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Chapter 26: Problem 59

(II) A \(45-\mathrm{V}\) battery of negligible internal resistance is to a \(44-\mathrm{k} \Omega\) and a \(27-\mathrm{k} \Omega\) resistor in series. What reading will a voltmeter, of internal resistance 95 \(\mathrm{k} \Omega\) , give when used to measure the voltage across each resistor? What is the percent inaccuracy due to meter resistance for each case?

Short Answer

Expert verified
The voltmeter reads 23.9 V across the 44 k惟 resistor with 14.3% inaccuracy and 14.4 V across the 27 k惟 resistor with 15.8% inaccuracy.

Step by step solution

01

Calculate Total Resistance Without Voltmeter

The total resistance of the circuit without considering the voltmeter is the sum of the resistances in series, i.e., \[ R_{total} = R_1 + R_2 = 44 \, \text{k}\Omega + 27 \, \text{k}\Omega = 71 \, \text{k}\Omega \] where \( R_1 = 44 \, \text{k}\Omega \) and \( R_2 = 27 \, \text{k}\Omega \).
02

Calculate Total Voltage Across Each Resistor Without Voltmeter

Using Ohm's Law \( V = IR \), we calculate the current in the circuit:\[ I = \frac{V}{R_{total}} = \frac{45 \, \text{V}}{71 \, \text{k}\Omega} \approx 0.634 \, \text{mA} \]Then, the voltage across each resistor:\[ V_1 = IR_1 = 0.634 \, \text{mA} \times 44 \, \text{k}\Omega = 27.9 \, \text{V} \]\[ V_2 = IR_2 = 0.634 \, \text{mA} \times 27 \, \text{k}\Omega = 17.1 \, \text{V} \]
03

Measure Voltage Across Each Resistor with Voltmeter

For each resistor, the voltmeter is put in parallel, thus affecting the total resistance of that part:For \( R_1 \): The combined resistance is \( R_{combined} = \frac{R_1 \times R_v}{R_1 + R_v} = \frac{44 \, \text{k}\Omega \times 95 \, \text{k}\Omega}{44 \, \text{k}\Omega + 95 \, \text{k}\Omega} \approx 30.6 \, \text{k}\Omega \).Using the series circuit total:\[ R_{new total} = R_{combined} + R_2 = 30.6 \, \text{k}\Omega + 27 \, \text{k}\Omega = 57.6 \, \text{k}\Omega \]For \( R_2 \): The combined resistance is \( R_{combined} = \frac{R_2 \times R_v}{R_2 + R_v} = \frac{27 \, \text{k}\Omega \times 95 \, \text{k}\Omega}{27 \, \text{k}\Omega + 95 \, \text{k}\Omega} \approx 20.8 \, \text{k}\Omega \).Using the series circuit total:\[ R_{new total} = R_1 + R_{combined} = 44 \, \text{k}\Omega + 20.8 \, \text{k}\Omega = 64.8 \, \text{k}\Omega \]
04

Calculate Voltage Reading Across Resistors With Voltmeter

For \( R_1 \):Using Ohm's Law, \[ I_{new} = \frac{V}{R_{new total}} = \frac{45 \, \text{V}}{57.6 \, \text{k}\Omega} \approx 0.781 \, \text{mA} \]The new voltage: \[ V_{new 1} = I_{new} \times R_{combined} = 0.781 \, \text{mA} \times 30.6 \, \text{k}\Omega \approx 23.9 \, \text{V} \]For \( R_2 \):\[ I_{new} = \frac{V}{R_{new total}} = \frac{45 \, \text{V}}{64.8 \, \text{k}\Omega} \approx 0.694 \, \text{mA} \]The new voltage: \[ V_{new 2} = I_{new} \times R_{combined} = 0.694 \, \text{mA} \times 20.8 \, \text{k}\Omega \approx 14.4 \, \text{V} \]
05

Calculate Percent Inaccuracy for Each Measurement

For \( R_1 \), the percent inaccuracy:\[ \text{Percent Inaccuracy} = \frac{|V_{1} - V_{new 1}|}{V_{1}} \times 100\% = \frac{|27.9 \, \text{V} - 23.9 \, \text{V}|}{27.9 \, \text{V}} \times 100\% \approx 14.3\% \]For \( R_2 \), the percent inaccuracy:\[ \text{Percent Inaccuracy} = \frac{|V_{2} - V_{new 2}|}{V_{2}} \times 100\% = \frac{|17.1 \, \text{V} - 14.4 \, \text{V}|}{17.1 \, \text{V}} \times 100\% \approx 15.8\% \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Series Circuit
In electrical circuits, the concept of a series circuit is foundational. A series circuit is where all components are connected end to end, forming a single path for the flow of current. In a series arrangement, the same current flows through each component, but the voltage across each component can be different. This happens because the total voltage of the circuit is divided among the components based on their resistances. When you add resistances in series, the total resistance is simply the sum of individual resistances.

For example, a circuit with two resistors of 44 k惟 and 27 k惟 connected in series would have a total resistance of 71 k惟. This is crucial because understanding the distribution of voltage and current in series circuits helps in analyzing the circuit's behavior under different conditions, such as changes in resistance or voltage.
Voltmeter Resistance
When measuring voltage in a circuit, a voltmeter is used. However, the concept of voltmeter resistance is important to understand to ensure accuracy in measurements. An ideal voltmeter should have infinitely high resistance so as not to disturb the circuit it is measuring. Practically, this is not possible, so voltmeters are designed with very high internal resistance, such as 95 k惟 in this exercise.

When a voltmeter is connected in parallel to a resistor, it alters the effective resistance across that part of the circuit. For instance, if a 44 k惟 resistor is measured with a 95 k惟 voltmeter, the combined resistance becomes approximately 30.6 k惟. This change affects the overall resistance of the circuit, influencing the current that flows and the voltage drop across the measured component. Understanding voltmeter resistance helps mitigate the errors introduced in voltage readings and ensures more precise measurements.
Percent Inaccuracy
In circuit measurements, accuracy is key. The concept of percent inaccuracy helps quantify how much a measurement deviates from the true value. This is particularly important when a measuring device, such as a voltmeter, impacts the circuit under measurement. Percent inaccuracy is calculated by comparing the difference between the ideally measured value and the actual measured value to the ideal.

For example, if the expected voltage across a resistor is 27.9 V but the voltmeter reads 23.9 V due to its effect on the circuit, the percent inaccuracy can be calculated using the formula:
  • Percent Inaccuracy = \( \left( \frac{|V_{ideal} - V_{measured}|}{V_{ideal}} \right) \times 100\% \)
This gives a percent inaccuracy of roughly 14.3% for one resistor and 15.8% for another in this scenario. By understanding and calculating percent inaccuracy, users can assess the reliability of their measurements and account for potential errors.

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Most popular questions from this chapter

In these Problems neglect the internal resistance of a battery unless the Problem refers to it. (I) \(\mathrm{A} 650-\Omega\) and a \(2200-\Omega\) resistor are connected in series with a 12-V battery. What is the voltage across the \(2200-\Omega\) resistor?

(II) Three \(1.70-\mathrm{k} \Omega\) resistors can be connected together in four different ways, making combinations of series and/or parallel circuits. What are these four ways, and what is the net resistance in each case?

(III) A \(2.8-\mathrm{k} \Omega\) and a 3.7 \(\mathrm{k} \Omega\) resistor are connected in parallel; this combination is connected in series with a \(1.8-\mathrm{k} \Omega\) resistor. If each resistor is rated at \(\frac{1}{2} \mathrm{W}\) (maximum without overheating), what is the maximum voltage that can be applied across the whole network?

Suppose that a person's body resistance is 950\(\Omega\) (a) What current passes through the body when the person accidentally is connected to 110 \(\mathrm{V} ?(b)\) If there is an alternative path to ground whose resistance is \(35 \Omega,\) what current passes through the person? \((c)\) If the voltage source can produce at most 1.5 \(\mathrm{A}\) , how much current passes through the person in case \((b) ?\)

IThe Problems in this Section are ranked I, II, or III according to estimated difficulty, with (I) Problems being easiest. Level (III) Problems are meant mainly as a challenge for the best students, for "extra credit." The Problems are arranged by Sections, meaning that the reader should have read up to and including that Section, but this Chapter also has a group of General Problems that are not arranged by Section and not ranked.] $$ \begin{array}{l}{\text { (I) Calculate the terminal voltage for a battery with an }} \\ {\text { internal resistance of } 0.900 \Omega \text { and an emf of } 6.00 \mathrm{V} \text { when the }} \\ {\text { battery is connected in series with }(a) \text { an } 81.0-\Omega \text { resistor, and }} \\ {\text { (b) an } 810-\Omega \text { resistor. }}\end{array} $$
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