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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

Expert verified
(a) 17.85 V, (b) 22.73 V, (c) 21.4% error

Step by step solution

01

Calculate the total resistance without the voltmeter

The circuit consists of two resistors in series, a 6.00-k\(\Omega\) resistor and a 5.00-k\(\Omega\) resistor. The total resistance \(R_{\text{total}}\) is given by the sum of these resistors: \[ R_{\text{total}} = 6.00\, \text{k}\Omega + 5.00\, \text{k}\Omega = 11.00\, \text{k}\Omega. \]
02

Calculate total current through the circuit without the voltmeter

Using Ohm's law, the total current \(I\) in the circuit can be calculated using the battery voltage \(V = 50.0\, \text{V}\) and the total resistance \(R_{\text{total}}\): \[ I = \frac{V}{R_{\text{total}}} = \frac{50.0}{11.00} \approx 4.545\, \text{mA}. \]
03

Calculate the potential difference across the 5.00-kΩ resistor without the voltmeter

Using Ohm's law for the 5.00-k\(\Omega\) resistor, calculate the true potential difference \(V_{\text{true}}\): \[ V_{\text{true}} = I \times 5.00\, \text{k}\Omega = 4.545\, \text{mA} \times 5.00\, \text{k}\Omega = 22.73 \text{ V}. \]
04

Calculate the equivalent resistance with the voltmeter

The voltmeter is connected across the 5.00-k\(\Omega\) resistor, forming a parallel combination. The equivalent resistance \(R_{\text{eq}}\) is given by: \[ \frac{1}{R_{\text{eq}}} = \frac{1}{5.00\, \text{k}\Omega} + \frac{1}{10.0\, \text{k}\Omega}. \] Solving for \(R_{\text{eq}}\): \[ R_{\text{eq}} = \frac{5.00 \times 10.0}{5.00 + 10.0} = \frac{50.0}{15.0} = 3.33\, \text{k}\Omega. \]
05

Calculate total resistance with the voltmeter

With the voltmeter in place, the total resistance becomes the sum of the 6.00-k\(\Omega\) resistor and the equivalent resistance \(R_{\text{eq}}\): \[ R'_{\text{total}} = 6.00\, \text{k}\Omega + 3.33\, \text{k}\Omega = 9.33\, \text{k}\Omega. \]
06

Calculate the total current with the voltmeter

Using Ohm's law again, calculate the total current \(I'\) using the modified total resistance:\[ I' = \frac{V}{R'_{\text{total}}} = \frac{50.0}{9.33} \approx 5.36\, \text{mA}. \]
07

Calculate measured potential difference across the 5.00-kΩ resistor with the voltmeter

The voltmeter measures the potential difference across the equivalent resistance \(R_{\text{eq}}\): \[ V_{\text{measured}} = I' \times R_{\text{eq}} = 5.36\, \text{mA} \times 3.33\, \text{k}\Omega = 17.85\, \text{V}. \]
08

Calculate the percentage error of the voltmeter reading

The percentage error is the difference between the measured and true potential differences, divided by the true potential difference, all multiplied by 100%: \[ \% \text{error} = \left( \frac{22.73 - 17.85}{22.73} \right) \times 100\% \approx 21.4\% \text{ error}. \]

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Key Concepts

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

Resistor Circuits
Resistor circuits are foundational elements in electrical engineering and electronics. Each resistor provides a certain amount of resistance to the flow of electrical current. In this exercise, we dealt with two resistors, 6.00-k\(\Omega\) and 5.00-k\(\Omega\), connected in series. The total resistance when resistors are connected in series is simply the sum of individual resistances. This leads to a situation where the current flowing through each resistor is the same, but the voltage across each one can differ. Shortening the path of current by additional resistors increases the total resistance, reducing the current flow overall based on Ohm's Law, which states: \[ V = IR \]where \( V \) is voltage, \( I \) is current, and \( R \) is resistance.
Parallel and Series Circuits
Understanding the difference between parallel and series circuits is crucial for solving circuits like the one in our exercise. In a series circuit, resistors are arranged in a chain, so the current has only one path to take. - **Series Circuits**: Total resistance is the sum of individual resistances. The current is the same through all components.On the other hand, the exercise also demonstrates a part where we have to consider components in parallel (voltmeter with a resistor). When resistors (or a load and a voltmeter, like in this case) are in parallel, they are connected such that one end of each component shares a common connection point, and the opposite ends do as well.- **Parallel Circuits**: Total resistance reduces, calculated by the formula: \[ \frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} + ... \frac{1}{R_n} \] This results in different currents through each component, but the voltage across each one remains the same.
Voltmeter Accuracy
A voltmeter is a device used to measure the potential difference (voltage) across a component. The accuracy of a voltmeter is extremely important when trying to measure true voltage because its internal resistance can affect the circuit it is measuring.
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
Percentage error quantifies how accurate your measurement is compared to the true value. It tells us the degree of discrepancy between the observed and true values. It's a handy tool for understanding the reliability of measurements and calculations.
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.

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Most popular questions from this chapter

A 12.4-\(\mu\)F capacitor is connected through a 0.895-M\(\Omega\) resistor to a constant potential difference of 60.0 V. (a) Compute the charge on the capacitor at the following times after the connections are made: 0, 5.0 s, 10.0 s, 20.0 s, and 100.0 s. (b) Compute the charging currents at the same instants. (c) Graph the results of parts (a) and (b) for \(t\) between 0 and 20 s.

A galvanometer having a resistance of 25.0 \(\Omega\) has a 1.00-\(\Omega\) shunt resistance installed to convert it to an ammeter. It is then used to measure the current in a circuit consisting of a 15.0-\(\Omega\) resistor connected across the terminals of a 25.0-V battery having no appreciable internal resistance. (a) What current does the ammeter measure? (b) What should be the \(true\) current in the circuit (that is, the current without the ammeter present)? (c) By what percentage is the ammeter reading in error from the \(true\) current?

The electronics supply company where you work has two different resistors, \(R_1\) and \(R_2\), in its inventory, and you must measure the values of their resistances. Unfortunately, stock is low, and all you have are \(R_1\) and \(R_2\) in parallel and in series-and you can't separate these two resistor combinations. You separately connect each resistor network to a battery with emf 48.0 V and negligible internal resistance and measure the power \(P\) supplied by the battery in both cases. For the series combination, \(P =\) 48.0 W; for the parallel combination, \(P =\) 256 W. You are told that \(R_1\) \(R_2\). (a) Calculate \(R_1\) and \(R_2\). (b) For the series combination, which resistor consumes more power, or do they consume the same power? Explain. (c) For the parallel combination, which resistor consumes more power, or do they consume the same power?

A 2.36-\(\mu\)F capacitor that is initially uncharged is connected in series with a 5.86-\(\Omega\) resistor and an emf source with \(\varepsilon =\) 120 V and negligible internal resistance. (a) Just after the connection is made, what are (i) the rate at which electrical energy is being dissipated in the resistor; (ii) the rate at which the electrical energy stored in the capacitor is increasing; (iii) the electrical power output of the source? How do the answers to parts (i), (ii), and (iii) compare? (b) Answer the same questions as in part (a) at a long time after the connection is made. (c) Answer the same questions as in part (a) at the instant when the charge on the capacitor is one-half its final value.

Two light bulbs have constant resistances of 400 \(\Omega\) and 800 \(\Omega\). If the two light bulbs are connected in series across a 120-V line, find (a) the current through each bulb; (b) the power dissipated in each bulb; (c) the total power dissipated in both bulbs. The two light bulbs are now connected in parallel across the 120-V line. Find (d) the current through each bulb; (e) the power dissipated in each bulb; (f) the total power dissipated in both bulbs. (g) In each situation, which of the two bulbs glows the brightest? (h) In which situation is there a greater total light output from both bulbs combined?

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