60.0 Ω 15.0 Ω 45.5 Ω 20.0 Ω 80.0 Ω An idealized voltmeter is connected across the terminals of a 20.0-V battery, and a 60.0-0 appliance is also connected across its terminals. If the voltmeter reads 15.0 V, what is the internal resistance of the battery?
60.0 Ω 15.0 Ω 45.5 Ω 20.0 Ω 80.0 Ω An idealized voltmeter is connected across the terminals of a 20.0-V battery, and a 60.0-0 appliance is also connected across its terminals. If the voltmeter reads 15.0 V, what is the internal resistance of the battery?
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![**Question:**
An idealized voltmeter is connected across the terminals of a 20.0-V battery, and a 60.0-Ω appliance is also connected across its terminals. If the voltmeter reads 15.0 V, what is the internal resistance of the battery?
**Options:**
- 60.0 Ω
- 15.0 Ω
- 45.5 Ω
- 20.0 Ω
- 80.0 Ω
**Explanation:**
This problem involves understanding how to calculate the internal resistance of a battery using the given readings and appliance characteristics. When the voltmeter reads a lower voltage than the actual battery voltage, it indicates that some of the voltage is dropped across the internal resistance of the battery. Using the formula:
\[ V_{\text{battery}} = V_{\text{terminal}} + I \times R_{\text{internal}} \]
where \( V_{\text{battery}} = 20.0 \, V \), \( V_{\text{terminal}} = 15.0 \, V \), and the current \( I \) can be determined from the appliance using \( I = \frac{V_{\text{terminal}}}{R_{\text{appliance}}} = \frac{15.0}{60.0} \).
You can solve for \( R_{\text{internal}} \) to find the correct answer.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8daeabfb-08fc-4e08-b5f3-009f3913f10d%2Fb866b046-fb19-4e2c-aa1a-64b0c66c668f%2Fg1e0is_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Question:**
An idealized voltmeter is connected across the terminals of a 20.0-V battery, and a 60.0-Ω appliance is also connected across its terminals. If the voltmeter reads 15.0 V, what is the internal resistance of the battery?
**Options:**
- 60.0 Ω
- 15.0 Ω
- 45.5 Ω
- 20.0 Ω
- 80.0 Ω
**Explanation:**
This problem involves understanding how to calculate the internal resistance of a battery using the given readings and appliance characteristics. When the voltmeter reads a lower voltage than the actual battery voltage, it indicates that some of the voltage is dropped across the internal resistance of the battery. Using the formula:
\[ V_{\text{battery}} = V_{\text{terminal}} + I \times R_{\text{internal}} \]
where \( V_{\text{battery}} = 20.0 \, V \), \( V_{\text{terminal}} = 15.0 \, V \), and the current \( I \) can be determined from the appliance using \( I = \frac{V_{\text{terminal}}}{R_{\text{appliance}}} = \frac{15.0}{60.0} \).
You can solve for \( R_{\text{internal}} \) to find the correct answer.
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