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 %?
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
To solve the question regarding the maximum value of the ratio R_b/R_v, we need to consider the percent error in the reading of the EMF of a battery. The percent error can be calculated using the formula:
Percent Error = (|Experimental Value - Accepted Value| / Accepted Value) * 100
In this case, the accepted value of the EMF of the battery is given as 90.0 V. Since the percent error is not to exceed 3.00%, we can set up the following equation:
3.00 = (|V_v - EMF| / EMF) * 100
We already know the value of V_v as 88.4 V (from the previous question), so we can substitute it into the equation:
3.00 = (|88.4 - 90.0| / 90.0) * 100
Simplifying this equation gives:
3.00 = (1.6 / 90.0) * 100
Now, to find the maximum value that the ratio R_b/R_v may have, we can rearrange the equation as follows:
1.6 = (3.00 / 100) * 90.0
1.6 = 0.03 * 90.0
1.6 = 2.7
However, this equation has no valid solution since 1.6 is not equal to 2.7. Therefore, there is no maximum value for the ratio R_b/R_v that satisfies the condition of keeping the percent error below 3.00% in this scenario.