91Ó°ÊÓ

A circuit consists of a series combination of \(6.00-\mathrm{k} \Omega\) and 5.00- \(\mathrm{k} \Omega\) resistors connected across a \(50.0-\mathrm{V}\) battery having negligible internal resistance. You want to measure the true potential difference (that is, the potential difference without the meter present) across the \(5.00-\mathrm{k} \Omega\) resistor using a voltmeter having an internal resistance of 10.0 \(\mathrm{k} \Omega\) (a) What potential difference does the voltmeter measure across the \(5.00-\mathrm{k} \Omega\) 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?

Step by step solution

01

Calculate Total Resistance with Voltmeter

First, calculate the total resistance in the circuit when the voltmeter is connected. The voltmeter and the 5.00-kΩ resistor are in parallel.The equivalent resistance of the parallel combination is given by:\[R_{parallel} = \left(\frac{1}{5.00\,k\Omega} + \frac{1}{10.0\,k\Omega}\right)^{-1}\]Calculating the above gives:\[R_{parallel} = \left(\frac{1}{5} + \frac{1}{10}\right)^{-1} \Omega = \left(\frac{2}{10} \right)^{-1} \Omega = 3.33\,k\Omega\]

02

Calculate Total Circuit Resistance

The total resistance in the circuit with the voltmeter is the sum of the 6.00-kΩ resistor and the parallel combination:\[R_{total} = 6.00\,k\Omega + 3.33\,k\Omega = 9.33\,k\Omega\]

03

Find Circuit Current

Now, we can find the current flowing through the circuit using Ohm's Law:\[I = \frac{V}{R_{total}} = \frac{50.0\,V}{9.33\,k\Omega} = 5.36\,mA\]

04

Measure Voltmeter Reading

The voltmeter reading is the voltage across the parallel combination which is the voltage across the 5.00-kΩ resistor.\[V_{meter} = I \times R_{parallel} = 5.36\,mA \times 3.33\,k\Omega = 17.87\,V\]

05

Calculate True Potential Difference

Without the voltmeter, the total resistance is the sum of the 6.00-kΩ and the 5.00-kΩ resistors:\[R_{true} = 6.00\,k\Omega + 5.00\,k\Omega = 11.00\,k\Omega\]The current through the circuit without the voltmeter is:\[I_{true} = \frac{50.0\,V}{11.00\,k\Omega} = 4.55\,mA\]The true potential difference across the 5.00-kΩ resistor:\[V_{true} = I_{true} \times 5.00\,k\Omega = 4.55\,mA \times 5.00\,k\Omega = 22.75\,V\]

06

Calculate Percentage Error

The error percentage in the voltmeter reading is calculated as follows:\[\text{Error Percentage} = \left( \frac{V_{true} - V_{meter}}{V_{true}} \right) \times 100\%\]\[\text{Error Percentage} = \left( \frac{22.75\,V - 17.87\,V}{22.75\,V} \right) \times 100\% = 21.43\%\]

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Potential Difference

Potential difference, often called voltage, is an essential aspect of any electrical circuit. Voltage measures the difference in electric potential energy between two points in a circuit. It's the force that pushes electric charges to flow through a circuit, much like how pressure pushes water through a pipe.

• Voltage is measured in volts (V).
• It can be measured directly across components using a voltmeter.
• Potential difference determines how much current will flow for a given resistance.

In our exercise, the true potential difference across the resistor is calculated by removing any external influences, such as the voltmeter, to measure the actual value as it exists in native conditions.
It's crucial to distinguish between the measured value (with the meter) and the true potential difference to ensure precise circuit analysis and functionality.

Series and Parallel Circuits

Understanding how resistors and other components are arranged in circuits affects how current and voltage behave. There are two primary ways components can be arranged: series and parallel.

• Series Circuits: Components are connected end-to-end. The same current flows through each component, and the total resistance is the sum of each component's resistance.
• Parallel Circuits: Components are connected across the same two points. The total current is divided among components, but they experience the same voltage.

In this exercise, the voltmeter and the 5.00-kΩ resistor are in parallel, affecting the potential difference measured across them. This arrangement changes how the total resistance and current are calculated, impacting the readings and the perceived potential difference.

Resistor Circuit Analysis

Resistor circuit analysis involves calculating circuit properties using Ohm's Law and understanding circuit configurations to determine values like total resistance, current flow, and voltages. Here's a quick breakdown:

• Ohm's Law: This fundamental law states that the current through a resistor is proportional to the voltage and inversely proportional to its resistance: \( I = \frac{V}{R} \).
• Calculating Equivalent Resistance: For series circuits, add all resistance values. For parallel circuits, use \( \left( \frac{1}{R_{1}} + \frac{1}{R_{2}} + \ldots \right)^{-1} \).

In our solution, steps involve finding the equivalent resistance when components are added in series or parallel and using those values with Ohm's Law to find current and voltage values.
This step-by-step process helps ensure all circuit components are analyzed correctly for precise circuit design and troubleshooting.