Find the Potential Difference

  • #1
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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?

Homework Equations



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.
 

Answers and Replies

  • #2
SammyS
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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?

Homework Equations



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?
 
  • #3
CWatters
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There are THREE resistors in the circuit. Two in series and the one in the voltmeter (appears in parallel) which moves depending on where it is used to measure the voltage.
 

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