Question

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

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

Step-by-step solution

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: \[ R_{\text{total}} = 6.00\, \text{k}\Omega + 5.00\, \text{k}\Omega = 11.00\, \text{k}\Omega. \]

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 \(R_{\text{total}}\): \[ I = \frac{V}{R_{\text{total}}} = \frac{50.0}{11.00} \approx 4.545\, \text{mA}. \]

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 \(V_{\text{true}}\): \[ V_{\text{true}} = I \times 5.00\, \text{k}\Omega = 4.545\, \text{mA} \times 5.00\, \text{k}\Omega = 22.73 \text{ V}. \]

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 \(R_{\text{eq}}\): \[ R_{\text{eq}} = \frac{5.00 \times 10.0}{5.00 + 10.0} = \frac{50.0}{15.0} = 3.33\, \text{k}\Omega. \]

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 \(R_{\text{eq}}\): \[ R'_{\text{total}} = 6.00\, \text{k}\Omega + 3.33\, \text{k}\Omega = 9.33\, \text{k}\Omega. \]

Calculate the total current with the voltmeter

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

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

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

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%: \[ \% \text{error} = \left( \frac{22.73 - 17.85}{22.73} \right) \times 100\% \approx 21.4\% \text{ error}. \]

Key concepts

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.