Problem 62 - Physics for Scientists and Engineers with Modern Physics

Question

(II) A 12.0-V battery (assume the internal resistance \(=0\) ) is connected to two resistors in series. A voltmeter whose internal resistance is \(18.0 \mathrm{k} \Omega\) measures \(5.5 \mathrm{~V}\) and \(4.0 \mathrm{~V}\) respectively, when connected across each of the resistors. What is the resistance of each resistor?

Step-by-Step Solution

Answer

Resistor values: \( R_1 \approx 8.1 \text{k} \Omega \), \( R_2 \approx 13.5 \text{k} \Omega \).

Step 1: Understand the Circuit Configuration

We have a series circuit with a 12.0 V battery connected to two resistors, let's name them \( R_1 \) and \( R_2 \), in series. A voltmeter measures the voltage across each resistor separately as 5.5 V and 4 V, respectively. The internal resistance of the voltmeter is 18 kΩ.

Step 2: Determine Total Voltage Across the Resistors

In a series circuit, the total voltage provided by the battery is distributed across the resistors. Therefore,\[ V_1 + V_2 = 12.0 ext{ V} \]where \( V_1 = 5.5 ext{ V} \) and \( V_2 = 4.0 ext{ V} \). This checks out with the battery voltage.

Step 3: Analyze Voltmeter Reading Across the First Resistor

When a voltmeter reads 5.5 V across \( R_1 \), it affects the effective resistance. The series resistance \( R' \) is used to calculate the resistance \( R_1 \). The relation given by\[ \frac{R_1 \cdot 18 \text{k} \Omega}{R_1 + 18 \text{k} \Omega} = 5.5 \text{ V}\]

Step 4: Derive Equation for First Resistor - \(R_1\)

From the above step, rearranging gives:\[ 5.5(R_1 + 18 \times 10^3) = R_1 \cdot 18 \times 10^3 \]Simplifying, \[ 18 \times 10^3 = 5.5 \times 18 \times 10^3 + 5.5 R_1 \]

Step 5: Solve for Resistance \(R_1\)

Solving the derived equation from previous step, we can isolate \( R_1 \) to find its value.

Step 6: Analyze Voltmeter Reading Across the Second Resistor

Similarly, for resistor \( R_2 \), the same principle applies:\[ \frac{R_2 \cdot 18 \text{k} \Omega}{R_2 + 18 \text{k} \Omega} = 4.0 \text{ V} \]

Step 7: Derive Equation for Second Resistor - \(R_2\)

Rearrange the equation to:\[ 4(R_2 + 18 \times 10^3) = R_2 \cdot 18 \times 10^3 \]Simplify and solve for \( R_2 \).

Step 8: Solve for Resistance \(R_2\)

Finish solving for \( R_2 \) to find its value.

Step 9: Check Calculations

Verify that the sum of the voltages across the resistors matches the battery voltage, and the calculated resistances uphold the condition of the circuit.

Key Concepts

Voltmeter Internal Resistance

Voltmeter internal resistance is a crucial concept in series circuit analysis. When you are measuring voltage across a component, the internal resistance of the voltmeter can affect the readings. In an ideal scenario, a voltmeter should have infinite resistance. This is because it should not draw any current from the circuit.

However, in reality, voltmeters have a finite internal resistance. This resistance can alter the effective resistance of the circuit when connected. In our problem, the voltmeter has an internal resistance of 18 kΩ. This means when it is connected across the resistors, it forms a parallel resistance with each resistor.

The equation that emerges from this scenario helps calculate the true resistance of the resistor being measured. Consider the formula: \[ \frac{R \cdot R_v}{R + R_v} = V_m \]where:

• \( R \) is the resistance being measured,
• \( R_v \) is the internal resistance of the voltmeter,
• \( V_m \) is the measured voltage.

This tells us how the voltmeter's internal resistance modifies the voltage reading, necessitating the need to account for it in our calculations.

Ohm's Law

Ohm's Law is fundamental to understanding how voltage, current, and resistance interplay in electrical circuits. It is given by the equation: \( V = I \times R \), where:

• \( V \) is the voltage across a resistor,
• \( I \) is the current flowing through the circuit,
• \( R \) is the resistance of the component.

Understanding Ohm's Law is critical in our exercise as it allows us to relate the voltage across each resistor to their respective resistances. In a series circuit, the same current flows through each component. This means that the voltage across a resistor is proportional to its resistance.

With the measured voltages available for each resistor (5.5 V and 4.0 V), we can use Ohm’s Law to establish relationships and solve for the resistances knowing the total current stays constant throughout the circuit.

This makes it easy to breakdown circuit elements into more manageable components, crucial for accurate calculations and understanding the overall circuit behavior.

Resistor Voltage Drop

In any series circuit, the voltage drop across a resistor is determined by the resistor's value and the current flowing through the circuit. As per Ohm's Law, the voltage drop \( V \) across a resistor \( R \) is \( V = I \times R \). This means more resistance leads to more voltage drop across that resistor.

In the given exercise, we identified a 12 V battery providing the total voltage. The sum of the voltages across both resistors will equal this total battery voltage due to the properties of a series circuit. Consider what we measured with the voltmeter across the resistors: 5.5 V and 4.0 V.

These measurements represent the voltage drops across the resistors R1 and R2, respectively. Understanding that these drops must add up to the battery voltage ensures we correctly calculate the resistances. By having this firm understanding of how voltage divides in a series circuit, one can analyze complex configurations faster and with greater accuracy.

The knowledge of voltage drop is essential when moving from theory to the practical analysis of circuits, ensuring you are prepared to handle various circuit designs.