Parallel Resistors One Over Formula Program

python

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#1 satchell

satchell

Newbie

• Members
• 3 posts

• Calculators:
fx-CG50, fx-991EX

Posted 25 September 2022 - 06:38 PM

Wrote a python program for the Casio fx-CG50 to do the One Over method for parallel resistors. Basically you'd enter the R1 and R2 values and 0 for R3 to get the simple || result. Or enter up to R5...

If you need more resistors you can iterate on the if statements.

```from math import*

a = float(input("R1 Value = "))
b = float(input("R2 Value = "))
c = float(input("R3 Value = "))

if  c == 0:
firstOneOver = (1/a)+(1/b)
solution = (1/firstOneOver)
print(solution)
if c > 0:
d = float(input("R4 Value = "))
if d == 0:
secondOneOver = (1/a)+(1/b)+(1/c)
solution = (1/secondOneOver)
print(solution)
if d > 0:
e = float(input("R5 Value = "))
if e == 0:
thirdOneOver = (1/a)+(1/b)+(1/c)+(1/d)
solution = (1/thirdOneOver)
print(solution)
if e > 0:
fourthOneOver = (1/a)+(1/b)+(1/c)+(1/d)+(1/e)
solution = (1/fourthOneOver)
print(solution)```

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