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5 what is the current Ia Ia {R=3 R₂ = 9 2 qu +1

5 what is the current Ia Ia {R=3 R₂ = 9 2 qu +1

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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### Problem 5: What is the Current \( I_a \)? In this problem, we are given an electrical circuit and are asked to find the current \( I_a \). Below is the detailed description and analysis of the provided circuit diagram. #### Circuit Description: 1. **Voltage Source:** The circuit contains a voltage source of 9V. 2. **Resistors:** - **Resistor \( R_1 \)**: This resistor has a resistance of 3 ohms. - **Resistor \( R_2 \)**: This resistor has a resistance of 9 ohms. #### Diagram Analysis: - The positive terminal of the 9V voltage source is connected to the top of the circuit. - Current \( I_a \) flows from the positive terminal of the voltage source through the resistors. - Both resistors \( R_1 \) and \( R_2 \) are connected in series with the voltage source. To find the current \( I_a \), we use Ohm's Law and the series resistor circuit rules: 1. **Total Resistance \( R_t \):** Since the resistors are in series, the total resistance \( R_t \) of the circuit is the sum of the resistances: \[ R_t = R_1 + R_2 = 3 \, \Omega + 9 \, \Omega = 12 \, \Omega \] 2. **Current \( I_a \):** Using Ohm's Law (\( V = I \times R \)), we can calculate the current \( I_a \) flowing through the circuit: \[ I_a = \frac{V}{R_t} = \frac{9 \, \text{V}}{12 \, \Omega} = 0.75 \, \text{A} \] Therefore, the current \( I_a \) flowing through the circuit is **0.75 A**.
Transcribed Image Text:### Problem 5: What is the Current \( I_a \)? In this problem, we are given an electrical circuit and are asked to find the current \( I_a \). Below is the detailed description and analysis of the provided circuit diagram. #### Circuit Description: 1. **Voltage Source:** The circuit contains a voltage source of 9V. 2. **Resistors:** - **Resistor \( R_1 \)**: This resistor has a resistance of 3 ohms. - **Resistor \( R_2 \)**: This resistor has a resistance of 9 ohms. #### Diagram Analysis: - The positive terminal of the 9V voltage source is connected to the top of the circuit. - Current \( I_a \) flows from the positive terminal of the voltage source through the resistors. - Both resistors \( R_1 \) and \( R_2 \) are connected in series with the voltage source. To find the current \( I_a \), we use Ohm's Law and the series resistor circuit rules: 1. **Total Resistance \( R_t \):** Since the resistors are in series, the total resistance \( R_t \) of the circuit is the sum of the resistances: \[ R_t = R_1 + R_2 = 3 \, \Omega + 9 \, \Omega = 12 \, \Omega \] 2. **Current \( I_a \):** Using Ohm's Law (\( V = I \times R \)), we can calculate the current \( I_a \) flowing through the circuit: \[ I_a = \frac{V}{R_t} = \frac{9 \, \text{V}}{12 \, \Omega} = 0.75 \, \text{A} \] Therefore, the current \( I_a \) flowing through the circuit is **0.75 A**.
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