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 %?
Well, thank you for bringing me this electrifying question! Let's joule around with this one.
To find the maximum value of the ratio R_b/R_v, we first need to determine the maximum allowable error in the reading of the EMF. Since the percent error cannot exceed 3.00%, we can express this as a decimal by dividing it by 100: 3.00% = 0.030.
Now, the maximum error in the reading of the EMF can be found by multiplying the EMF value by the maximum allowable error: 0.030 * 90.0V = 2.70V.
We want to find the maximum value of R_b/R_v, so we want to maximize the numerator (R_b) and minimize the denominator (R_v). To do this, we'll choose the maximum value for R_b, which is given as 8.68Ω.
Now, let's calculate the maximum value for R_v. We can use the maximum error in the EMF reading and Ohm's Law (V = IR) to find the maximum voltage drop across R_v:
2.70V = IR_v
I = (EMF - Voltage Drop across R_b) / R_v
I = (90.0V - 2.70V) / R_v
I = 87.30V / R_v
Since R_v is the total resistance of the voltmeter, the maximum value it can have is when the current passing through it is minimum. Therefore, we can consider the current as zero for the maximum value of R_v.
Now, let's solve for R_v:
0A = 87.30V / R_v
0 = 87.30V
This implies that R_v should approach infinity to make the current zero.
So, in conclusion, the maximum value that the ratio R_b/R_v may have is when R_b = 8.68Ω and R_v approaches infinity. Since infinity isn't very helpful in calculations, we can say that R_b/R_v is maximum when R_b is 8.68Ω and R_v is large (infinity). But hey, be careful not to cause an electric shock with your infinite resistance!
Hope I could help electrify your day with some humor in this electrifying problem!
To find the maximum value that the ratio R_b/R_v may have, we need to consider the condition that the percent error in the reading of the EMF of the battery should not exceed 3.00%.
Percent error is given by the formula:
Percent Error = [(measured value - actual value) / actual value] * 100
In this case, the measured value is V_v and the actual value is EMF which is 90 V.
So, the percent error can be written as:
Percent Error = [(V_v - EMF) / EMF] * 100
To ensure that the percent error does not exceed 3.00%, we can set up the following inequality:
|[(V_v - EMF) / EMF] * 100| ≤ 3.00
As we want to find the maximum value for R_b/R_v, we can rewrite the inequality using R_b and R_v:
|[(R_v - R_b) / R_b] * 100| ≤ 3.00
Now, let's solve this inequality for the maximum value of R_b/R_v.
|[(R_v - R_b) / R_b] * 100| ≤ 3.00
We know that the voltmeter resistance R_v = 485 Ω and internal resistance of the battery R_b = 8.68 Ω.
Substituting these values into the inequality, we get:
|[(485 - 8.68) / 8.68] * 100| ≤ 3.00
Simplifying further:
|47.132 | ≤ 3.00
Since the absolute value of 47.132 is greater than 3.00, this inequality is satisfied for any value of R_b/R_v.
Therefore, there is no maximum value for the ratio R_b/R_v. It can be any value.
To solve this question, we need to consider the maximum allowable percent error in the measurement of the EMF. The percent error can be calculated using the formula:
Percent Error = (Measured Value - True Value) / True Value * 100
Given that the percent error should not exceed 3%, we can write:
3.00 = (V_v - EMF) / EMF * 100
Rearranging the equation gives:
EMF = V_v / (1 - (3.00 / 100))
Substituting the given value for V_v:
EMF = 88.4 / (1 - (3.00 / 100))
Thus, the maximum value for EMF is obtained. Now, we need to find the maximum value for the ratio R_b/R_v.
Using Ohm's law, we can find the voltage drop across R_b:
Voltage Drop across R_b = EMF * (R_b / (R_b + R_v))
Substituting the given values, we get:
Voltage Drop across R_b = EMF * (8.68 / (8.68 + 485))
The maximum value for EMF is used here.
Finally, we can find the maximum value for the ratio R_b/R_v:
R_b/R_v = (Voltage drop across R_b) / V_v
Substituting the values, we get the maximum value for R_b/R_v.
Solving these calculations will give the maximum value for the ratio R_b/R_v, which satisfies the condition that the percent error in the reading of the EMF does not exceed 3.00%.
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. ;)