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A battery with EMF 90.0 V has internal resistance R_b = 8.68 Omega.
What is the reading V_v of a voltmeter having total resistance R_v = 485 \Omega when it is placed across the terminals of the battery?
V_v =88.4V
This is the value I found is correct. I don't see how to solve this next question.

What is the maximum value that the ratio R_b/R_v may have if the percent error in the reading of the EMF of a battery is not to exceed 3.00 %?

• repost physics E&M please -

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

• repost physics E&M please -

very interesting i get it. Thanks drwls

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