A circuit consists of a series combination of 6.50 ? k ? and 5.00 ? k ? resistor

Question

A circuit consists of a series combination of 6.50?k?  and 5.00?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 5.00?k?  resistor using a voltmeter having an internal resistance of 10.0 k?.

Part A

What potential difference does the voltmeter measure across the 5.00?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?

Explanation / Answer

equivalent resistance across voltmeter = 5*10 / 15 = 3.33 K ohm

so voltage = 50 * 3.33 / ( 3.33 + 6.5) = 16.90 = 17 Volts

True potential = 50 * 5 / ( 6.5 + 5 ) = 21.74 Volts

%age error = 100 * ( 21.74 - 17) / 21.74 = 22.26 %

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