**Measuring the Potential of a Nonideal Battery**

Part A = 88.4 V

Part B = 0.025641 V

**Solution Below:**

_{b}= 8.48 Ω.

Part A

_{v}of a voltmeter having total resistance R

_{v}= 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 = R_{internal} + R_{voltmeter}

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

_{b}/ R

_{v}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 – V_{voltmeter}) / V

We have:

0.03 = (V – V_{voltmeter}) / V

And we know that V = IR, so:

V(total) = I * (R_{internal} + R_{voltmeter})

And:

V_{voltmeter} = I * R_{voltmeter}

Plug these into the equation for error:

Error = (V – V_{voltmeter}) / V

Error = ([I * (R_{internal} + R_{voltmeter}] – [I * R_{voltmeter}]) / [I * (R_{internal} + R_{voltmeter})]

Current cancels, so:

Error = (R_{internal} + R_{voltmeter} – R_{voltmeter}) / (R_{internal} + R_{voltmeter})

Error = (R_{internalinternal} + R_{voltmeter})

(R_{internal} + R_{voltmeter}) * Error = R_{internal}

R_{internal} + R_{voltmeter} = R_{internal} / Error

(R_{internal} + R_{voltmeter}) / R_{internal} = 1 / Error

R_{internal} cancels, leaving:

1 + (R_{voltmeter} / R_{internal}) = 1 / Error

And:

(R_{voltmeter} / R_{internal}) = (1 / Error) – 1

Now just take the reciprocal of both sides:

(R_{internal} / R_{voltmeter}) = 1 / ((1 / Error) – 1)

Since the maximum error is 2.50%:

(R_{internal} / R_{voltmeter}) = 1 / ((1 / 0.025) – 1)

(R_{internal} / R_{voltmeter}) = 1 / (40 – 1)

(R_{internal} / R_{voltmeter}) = 1 / 39

(R_{internal} / R_{voltmeter}) = 0.025641

0.025641

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

Glad to help!