A circuit consists of a series combination of 6.00 k and 5.00 k resistors connec

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.

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

What is Voltage V?

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

What is Voltage V?

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

%error = ? %

Explanation / Answer

a) equivalent resistance of 5 kohm and 10 kOhm .

1/R' = 1/5 + 1/10

R' = 3.33 kOhm

potential across 5 kOhm resistor = 50 x 3.33 / ( 6 + 3.33) = 17.86 Volt

b) when meter is not present,

V = 50 x 5 / (5 + 6) = 22.73 volt

c) % error = ( 22.73 - 17.86 ) x 100 / 22.73   = 21.41 %

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