Find the Potential Difference

Homework Statement

A circuit consists of a series combination of 5.50-kΩ and 4.50-kΩ resistors connected across a 50.0 V battery having negligible internal resistance. You want to measure the true potential difference (that is, the potential difference without the meter present) across the 4.50-kΩ resistor using a voltmeter having an internal resistance of 10.0 kΩ.

A) What potential difference does the voltmeter measure across the 4.50-kΩ resistor?

B) What is the true potential difference across this resistor when the meter is not present?

C) By what percentage is the voltmeter reading in error from the true potential difference?

V = IR

The Attempt at a Solution

I'm stuck on part A. From what I understand, the question says there is a battery with resistors on either side of it.

So, knowing that in a loop the voltage is equal to 0, I did my calculations as follow:

0 = 50V - 5500I - 4500I
Thus this tells me that the current is 0.005 A

Then, using V = IR

V = 0.005*4500
V = 22.5 V

It says my answer is wrong and I'm not sure what else to try. Help would be appreciated.

SammyS
Staff Emeritus
Homework Helper
Gold Member

Homework Statement

A circuit consists of a series combination of 5.50-kΩ and 4.50-kΩ resistors connected across a 50.0 V battery having negligible internal resistance. You want to measure the true potential difference (that is, the potential difference without the meter present) across the 4.50-kΩ resistor using a voltmeter having an internal resistance of 10.0 kΩ.

A) What potential difference does the voltmeter measure across the 4.50-kΩ resistor?

B) What is the true potential difference across this resistor when the meter is not present?

C) By what percentage is the voltmeter reading in error from the true potential difference?

V = IR

The Attempt at a Solution

I'm stuck on part A. From what I understand, the question says there is a battery with resistors on either side of it.

So, knowing that in a loop the voltage is equal to 0, I did my calculations as follow:

0 = 50V - 5500I - 4500I
Thus this tells me that the current is 0.005 A

Then, using V = IR

V = 0.005*4500
V = 22.5 V

It says my answer is wrong and I'm not sure what else to try. Help would be appreciated.
There are two resistors, 5.50-kΩ and 4.50-kΩ, in series with a 50 Volt ideal battery.

When you use a volt-meter to measure the voltage drop across the 4.50-kΩ resistor, how is the volt-meter connected in the circuit? ... in parallel or series? ... and with which device?

CWatters