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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a) Calculate the equivalent resistance of the three resistors connected in series and calculate the equivalent resistance for the entire circuit.
![**Understanding Potential Difference in a Series Circuit**
**Objective:**
To calculate the potential difference \( V_A - V_B \) and \( V_C - V_D \) in the given circuit. Note that the battery has an internal resistance.
**Circuit Description:**
The circuit comprises the following components:
1. A battery with an electromotive force (emf) of \( 60 \text{ V} \) and an internal resistance of \( 2 \Omega \).
2. Three resistors connected in series:
- \( 20 \Omega \) resistor between points A and B,
- \( 30 \Omega \) resistor between points B and C,
- \( 25 \Omega \) resistor between points C and D.
**Detailed Description:**
- The series circuit begins with the positive terminal of the battery.
- The negative terminal of the battery is connected through an internal resistance of \( 2 \Omega \).
- Point A to B consists of a \( 20 \Omega \) resistor.
- Point B to C consists of a \( 30 \Omega \) resistor.
- Point C to D consists of a \( 25 \Omega \) resistor.
- Points A to D are linearly arranged in a series connection.
**Diagrams and Explanation:**
The diagram is a series electrical circuit consisting of a single path for current flow:
- Battery of \( 60 \text{ V} \) with \( 2 \Omega \) internal resistance.
- \( 20 \Omega \rightarrow 30 \Omega \rightarrow 25 \Omega \)
**Steps to Calculate Potential Difference:**
To calculate the potential difference across specific points in the circuit:
1. **Find the total resistance in the circuit:**
\[
R_{total} = R_{internal} + R_1 + R_2 + R_3 = 2 \Omega + 20 \Omega + 30 \Omega + 25 \Omega = 77 \Omega
\]
2. **Calculate the total current flowing through the circuit:**
\[
I = \frac{V_{battery}}{R_{total}} = \frac{60 \text{ V}}{77 \Omega} \approx 0.779 \text{ A}
\]
3. **Calculate the potential difference across each resistor using Ohm’s Law (V = IR):**
- For the \( 20 \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F00e5c2ed-ec3e-4354-a60a-b0d8fa7c3143%2F5b020519-79f9-4abf-8711-f4846934e4ce%2Fzioo55_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Understanding Potential Difference in a Series Circuit**
**Objective:**
To calculate the potential difference \( V_A - V_B \) and \( V_C - V_D \) in the given circuit. Note that the battery has an internal resistance.
**Circuit Description:**
The circuit comprises the following components:
1. A battery with an electromotive force (emf) of \( 60 \text{ V} \) and an internal resistance of \( 2 \Omega \).
2. Three resistors connected in series:
- \( 20 \Omega \) resistor between points A and B,
- \( 30 \Omega \) resistor between points B and C,
- \( 25 \Omega \) resistor between points C and D.
**Detailed Description:**
- The series circuit begins with the positive terminal of the battery.
- The negative terminal of the battery is connected through an internal resistance of \( 2 \Omega \).
- Point A to B consists of a \( 20 \Omega \) resistor.
- Point B to C consists of a \( 30 \Omega \) resistor.
- Point C to D consists of a \( 25 \Omega \) resistor.
- Points A to D are linearly arranged in a series connection.
**Diagrams and Explanation:**
The diagram is a series electrical circuit consisting of a single path for current flow:
- Battery of \( 60 \text{ V} \) with \( 2 \Omega \) internal resistance.
- \( 20 \Omega \rightarrow 30 \Omega \rightarrow 25 \Omega \)
**Steps to Calculate Potential Difference:**
To calculate the potential difference across specific points in the circuit:
1. **Find the total resistance in the circuit:**
\[
R_{total} = R_{internal} + R_1 + R_2 + R_3 = 2 \Omega + 20 \Omega + 30 \Omega + 25 \Omega = 77 \Omega
\]
2. **Calculate the total current flowing through the circuit:**
\[
I = \frac{V_{battery}}{R_{total}} = \frac{60 \text{ V}}{77 \Omega} \approx 0.779 \text{ A}
\]
3. **Calculate the potential difference across each resistor using Ohm’s Law (V = IR):**
- For the \( 20 \
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