4. Three resistors, 60.0 N, 30.0 N, and 20.0 N, are connected in parallel across a 90.0-V battery. a. Find the current through each branch of the circuit. b. Find the equivalent resistance of the circuit. 90.0 V. c. Find the current through the battery. 60.0 0 30.0 n 20.0 0
4. Three resistors, 60.0 N, 30.0 N, and 20.0 N, are connected in parallel across a 90.0-V battery. a. Find the current through each branch of the circuit. b. Find the equivalent resistance of the circuit. 90.0 V. c. Find the current through the battery. 60.0 0 30.0 n 20.0 0
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...
Related questions
Question
![**Parallel Circuit with Three Resistors**
**Problem:**
Three resistors, 60.0 Ω, 30.0 Ω, and 20.0 Ω, are connected in parallel across a 90.0 V battery.
**Tasks:**
a. Find the current through each branch of the circuit.
b. Find the equivalent resistance of the circuit.
c. Find the current through the battery.
**Diagram Explanation:**
The diagram shows a parallel circuit with three resistors. A 90.0 V battery is connected to three parallel branches, each containing a resistor and an ammeter. The resistors have values of 60.0 Ω, 30.0 Ω, and 20.0 Ω, respectively. The current through each resistor is labeled as \(I_a\), \(I_b\), and \(I_c\).
**Solution:**
To find the current through each branch, use Ohm's Law (\(I = \frac{V}{R}\)).
1. For the 60.0 Ω resistor:
\[
I_a = \frac{90.0 \, \text{V}}{60.0 \, \Omega} = 1.5 \, \text{A}
\]
2. For the 30.0 Ω resistor:
\[
I_b = \frac{90.0 \, \text{V}}{30.0 \, \Omega} = 3.0 \, \text{A}
\]
3. For the 20.0 Ω resistor:
\[
I_c = \frac{90.0 \, \text{V}}{20.0 \, \Omega} = 4.5 \, \text{A}
\]
To find the equivalent resistance (\(R_{\text{eq}}\)) of the parallel circuit:
\[
\frac{1}{R_{\text{eq}}} = \frac{1}{60.0} + \frac{1}{30.0} + \frac{1}{20.0}
\]
\[
\frac{1}{R_{\text{eq}}} = \frac{1}{60} + \frac{1}{30} + \frac{1}{20}
\]
\[
\frac{1}{R_{\text{eq}}} = \frac{1 + 2 + 3}{60](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9924bec9-170c-4939-801e-fa6aed25e449%2Fb848ba34-c562-48f9-8ca0-16d9567e7e41%2Fen6t4h8_processed.png&w=3840&q=75)
Transcribed Image Text:**Parallel Circuit with Three Resistors**
**Problem:**
Three resistors, 60.0 Ω, 30.0 Ω, and 20.0 Ω, are connected in parallel across a 90.0 V battery.
**Tasks:**
a. Find the current through each branch of the circuit.
b. Find the equivalent resistance of the circuit.
c. Find the current through the battery.
**Diagram Explanation:**
The diagram shows a parallel circuit with three resistors. A 90.0 V battery is connected to three parallel branches, each containing a resistor and an ammeter. The resistors have values of 60.0 Ω, 30.0 Ω, and 20.0 Ω, respectively. The current through each resistor is labeled as \(I_a\), \(I_b\), and \(I_c\).
**Solution:**
To find the current through each branch, use Ohm's Law (\(I = \frac{V}{R}\)).
1. For the 60.0 Ω resistor:
\[
I_a = \frac{90.0 \, \text{V}}{60.0 \, \Omega} = 1.5 \, \text{A}
\]
2. For the 30.0 Ω resistor:
\[
I_b = \frac{90.0 \, \text{V}}{30.0 \, \Omega} = 3.0 \, \text{A}
\]
3. For the 20.0 Ω resistor:
\[
I_c = \frac{90.0 \, \text{V}}{20.0 \, \Omega} = 4.5 \, \text{A}
\]
To find the equivalent resistance (\(R_{\text{eq}}\)) of the parallel circuit:
\[
\frac{1}{R_{\text{eq}}} = \frac{1}{60.0} + \frac{1}{30.0} + \frac{1}{20.0}
\]
\[
\frac{1}{R_{\text{eq}}} = \frac{1}{60} + \frac{1}{30} + \frac{1}{20}
\]
\[
\frac{1}{R_{\text{eq}}} = \frac{1 + 2 + 3}{60
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
This is a popular solution!
Trending now
This is a popular solution!
Step by step
Solved in 3 steps with 3 images
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, electrical-engineering and related others by exploring similar questions and additional content below.Recommended textbooks for you
Introductory Circuit Analysis (13th Edition)
Electrical Engineering
ISBN:
9780133923605
Author:
Robert L. Boylestad
Publisher:
PEARSON
Delmar's Standard Textbook Of Electricity
Electrical Engineering
ISBN:
9781337900348
Author:
Stephen L. Herman
Publisher:
Cengage Learning
Programmable Logic Controllers
Electrical Engineering
ISBN:
9780073373843
Author:
Frank D. Petruzella
Publisher:
McGraw-Hill Education
Introductory Circuit Analysis (13th Edition)
Electrical Engineering
ISBN:
9780133923605
Author:
Robert L. Boylestad
Publisher:
PEARSON
Delmar's Standard Textbook Of Electricity
Electrical Engineering
ISBN:
9781337900348
Author:
Stephen L. Herman
Publisher:
Cengage Learning
Programmable Logic Controllers
Electrical Engineering
ISBN:
9780073373843
Author:
Frank D. Petruzella
Publisher:
McGraw-Hill Education
Fundamentals of Electric Circuits
Electrical Engineering
ISBN:
9780078028229
Author:
Charles K Alexander, Matthew Sadiku
Publisher:
McGraw-Hill Education
Electric Circuits. (11th Edition)
Electrical Engineering
ISBN:
9780134746968
Author:
James W. Nilsson, Susan Riedel
Publisher:
PEARSON
Engineering Electromagnetics
Electrical Engineering
ISBN:
9780078028151
Author:
Hayt, William H. (william Hart), Jr, BUCK, John A.
Publisher:
Mcgraw-hill Education,