A circuit consists of a series combination of 6.50-k\\Omega and 5.00-k\\Omega re

Question

A circuit consists of a series combination of 6.50-kOmega and 5.00-kOmega resistors connected across a 50.0-{ m 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-kOmega resistor using a voltmeter having an internal resistance of 10.0 m k Omega.
What potential difference does the voltmeter measure across the 5.00-kOmega 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

a) Req = 5.5+(4.5||10) = 8.6 kohms

I = 5.81 mA

I in 4.5 kohms resistor = 4.006 mA

so V across 4.5 kohms resistor = 18.03 V

b) true potential difference = 4.5*50/10 = 22.5 V

c)% error is 19.86%

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