(II) What is the internal resistance of a 12.0-V car battery whose terminal voltage drops to 8.4 V when the starter draws 95 A? What is the resistance of the starter?
(II) What is the internal resistance of a 12.0-V car battery whose terminal voltage drops to 8.4 V when the starter draws 95 A? What is the resistance of the starter?
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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![**Physics Problem: Internal Resistance and Starter Resistance Calculation**
**Problem 4:**
What is the internal resistance of a 12.0-V car battery whose terminal voltage drops to 8.4 V when the starter draws 95 A? What is the resistance of the starter?
**Explanation:**
To solve this problem, we'll use the concepts of internal resistance and Ohm's law.
First, we need to determine the internal resistance of the battery.
Given:
- The emf (electromotive force) of the battery \( V_{\text{emf}} \) = 12.0 V
- The terminal voltage \( V_{\text{terminal}} \) when the starter is drawing current is 8.4 V.
- The current \( I \) drawn by the starter is 95 A.
The internal resistance \( r \) of the battery can be calculated using the equation:
\[ V_{\text{emf}} = V_{\text{terminal}} + Ir \]
Rearrange this formula to solve for internal resistance \( r \):
\[ r = \frac{V_{\text{emf}} - V_{\text{terminal}}}{I} \]
Substitute the given values into the equation:
\[ r = \frac{12.0\ \text{V} - 8.4\ \text{V}}{95\ \text{A}} \]
\[ r = \frac{3.6\ \text{V}}{95\ \text{A}} \]
\[ r = 0.0379\ \Omega \]
So, the internal resistance of the battery is approximately \( 0.0379\ \Omega \).
Next, we need to determine the resistance of the starter. The total voltage provided by the battery minus the voltage drop due to internal resistance provides the voltage across the starter. The resistance \( R_{\text{starter}} \) can be calculated using Ohm's Law:
\[ V_{\text{starter}} = V_{\text{terminal}} \]
\[ R_{\text{starter}} = \frac{V_{\text{starter}}}{I} \]
Substitute the known values:
\[ R_{\text{starter}} = \frac{8.4\ \text{V}}{95\ \text{A}} \]
\[ R_{\text{starter}} = 0.0884\ \Omega \]
So, the resistance of the starter is approximately](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F71886d56-a83a-4d7b-8e18-9a0e9bdd4c82%2F4af5e0d2-158a-429b-82b6-487608c75059%2Fipejocl_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Physics Problem: Internal Resistance and Starter Resistance Calculation**
**Problem 4:**
What is the internal resistance of a 12.0-V car battery whose terminal voltage drops to 8.4 V when the starter draws 95 A? What is the resistance of the starter?
**Explanation:**
To solve this problem, we'll use the concepts of internal resistance and Ohm's law.
First, we need to determine the internal resistance of the battery.
Given:
- The emf (electromotive force) of the battery \( V_{\text{emf}} \) = 12.0 V
- The terminal voltage \( V_{\text{terminal}} \) when the starter is drawing current is 8.4 V.
- The current \( I \) drawn by the starter is 95 A.
The internal resistance \( r \) of the battery can be calculated using the equation:
\[ V_{\text{emf}} = V_{\text{terminal}} + Ir \]
Rearrange this formula to solve for internal resistance \( r \):
\[ r = \frac{V_{\text{emf}} - V_{\text{terminal}}}{I} \]
Substitute the given values into the equation:
\[ r = \frac{12.0\ \text{V} - 8.4\ \text{V}}{95\ \text{A}} \]
\[ r = \frac{3.6\ \text{V}}{95\ \text{A}} \]
\[ r = 0.0379\ \Omega \]
So, the internal resistance of the battery is approximately \( 0.0379\ \Omega \).
Next, we need to determine the resistance of the starter. The total voltage provided by the battery minus the voltage drop due to internal resistance provides the voltage across the starter. The resistance \( R_{\text{starter}} \) can be calculated using Ohm's Law:
\[ V_{\text{starter}} = V_{\text{terminal}} \]
\[ R_{\text{starter}} = \frac{V_{\text{starter}}}{I} \]
Substitute the known values:
\[ R_{\text{starter}} = \frac{8.4\ \text{V}}{95\ \text{A}} \]
\[ R_{\text{starter}} = 0.0884\ \Omega \]
So, the resistance of the starter is approximately
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