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(12C) Resistors in Parallel - Gamo - 10-25-2018 10:18 AM
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:
Gamo RE: (12C) Resistors in Parallel - Albert Chan - 10-25-2018 12:50 PM
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 RE: (12C) Resistors in Parallel - Thomas Klemm - 10-25-2018 02:57 PM
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: \(R_1 - \frac{R_1^2}{R_1 + R_2} = \frac{R_1 R_2}{R_1 + R_2}\) RE: (12C) Resistors in Parallel - Gamo - 10-26-2018 05:01 AM
Thanks to Albert and Thomas Many good varieties of ways to solve this type of problem. Gamo RE: (12C) Resistors in Parallel - Dieter - 10-28-2018 09:23 AM
(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` 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 |