A circuit consists of a series combination of 6.00 ? k ? and 5.00 ? k ? resistor

Question

A circuit consists of a series combination of 6.00?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?.


What potential difference does the voltmeter measure across the 5.00

?k? resistor?


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


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

Explanation / Answer

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

10k in parallel with 5k = 5*10/5+10 = 50/15 = 3,333 ohms

the circuit is effectively 3.33k in series with 6k across 50v

voltage divider operation is 50* 3.33/(3.33+6) = 17.855Volts

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

voltage divider action is 50*5/(5+6) = 22.73 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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