A battery with EMF 90.0 V has internal resistance R_b = 8.68 Omega.

Question

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?
Express your answer with three significant figures.
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 %?

Answer 1

Consider the fractional error as
a = (V-V_v)/V

where a is equal to the 0.03 (because in this question the percent error is 3.00%)

You should know from Kirchhoff's rule that the Electric potential difference is

V=I(R_b+R_v)

You also found out that Elec. Poten. diff. with only the voltmeter as resistance is

V_v =IR_v

You plug in the Voltage equations to the first equation for the error and play a little with algebra you get the ratio as

R_b/R_v = (1/(1-a))-1

Got me the right answer. ;)