How do I learn RPL and solve this problem with it?
09-25-2017, 08:04 PM
Post: #21
 emece67 Senior Member Posts: 379 Joined: Feb 2015
RE: How do I learn RPL and solve this problem with it?
I have not read the whole thread, but I'm wondering if the fact that the base resistor values form a, more or less, geometric series is used in some way.

Thus, the E12 series uses 12 values on each decade, this way the ratio between one value and the next is $$\sqrt[12]{10}=1.2115\ldots$$. This way the E12 values are:
• 1
• $$\sqrt[12]{10}=1.2115\ldots\approx1.2$$
• $$(\sqrt[12]{10})^2=1.4677\ldots\approx1.5$$
• $$(\sqrt[12]{10})^3=1.7782\ldots\approx1.8$$
• $$(\sqrt[12]{10})^4=2.1544\ldots\approx2.2$$
• $$(\sqrt[12]{10})^5=2.6101\ldots\approx2.6$$ (E24 value is 2.7)
• $$(\sqrt[12]{10})^6=3.1622\ldots\approx3.2$$ (E24 value is 3.3)
• $$(\sqrt[12]{10})^7=3.8311\ldots\approx3.8$$ (E24 value is 3.9)
• $$(\sqrt[12]{10})^8=4.6415\ldots\approx4.6$$ (E24 value is 4.7)
• $$(\sqrt[12]{10})^9=5.6234\ldots\approx5.6$$
• $$(\sqrt[12]{10})^{10}=6.8129\ldots\approx6.8$$
• $$(\sqrt[12]{10})^{11}=8.2540\ldots\approx8.3$$ (E24 value is 8.2)

I'm not yet worked this, but perhaps it would be possible to only search for resistor pairs that are zero steps apart, then one, then two,... and so on.

Regards.
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