Posted by Spencer on Tuesday, March 11, 2008 at 7:31am.
The battery current must be less than that required to cause a drop of 0.03 V
I*R_b < 0.03V
I = V/(R_v + R_b)
0.03V/R_b < V/(R_v + R_b)
Divide out the V/R_b
0.03 < R_b/(R_v + R_b)
0.03 < 1/[1 + (R_v/R_b)]
[1 + (R_v/R_b)] > 33.3
R_v/R_b > 32.3
R_b/R_v < 1/32.3 = 0.031
very interesting i get it. Thanks drwls
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