Chapter 26: Problem 37
A circuit consists of a series combination of 6.00-k\(\Omega\) and 5.00-k\(\Omega\) 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\(\Omega\) resistor using a voltmeter having an internal resistance of 10.0 k\(\Omega\). (a) What potential difference does the voltmeter measure across the 5.00-k\(\Omega\) resistor? (b) What is the \(true\) potential difference across this resistor when the meter is not present? (c) By what percentage is the voltmeter reading in error from the true potential difference?
Short Answer
Step by step solution
Calculate the total resistance without the voltmeter
Calculate total current through the circuit without the voltmeter
Calculate the potential difference across the 5.00-kΩ resistor without the voltmeter
Calculate the equivalent resistance with the voltmeter
Calculate total resistance with the voltmeter
Calculate the total current with the voltmeter
Calculate measured potential difference across the 5.00-kΩ resistor with the voltmeter
Calculate the percentage error of the voltmeter reading
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Resistor Circuits
Parallel and Series Circuits
Voltmeter Accuracy
In this situation, inserting a voltmeter across the 5.00-k\(\Omega\) resistor forms a parallel circuit, altering the total resistance of the circuit. The internal resistance of the voltmeter should ideally be very high to minimize its impact, but as seen in this exercise, it had a finite resistance (10.0-k\(\Omega\)), impacting the overall measurement accuracy.
For the most accurate reading, the voltmeter should not appreciably change the circuit conditions, which is why a higher internal resistance is desirable. However, practical voltmeters have a finite resistance, which should be taken into account during analysis.
Percentage Error Calculation
The formula used is:\[\% \text{ error} = \left( \frac{\text{True Value} - \text{Measured Value}}{\text{True Value}} \right) \times 100\% \]In our exercise, calculating the percentage error between the voltmeter reading and the actual voltage helps us understand the influence of the voltmeter's internal resistance on measurement accuracy.
A smaller percentage error indicates a more accurate measurement, which is the ideal situation for precise volt calculations in any resistor circuit. Understanding this concept aids in designing circuits with minimal error, ensuring accurate voltage readings.