\) 

10262018, 05:01 AM
Post: #4




RE: (12C) Resistors in Parallel
Thanks to Albert and Thomas
Many good varieties of ways to solve this type of problem. Gamo 

10282018, 09:23 AM
Post: #5




RE: (12C) Resistors in Parallel
(10252018 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 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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