Problem 83 - Physics for Scientists and Engineers with Modern Physics

Question

A flashlight bulb rated at \(2.0 \mathrm{~W}\) and \(3.0 \mathrm{~V}\) is operated by a 9.0-V battery. To light the bulb at its rated voltage and power, a resistor \(R\) is connected in series as shown in Fig. 26-72. What value should the resistor have?

Step-by-Step Solution

Answer

Resistor should have a value of \( 9 \Omega \).

Step 1: Understand the Problem

We need to find the value of resistor \( R \) so that a bulb rated at \( 2.0 \mathrm{~W} \) and \( 3.0 \mathrm{~V} \) can be operated using a \( 9.0 \mathrm{~V} \) battery. This means that the bulb should have a voltage drop of \( 3.0 \mathrm{~V} \) across it.

Step 2: Calculate the Bulb's Current

To find the current running through the bulb, we use the power formula \( P = V \cdot I \). Rearrange it to find \( I \): \[I = \frac{P}{V} = \frac{2.0}{3.0} = 0.67 \mathrm{~A}. \]

Step 3: Calculate the Voltage Across the Resistor

The total voltage provided by the battery is \( 9.0 \mathrm{~V} \). Since the bulb uses \( 3.0 \mathrm{~V} \), the voltage drop across the resistor \( R \) is \[V_R = 9.0 - 3.0 = 6.0 \mathrm{~V}. \]

Step 4: Apply Ohm's Law to Find the Resistor Value

Using Ohm's law, \( V = I \cdot R \), and knowing the current \( I \) through the circuit (same as through the bulb), we rearrange for \( R \): \[R = \frac{V_R}{I} = \frac{6.0}{0.67} \approx 8.96 \Omega. \]

Step 5: Round the Result

The calculated resistor value is approximately \( 8.96 \Omega \). For practical purposes, we round this to \( 9 \Omega \).

Step 6: Conclusion: State the Resistor Value

The value of the resistor \( R \) should be \( 9 \Omega \) to operate the bulb at its rated voltage and power.

Key Concepts

Resistor Value Calculation

To find the correct resistor value for a circuit, we must determine the voltage drop and current requirements first. Let's take a flashlight bulb example, where we know the bulb's power (2.0 W) and voltage (3.0 V). Since we plan to operate it with a 9.0 V battery, the extra voltage needs to be handled by a resistor.The first step is calculating the current through the bulb using the power formula: \( P = V \cdot I \).

• P is the power (2.0 W)
• V is the voltage across the bulb (3.0 V)

Rearranging, we find the current \( I = \frac{P}{V} \). This results in \( 0.67 \, \mathrm{A} \).Now, determine the voltage drop across the resistor: the total voltage minus the bulb's voltage (9.0 V - 3.0 V = 6.0 V). Next, using Ohm's Law \( V = I \cdot R \), solve for the resistor value:\[ R = \frac{V_R}{I} = \frac{6.0}{0.67} \approx 8.96 \, \Omega \]Rounding up, we choose a 9 Ω resistor for a practical solution.

Series Circuit

A series circuit is one where components are connected end-to-end in a single path for the electric current to flow. In our flashlight example, both the bulb and the resistor are connected in series with the battery.
Here's what happens in a series circuit:

• The same current flows through all components.
• The total voltage supply is divided between the components.
• The total resistance is the sum of all individual resistances.

This knowledge helps us understand why the current through the resistor must equal that through the bulb. Whenever you insert a resistor in series, it manipulates the voltage drop across each component, contributing to optimal performance of devices like light bulbs.
The total resistance encountered in the circuit ensures that each element functions within its designed parameters.

Voltage Drop

Voltage drop refers to the reduction in voltage across a component in a circuit. In a series circuit, as with our flashlight setup, the voltage drop plays a crucial role in ensuring each component receives the correct amount of voltage.
Let's break it down:

• The total supplied voltage must equal the sum of voltage drops across all components in the series.
• Correctly sized resistors ensure the appropriate voltage drop across sensitive components like a bulb, which requires exactly 3.0 V.

In our example, the battery's 9.0 V is fully allocated by dropping 3.0 V across the bulb and 6.0 V across the resistor. This controlled voltage drop ensures proper operation. If not managed, excessive voltage could harm the bulb, leading to a short lifespan or damage.