Problem 59 - Physics for Scientists and Engineers with Modern Physics

Question

(II) A 45-V battery of negligible internal resistance is connected 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?

Step-by-Step Solution

Answer

With the voltmeter, 44 kΩ reads 24 V, 13.97% inaccuracy; 27 kΩ reads 14.4 V, 15.79% inaccuracy.

Step 1: Understand the Circuit Configuration

In this exercise, we have a 45-V battery connected to two resistors, one with 44 kΩ and the other with 27 kΩ, in series. A voltmeter with an internal resistance of 95 kΩ is used to measure the voltage across each of these resistors.

Step 2: Calculate the Total Resistance of the Circuit

Since the resistors are in series, the total resistance of the circuit, without considering the voltmeter, is the sum of the individual resistances: \[ R_{ ext{total}} = 44 ext{k}Ω + 27 ext{k}Ω = 71 ext{k}Ω. \]

Step 3: Calculate Current in the Circuit

Use Ohm's law to calculate the current flowing through the circuit: \[ I = \frac{V}{R_{ ext{total}}} = \frac{45 ext{ V}}{71 ext{k}Ω} = 0.634 ext{ mA}. \]

Step 4: Calculating Actual Voltage Across Each Resistor

Calculate the actual voltage across each resistor using Ohm's law:- For the 44 kΩ resistor: \[ V_{44} = I \times 44 ext{k}Ω = 0.634 ext{ mA} \times 44 ext{k}Ω = 27.9 ext{ V}. \] - For the 27 kΩ resistor: \[ V_{27} = I \times 27 ext{k}Ω = 0.634 ext{ mA} \times 27 ext{k}Ω = 17.1 ext{ V}. \]

Step 5: Analyze the Effect of the Voltmeter when Measuring Voltage Across the 44 kΩ Resistor

When measuring voltage across the 44 kΩ resistor, the voltmeter is in parallel with it. The parallel resistance of the voltmeter and 44 kΩ resistor is:\[ R_{ ext{parallel}} = \frac{44 ext{k}Ω \times 95 ext{k}Ω}{44 ext{k}Ω + 95 ext{k}Ω} = 30.85 ext{k}Ω. \] The new total resistance becomes:\[ R'_{ ext{total}} = 30.85 ext{k}Ω + 27 ext{k}Ω = 57.85 ext{k}Ω. \] The new current is:\[ I' = \frac{45 ext{ V}}{57.85 ext{k}Ω} = 0.778 ext{ mA}. \] The voltage across the 44 kΩ when the voltmeter is connected is:\[ V'_{44} = I' \times 30.85 ext{k}Ω = 0.778 ext{ mA} \times 30.85 ext{k}Ω = 24 ext{ V}. \]

Step 6: Calculate Percent Inaccuracy for the 44 kΩ Resistor

The percent inaccuracy due to the voltmeter resistance for the 44 kΩ resistor is:\[ \text{Percent inaccuracy} = \frac{V_{44} - V'_{44}}{V_{44}} \times 100\% = \frac{27.9 ext{ V} - 24 ext{ V}}{27.9 ext{ V}} \times 100\% \approx 13.97\%. \]

Step 7: Analyze the Effect of the Voltmeter when Measuring Voltage Across the 27 kΩ Resistor

When measuring voltage across the 27 kΩ resistor, the voltmeter is in parallel with it. The parallel resistance of the voltmeter and 27 kΩ resistor is:\[ R_{ ext{parallel}} = \frac{27 ext{k}Ω \times 95 ext{k}Ω}{27 ext{k}Ω + 95 ext{k}Ω} = 20.77 ext{k}Ω. \] The new total resistance becomes:\[ R'_{ ext{total}} = 44 ext{k}Ω + 20.77 ext{k}Ω = 64.77 ext{k}Ω. \] The new current is:\[ I' = \frac{45 ext{ V}}{64.77 ext{k}Ω} = 0.695 ext{ mA}. \] The voltage across the 27 kΩ when the voltmeter is connected is:\[ V'_{27} = I' \times 20.77 ext{k}Ω = 0.695 ext{ mA} \times 20.77 ext{k}Ω = 14.4 ext{ V}. \]

Step 8: Calculate Percent Inaccuracy for the 27 kΩ Resistor

The percent inaccuracy due to the voltmeter resistance for the 27 kΩ resistor is:\[ \text{Percent inaccuracy} = \frac{V_{27} - V'_{27}}{V_{27}} \times 100\% = \frac{17.1 ext{ V} - 14.4 ext{ V}}{17.1 ext{ V}} \times 100\% \approx 15.79\%. \]

Key Concepts

Ohm's Law

Ohm's law is a fundamental principle in the field of electronics and electrical engineering. It shows us how voltage, current, and resistance relate in an electrical circuit. This relationship is expressed with the formula:

• \( V = I \times R \)

Where:

• \( V \) is the voltage measured in volts (V).
• \( I \) is the current measured in amperes (A).
• \( R \) is the resistance measured in ohms (Ω).

In our exercise, Ohm’s law helps determine the current flowing through a circuit. Given a total resistance and a known voltage (like our 45-V battery), we calculate the current (

• \( I = \frac{45 \text{ V}}{71 \text{k}Ω} = 0.634 \text{ mA} \)

). This calculation allows us to determine the voltage drop across each resistor.
When a resistance is known (e.g., 44 kΩ or 27 kΩ in the series circuit), the voltage drop can be calculated by multiplying the current by the resistance of the specific resistor:

• For the 44 kΩ resistor, \( V_{44} = 27.9 \text{ V} \).
• For the 27 kΩ resistor, \( V_{27} = 17.1 \text{ V} \).

Ohm's Law is crucial because it sets the foundation for understanding how voltages and currents behave in any given resistor or series of resistors.

Series Circuit

In a series circuit, multiple components are connected end-to-end, forming a single path for the electrical current to flow. This type of circuit has some distinct characteristics:

• All components share the same current.
• The total resistance is the sum of the individual resistances.
• The voltage across the entire circuit is the sum of the voltages across each component.

In the exercise, we have two resistors, 44 kΩ and 27 kΩ, connected in series. The total resistance is calculated as:

• \( R_{\text{total}} = 44 \text{k}Ω + 27 \text{k}Ω = 71 \text{k}Ω \)

The current that flows through each component is the same, making it possible to apply Ohm's Law effectively to find the voltage drop across each resistor.
The series circuit setup means that the voltage source (45 V) is divided among the resistors in the circuit. Understanding how these divisions occur is essential for analyzing series circuits, allowing you to determine how each component affects the overall circuit behavior.

Voltmeter Resistance Effect

The voltmeter, an essential tool for measuring voltage, ideally should not affect the circuit. However, real voltmeters have internal resistance, like the 95 kΩ voltmeter in our exercise. When measuring a component in a circuit, the voltmeter is connected in parallel with that component:

• This parallel setup effectively changes the total resistance measured in the circuit.
• It introduces some error in the voltage reading due to its own resistance adding into the parallel combination.

For the 44 kΩ resistor, the voltmeter causes the total parallel resistance to be:

• \( R_{\text{parallel}} = \frac{44 \text{k}Ω \times 95 \text{k}Ω}{44 \text{k}Ω + 95 \text{k}Ω} = 30.85 \text{k}Ω \)

This change affects the overall current and thus the voltage reading. The corrected voltage across the resistor when the voltmeter is attached becomes only 24 V instead of 27.9 V, resulting in a notable percent inaccuracy.
Understanding the effect of voltmeter resistance is crucial because it helps identify discrepancies between the true voltage and what is read by the voltmeter. This understanding enables practitioners to calculate the error and adjust measurements properly, crucial for precise electrical diagnostics and repairs.