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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.