Posted by Spencer on .
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 %?

Physics EM 
Steve,
Give up  this question is impossible.

Physics EM 
Steve,
roflcopter

Physics EM 
JB,
Physics2212 students... lets all protest against mastering physics.

Physics EM 
Alina,
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. ;)