The Electrician
- Joined Oct 9, 2007
- 2,971
In post #16, you have a value of 1900.2534 Ω for R345||R2. If I work the problem of finding the equivalent value of parallel resistors "backwards" I can find the value of R345 that you used in the calculation of R345||R2. The result is that you used a value of 3190 Ω.
That is, 3190 Ω in parallel with 4700 Ω gives 1900.2534 Ω. But in post #12, it was pointed out that R3 + R4 + R5 = 3130 Ω. You have apparently misread your own handwriting and taken your handwritten value of 3130 Ω to be 3190 Ω somewhere on one of your sheets of paper.
WBahn mentioned that "your 3's and your 9's look an awful lot alike"; so much alike that you have fooled yourself somewhere in your handwritings.
So, to get a correct final result you need to correct your value for R345||R2.
That is, 3190 Ω in parallel with 4700 Ω gives 1900.2534 Ω. But in post #12, it was pointed out that R3 + R4 + R5 = 3130 Ω. You have apparently misread your own handwriting and taken your handwritten value of 3130 Ω to be 3190 Ω somewhere on one of your sheets of paper.
WBahn mentioned that "your 3's and your 9's look an awful lot alike"; so much alike that you have fooled yourself somewhere in your handwritings.
So, to get a correct final result you need to correct your value for R345||R2.