Physics EM
posted by Spencer .
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?
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 %?

Give up  this question is impossible.

roflcopter

Physics2212 students... lets all protest against mastering physics.

Consider the fractional error as
a = (VV_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/(1a))1
Got me the right answer. ;)