Exercise 26.37 A circuit consists of a series combination of 6.50?k? and 4.50?k?

Question

Exercise 26.37

A circuit consists of a series combination of 6.50?k? and 4.50?k? 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?k? resistor using a voltmeter having an internal resistance of 10.0 k?.

Part A

What potential difference does the voltmeter measure across the 4.50?k? resistor?

Part B

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

Part C

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

Exercise 26.37

A circuit consists of a series combination of 6.50?k? and 4.50?k? 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?k? resistor using a voltmeter having an internal resistance of 10.0 k?.

Part A

What potential difference does the voltmeter measure across the 4.50?k? resistor?

_________V  

Part B

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

________V  

Part C

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

%error = ______ %

Explanation / Answer

A circuit consists of a series combination of 6.500?k? and 4.500?k? 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 4.500?k? resistor using a voltmeter having an internal resistance of 10.0 k?.

What potential difference does the voltmeter measure across the 4.500?k? resistor?
10k in parallel with 4.5k = 4.5*10/4.5+10 = 50/15 = 3,103 ohms
the circuit is effectively 3.1k in series with 6.5k across 50v
voltage divider operation is 50* 3.1/(3.1+6.5) = 16.145Volts

What is the true potential difference across this resistor when the meter is not present?
voltage divider action is 50*4.5/(4.5+6.5) = 22.5 Volts

By what percentage is the voltmeter reading in error from the true potential difference?
(22.73 - 17.86)/22.73 = .214 = 21.4%

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