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

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Chapter10: Direct-current Circuits
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Problem 72AP: The rather simple circuit shown below is known as a voltage divider. The symbol consisting of three...
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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}{
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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