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

Question

A circuit consists of a series combination of 6.50-k Ohm and 5.00-k Ohm 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 Ohm resistor using a voltmeter having an internal resistance of 10.0 k Ohm. What potential difference does the voltmeter measure across the 5.00-k Ohm 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

Total resistance with pot meter R = 1/(1/10 + 1/5) + 6.5 = 9.83 k

Total current = 50/9.83 = 5.08 mA

a) 5*I1 = 10*I2 and I1 + I2 = 5.08 mA

Solving these, I1 = 3.39 mA and I2 = 1.685 mA

so, V = 3.39*5 = 16.95 V

b) Total resistance without pot meter = 6.5 + 5 = 11.5 k

Total current = 50/11.5 = 4.348 mA

V = 4.348*5 = 21.74 V

c) % error = (21.74-16.95)/21.74 *100 = 22.03 %

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