Mastering Physics Solutions: Measuring the Potential of a Nonideal Battery

Measuring the Potential of a Nonideal Battery

Solution Below:

A battery with EMF 90.0 V has internal resistance Rb = 8.48 Ω.

Part A

What is the reading Vv of a voltmeter having total resistance Rv = 475 Ω when it is placed across the terminals of the battery?

The EMF of the battery includes the force to to drive across its internal resistance. We need to find the voltage of the battery net of its internal resistance. Start by finding the total resistance:

R = Rinternal + Rvoltmeter
R = 8.48 + 475
R = 483.48

Now find the current:

V = IR
90 = I * 483.48
I = 0.18615

Since the current is 0.18615 A, we can find the voltage at the battery terminal (which is net of internal resistance):

V = IR
V = 0.18615 * 8.48
V = 1.578552

So the voltage at the terminal is:

V = 90 – 1.578552
V = 88.421448

This is what the voltmeter will read

88.4 V

Part B

What is the maximum value that the ratio Rb / Rv may have if the percent error in the reading of the EMF of a battery is not to exceed 2.50%?

Since the error is given by:

Error = (V – Vvoltmeter) / V

We have:

0.03 = (V – Vvoltmeter) / V

And we know that V = IR, so:

V(total) = I * (Rinternal + Rvoltmeter)

And:

Vvoltmeter = I * Rvoltmeter

Plug these into the equation for error:

Error = (V – Vvoltmeter) / V

Error = ([I * (Rinternal + Rvoltmeter] – [I * Rvoltmeter]) / [I * (Rinternal + Rvoltmeter)]

Current cancels, so:

Error = (Rinternal + Rvoltmeter – Rvoltmeter) / (Rinternal + Rvoltmeter)
Error = (Rinternalinternal + Rvoltmeter)
(Rinternal + Rvoltmeter) * Error = Rinternal
Rinternal + Rvoltmeter = Rinternal / Error
(Rinternal + Rvoltmeter) / Rinternal = 1 / Error

Rinternal cancels, leaving:

1 + (Rvoltmeter / Rinternal) = 1 / Error

And:

(Rvoltmeter / Rinternal) = (1 / Error) – 1

Now just take the reciprocal of both sides:

(Rinternal / Rvoltmeter) = 1 / ((1 / Error) – 1)

Since the maximum error is 2.50%:

(Rinternal / Rvoltmeter) = 1 / ((1 / 0.025) – 1)
(Rinternal / Rvoltmeter) = 1 / (40 – 1)
(Rinternal / Rvoltmeter) = 1 / 39
(Rinternal / Rvoltmeter) = 0.025641

0.025641

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2 Responses to Mastering Physics Solutions: Measuring the Potential of a Nonideal Battery

1. Anonymous says:

Great help when it comes to crunchtime, and good written explanation. Thanks!

• Mastering Physics Solutions says: