Use Kirchhoff's Voltage Law to solve for the voltage across resistor R₂ in the circuit, given: Ę₁ - = 35 V and Er1 = 11 V (Round the FINAL answer to two decimal places.) ET R₁ P1₂
Use Kirchhoff's Voltage Law to solve for the voltage across resistor R₂ in the circuit, given: Ę₁ - = 35 V and Er1 = 11 V (Round the FINAL answer to two decimal places.) ET R₁ P1₂
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
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![### Educational Content on Kirchhoff's Voltage Law
#### Problem Statement:
Use Kirchhoff's Voltage Law to solve for the voltage across resistor \( R_2 \) in the circuit, given:
- \( E_T = 35 \, \text{V} \)
- \( E_{R_1} = 11 \, \text{V} \)
*Round the FINAL answer to two decimal places.*
#### Circuit Diagram Explanation:
The diagram shows a simple series circuit consisting of a power source with voltage \( E_T \) and two resistors labeled \( R_1 \) and \( R_2 \). The total voltage across the circuit is the sum of the voltages across each resistor.
#### Solution Process:
Using **Kirchhoff's Voltage Law**, which states that the sum of the electromotive forces (voltages) in any closed loop is equivalent to the sum of the potential drops (voltages across resistors), we can calculate:
\[
E_T = E_{R_1} + E_{R_2}
\]
Plugging in the known values:
\[
35 \, \text{V} = 11 \, \text{V} + E_{R_2}
\]
To find \( E_{R_2} \):
\[
E_{R_2} = 35 \, \text{V} - 11 \, \text{V} = 24 \, \text{V}
\]
Thus, the voltage across resistor \( R_2 \) is:
\[
E_{R_2} = 24.00 \, \text{V}
\]
This confirms the application of Kirchhoff's Voltage Law to solve for unknown voltages in a series circuit.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3864bbe0-e2c9-459a-9920-261a79870337%2F01cba764-fc05-417c-a815-b6ffcc9dbba4%2Fdvln9n8_processed.png&w=3840&q=75)
Transcribed Image Text:### Educational Content on Kirchhoff's Voltage Law
#### Problem Statement:
Use Kirchhoff's Voltage Law to solve for the voltage across resistor \( R_2 \) in the circuit, given:
- \( E_T = 35 \, \text{V} \)
- \( E_{R_1} = 11 \, \text{V} \)
*Round the FINAL answer to two decimal places.*
#### Circuit Diagram Explanation:
The diagram shows a simple series circuit consisting of a power source with voltage \( E_T \) and two resistors labeled \( R_1 \) and \( R_2 \). The total voltage across the circuit is the sum of the voltages across each resistor.
#### Solution Process:
Using **Kirchhoff's Voltage Law**, which states that the sum of the electromotive forces (voltages) in any closed loop is equivalent to the sum of the potential drops (voltages across resistors), we can calculate:
\[
E_T = E_{R_1} + E_{R_2}
\]
Plugging in the known values:
\[
35 \, \text{V} = 11 \, \text{V} + E_{R_2}
\]
To find \( E_{R_2} \):
\[
E_{R_2} = 35 \, \text{V} - 11 \, \text{V} = 24 \, \text{V}
\]
Thus, the voltage across resistor \( R_2 \) is:
\[
E_{R_2} = 24.00 \, \text{V}
\]
This confirms the application of Kirchhoff's Voltage Law to solve for unknown voltages in a series circuit.
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