If the battery in (Figure 1) were ideal, lightbulb A would not dim when the switch is closed. However, real batteries have a small internal resistance, which we can model as the 0.5 resistor shown inside the battery. In this case, the brightness of bulb A changes when the switch is closed. Figure 1 of 1 16.0 Ω 0.5 Ω 3.0 V www.i A 06.02 ▼ Part A How much power does bulb A dissipate when the switch is open? Express your answer with the appropriate units. μA ? Popen = Value Units Submit Request Answer Part B How much power does bulb A dissipate when the switch is closed? Express your answer with the appropriate units. μA ? Pelosed= Value Units Submit Request Answer

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### Understanding Power Dissipation in a Circuit with Internal Resistance

**Context:**
If the battery in the figure below were ideal, lightbulb \( A \) would not dim when the switch is closed. However, real batteries have a small internal resistance, which we can model as the \( 0.5 \, \Omega \) resistor shown inside the battery. In this case, the brightness of bulb \( A \) changes when the switch is closed.

**Circuit Diagram:**

![Circuit Diagram](image-url)

1. The diagram shows a battery with a voltage of \( 3.0 \, V \), a small internal resistor of \( 0.5 \, \Omega \), and two light bulbs labeled \( A \) and \( B \), each with a resistance of \( 6.0 \, \Omega \).
2. Bulb \( A \) is in series with the internal resistor and parallel to the switch and bulb \( B \).

### Part A
**Question:**
How much power does bulb \( A \) dissipate when the switch is open?

**Answer Box:**
\( P_{\text{open}} = \_\_\_\_\_ \, \text{(Value)} \, \_\_\_\_\_ \, \text{(Units)} \)

**Submission Box:**
[Submit] [Request Answer]

### Part B
**Question:**
How much power does bulb \( A \) dissipate when the switch is closed?

**Answer Box:**
\( P_{\text{closed}} = \_\_\_\_\_ \, \text{(Value)} \, \_\_\_\_\_ \, \text{(Units)} \)

**Submission Box:**
[Submit] [Request Answer]

**Explanation:**
- **With the switch open**: Only bulb \( A \) conducts current, forming a series circuit with the internal resistance of the battery.
- **With the switch closed**: Bulbs \( A \) and \( B \) are in parallel, which changes the total resistance in the circuit and affects the power dissipation across each bulb.

Understanding the changes in total resistance and how they affect power dissipation is crucial for comprehending real-world circuits and the practical limitations of batteries due to internal resistance.
Transcribed Image Text:### Understanding Power Dissipation in a Circuit with Internal Resistance **Context:** If the battery in the figure below were ideal, lightbulb \( A \) would not dim when the switch is closed. However, real batteries have a small internal resistance, which we can model as the \( 0.5 \, \Omega \) resistor shown inside the battery. In this case, the brightness of bulb \( A \) changes when the switch is closed. **Circuit Diagram:** ![Circuit Diagram](image-url) 1. The diagram shows a battery with a voltage of \( 3.0 \, V \), a small internal resistor of \( 0.5 \, \Omega \), and two light bulbs labeled \( A \) and \( B \), each with a resistance of \( 6.0 \, \Omega \). 2. Bulb \( A \) is in series with the internal resistor and parallel to the switch and bulb \( B \). ### Part A **Question:** How much power does bulb \( A \) dissipate when the switch is open? **Answer Box:** \( P_{\text{open}} = \_\_\_\_\_ \, \text{(Value)} \, \_\_\_\_\_ \, \text{(Units)} \) **Submission Box:** [Submit] [Request Answer] ### Part B **Question:** How much power does bulb \( A \) dissipate when the switch is closed? **Answer Box:** \( P_{\text{closed}} = \_\_\_\_\_ \, \text{(Value)} \, \_\_\_\_\_ \, \text{(Units)} \) **Submission Box:** [Submit] [Request Answer] **Explanation:** - **With the switch open**: Only bulb \( A \) conducts current, forming a series circuit with the internal resistance of the battery. - **With the switch closed**: Bulbs \( A \) and \( B \) are in parallel, which changes the total resistance in the circuit and affects the power dissipation across each bulb. Understanding the changes in total resistance and how they affect power dissipation is crucial for comprehending real-world circuits and the practical limitations of batteries due to internal resistance.
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