26.37 A circuit consists of a series combination of 6.00 k and 4.50 k resistors

Question

26.37

A circuit consists of a series combination of 6.00 k and 4.50 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 4.50 k resistor using a voltmeter having an internal resistance of 10.0 k.

Part A

What potential difference does the voltmeter measure across the 4.50 k resistor?

17.9

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Part B

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

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Part C

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

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26.37

A circuit consists of a series combination of 6.00 k and 4.50 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 4.50 k resistor using a voltmeter having an internal resistance of 10.0 k.

Part A

What potential difference does the voltmeter measure across the 4.50 k resistor?

17.9

  V  

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Incorrect; Try Again; 9 attempts remaining

Part B

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

  V  

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Part C

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

%error = %

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Explanation / Answer

(A)
When Meter is present,
Req = R1 + R2*R3/(R2+R3)
Req = 6 + 4.5*10/ (4.5+10)
Req = 9.10 k ohm

I = V/R
I = 50.0/9.10
I = 5.49 mA

I1* 4.5 = I2*10.0
I2 = 0.45 * I1

I1 + I2 = 5.49 mA
I1 = 5.49/1.45 mA
I1 = 3.79 mA

Potential Difference measure, V = I1*R
V = 3.79 * 4.5
V = 17.06 V

(B)
When Meter is not present,
Req = R1 + R2
Req = 6 + 4.5 = 10.5 k ohm

I = 50.0/10.5
I = 4.76 mA

True potential difference across this resistor, V = I*R
V = 4.76 * 4.5
V = 21.42 V

(C)
Error % = (21.42 -17.06)/21.42 * 100%
Error % = 20.35 %

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