
Glencoe Physics: Principles and Problems, Student Edition
1st Edition
ISBN: 9780078807213
Author: Paul W. Zitzewitz
Publisher: Glencoe/McGraw-Hill
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![### Series-Parallel Circuit Analysis
**Objective:**
To analyze a circuit consisting of a 9.0 V battery connected to three resistors, and to carry out the following tasks:
a. Draw the equivalent simplified circuit diagram.
b. Determine the current drawn from the battery.
c. Calculate the voltage across and the current through each resistor, and present the values in a table.
**Circuit Description:**
The circuit consists of three resistors:
1. \( R1 = 50 \Omega \)
2. \( R2 = 100 \Omega \)
3. \( R3 = 75 \Omega \)
**Diagram Analysis:**
The provided circuit diagram shows the 9.0 V battery connected to a combination of resistors arranged in a mixed series-parallel configuration.
- Resistor \( R1 \) and \( R2 \) are arranged in parallel.
- The combination of \( R1 \) and \( R2 \) is then connected in series with \( R3 \).
#### a. Simplified Circuit Diagram
To simplify the circuit:
1. Calculate the equivalent resistance \( R_{12} \) of the parallel combination of \( R1 \) and \( R2 \):
\[
\frac{1}{R_{12}} = \frac{1}{R1} + \frac{1}{R2}
\]
\[
\frac{1}{R_{12}} = \frac{1}{50 \Omega} + \frac{1}{100 \Omega}
\]
\[
R_{12} = \left( \frac{1}{\frac{1}{50} + \frac{1}{100}} \right) \Omega
\]
\[
R_{12} = \left( \frac{1}{0.03} \right) \Omega = 33.33 \Omega
\]
2. The equivalent resistance (\( R_{Total} \)) for the entire circuit, which is \( R_{12} \) in series with \( R3 \):
\[
R_{Total} = R_{12} + R3 = 33.33 \Omega + 75 \Omega = 108.33 \Omega
\]
#### b. Current Drawn from the Battery
Using Ohm's Law:
\[
I = \frac{V}{R_{Total}}
\]
\[
I = \frac{9.0 V}{](https://content.bartleby.com/qna-images/question/4a6e4794-5644-431d-9e96-b37fbaf0b60a/ee8603fd-6995-459a-a80c-e37ef8e86972/90litfh_thumbnail.jpeg)
Transcribed Image Text:### Series-Parallel Circuit Analysis
**Objective:**
To analyze a circuit consisting of a 9.0 V battery connected to three resistors, and to carry out the following tasks:
a. Draw the equivalent simplified circuit diagram.
b. Determine the current drawn from the battery.
c. Calculate the voltage across and the current through each resistor, and present the values in a table.
**Circuit Description:**
The circuit consists of three resistors:
1. \( R1 = 50 \Omega \)
2. \( R2 = 100 \Omega \)
3. \( R3 = 75 \Omega \)
**Diagram Analysis:**
The provided circuit diagram shows the 9.0 V battery connected to a combination of resistors arranged in a mixed series-parallel configuration.
- Resistor \( R1 \) and \( R2 \) are arranged in parallel.
- The combination of \( R1 \) and \( R2 \) is then connected in series with \( R3 \).
#### a. Simplified Circuit Diagram
To simplify the circuit:
1. Calculate the equivalent resistance \( R_{12} \) of the parallel combination of \( R1 \) and \( R2 \):
\[
\frac{1}{R_{12}} = \frac{1}{R1} + \frac{1}{R2}
\]
\[
\frac{1}{R_{12}} = \frac{1}{50 \Omega} + \frac{1}{100 \Omega}
\]
\[
R_{12} = \left( \frac{1}{\frac{1}{50} + \frac{1}{100}} \right) \Omega
\]
\[
R_{12} = \left( \frac{1}{0.03} \right) \Omega = 33.33 \Omega
\]
2. The equivalent resistance (\( R_{Total} \)) for the entire circuit, which is \( R_{12} \) in series with \( R3 \):
\[
R_{Total} = R_{12} + R3 = 33.33 \Omega + 75 \Omega = 108.33 \Omega
\]
#### b. Current Drawn from the Battery
Using Ohm's Law:
\[
I = \frac{V}{R_{Total}}
\]
\[
I = \frac{9.0 V}{
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