+- HP Forums (https://www.hpmuseum.org/forum)
+-- Forum: HP Software Libraries (/forum-10.html)
+--- Forum: General Software Library (/forum-13.html)
+--- Thread: (15C) Parallel Resistance, two resistors (/thread-20376.html)
(15C) Parallel Resistance, two resistors - SlideRule - 08-23-2023 11:52 PM
An excerpt from EDN magazine, HP-15 program needs fewest strokes, Jon Vicklund, 11/OCT/1990, page 242.
"The program in LISTING 1 takes advantage of a Hewlett-Packard RPN calculator's stack operators to calculate the parallel resistance of two resistors using a minimum number of keystrokes.
Listing 1―Minimal-keystroke parallel-resistance
calculation for HP-15 calculator
R1 ; R1 value
ENTER↑ ENTER↑ ; press enter twice
R2 ; R2 value
ENTER↑ ; enter
R↓ ; roll down
+ ; add
X↔Y ;interchange X, Y
÷ ; divide Y by X
÷ ; divide Y by X
Display = REQ"
BEST!
SlideRule
RE: (15C) Parallel Resistance, two resistors - Steve Simpkin - 08-24-2023 12:31 AM
It seems like the typical steps shown (4) below to calculate parallel resistance use less keystrokes than the EDN example (8).
R1
1/x
R2
1/x
+
1/x
RE: (15C) Parallel Resistance, two resistors - Joe Horn - 08-24-2023 01:36 AM
Both programs above return 5.999999999 for values of 15 and 10 in R1 and R2 respectively. The program below returns the correct result of 6.
RCL 1
RCL × 2
RCL 1
RCL + 2
÷
RTN
Optimizing this code is left as an exercise for old PPC members.
RE: (15C) Parallel Resistance, two resistors - Steve Simpkin - 08-24-2023 03:02 AM
This is about the shortest I can come up using the above formula and only the stack.
To use:
R1
Enter
R2
----------
Enter
Rdn
X<>Y (swap X-Y)
X
Last X
Rup
+
/
RTN
RE: (15C) Parallel Resistance, two resistors - Thomas Klemm - 08-25-2023 03:30 AM
It's interesting that it was declared as the minimal solution when the obvious solution is shorter:
R1
ENTER ENTER
R2
×
x↔y
LST x
+
÷
On the HP-41C we can bring it down to:
ST* Z
+
/
And even Joe can be happy with the result.
RE: (15C) Parallel Resistance, two resistors - J-F Garnier - 08-26-2023 02:18 PM
(08-24-2023 01:36 AM)Joe Horn Wrote: Both programs above return 5.999999999 for values of 15 and 10 in R1 and R2 respectively. The program below returns the correct result of 6.
RCL 1
RCL × 2
RCL 1
RCL + 2
÷
RTN
Beside the two mentioned equivalent formulae:
Req=R1.R2/(R1+R2)
and
Req=1/(1/R1+1/R2)
there is a third (also equivalent) one that I'm using occasionally, even without programming:
Req=R1/(1+R1/R2)
which is quite convenient when R1 is a multiple of R2, e.g.
R1=100k, R2=10k, Req=100k/(1+10) = 9.09k
The benefit of the two last formulae is that they can be easily extended to handle multiple parallel resistors:
Req=1/(1/R1+1/R2+1/R3+...+1/Rn)
or
Req=R1/(1+R1/R2+R1/R3+...+R1/Rn)
J-F
RE: (15C) Parallel Resistance, two resistors - Thomas Klemm - 08-26-2023 05:20 PM
(08-26-2023 02:18 PM)J-F Garnier Wrote: Req=R1/(1+R1/R2)
Nice. With this formula we can make it even shorter:
R1
ENTER ENTER
R2
÷
1
+
÷
If we use it with R1=15 and R2=10 Joe is still happy.
Not so much if we reverse the entries.