### 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 DiagramTo 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 BatteryUsing Ohm's Law:\[I = \frac{V}{R_{Total}}\]\[I = \frac{9.0 V}{

Answered: Image /qna-images/answer/ee8603fd-6995-459a-a80c-e37ef8e86972.jpg

A 9.0 V battery is connected to three resistors as shown in the figure below with R₁ = 502, R₂ 100 , and R3 = 75 Q = a. Draw the equivalent simplified circuit diagram. b. Determine the current drawn from the battery. C. With the value of current from the battery, find the voltage across and current through each resistor. Enter the values in a table 9.0 V 50 Ω 100 Ω www ww 75 02

A 9.0 V battery is connected to three resistors as shown in the figure below with R₁ = 502, R₂ 100 , and R3 = 75 Q = a. Draw the equivalent simplified circuit diagram. b. Determine the current drawn from the battery. C. With the value of current from the battery, find the voltage across and current through each resistor. Enter the values in a table 9.0 V 50 Ω 100 Ω www ww 75 02

Glencoe Physics: Principles and Problems, Student Edition

1st Edition

ISBN:9780078807213

Author:Paul W. Zitzewitz

Publisher:Paul W. Zitzewitz

Chapter23: Series And Parallel Circuits

Section: Chapter Questions

Problem 48A

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Question

### 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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