Sample Problem: Determine I1, I2 and I3 given that E1 =5.00 V, E2= 10.0 V, E3= 12.00 V, R1 =2.00 N, R2 =3.00 2, and R3 =4.002. Solutions: The current for each branch is labeled and a direction for each current is assumed. There are only two junctions in the given circuit, A and B. The junction equation for points A and B are as follows: For junction A: I1-l2-l3 = 0 For junction B: -I1+ I2 + I3 = 0 (1) (2) The three loops that may be considered for the given circuit are shown on the right Starting with points A and going clockwise around each loop will yield the following loop equation. For loop 1: -€1 + I¡R1 + I2R2 + E2 = 0 -5.00V + I1(2.002) + 12(3.000) + 10.00V = 0 (3)

College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
icon
Related questions
Question

Can you explain to me how we come up with these final answers in the sample problem:

for Current 1= 1.19A

Current 2= -2.46A

Current 3= 3.65A

E 5.00 V
R 2.00N
E, - 10.00 V
R 3.00N
loop 1
loop 2
E7
10.00 V
R 3.00n
R,-4.00N
E, - 12 0 V
E, - 5.00 V
R,
loop 3
R-4.00 n E,-12.00 v T.
For loop 2:
-E1
- IR2 - E3 + I3R3 = 0
(4)
-10.00V + 12(3.002) – 12.00 + 13(4.002) = 0
%3D
For loop 3:
-E1 + IR1 – E3 + I3R3 = 0
(5)
-5.00 V + I1(2.00N) – 12.00 V + I3(4.002) = 0
%3!
From (1), I1 = I2 + I3
Substituting (6) to (3)
(-5.00 V) + (12 + I3)(2.002) + I½(3.002) + 10.00V = 0
-5.00V + (2.00N) L2 + (2.00 N) 3+ (3.00N) I2+ 10.00V = 0
I3 = (-5.00 Q) I2 - 5.00 V
(6)
(7)
2.00 N
Substituting (7) for I3 in (4)
(-5.00Ω 12 )-(5.00 ν)
(-10.00 V) – 12(3.00 N) – (12.00 V) +
(4.00 2) = 0
2.00 N
(-13.00 N) I2 = 32 V
I2 = -2.46 A
Using (7)
I3 = (-5.00 N) ( -2.46A) – 5.00V
3.65 A
%3D
2.00 N
Using (6),
I = 3.65 A - 2.46 A = 1.19 A
Transcribed Image Text:E 5.00 V R 2.00N E, - 10.00 V R 3.00N loop 1 loop 2 E7 10.00 V R 3.00n R,-4.00N E, - 12 0 V E, - 5.00 V R, loop 3 R-4.00 n E,-12.00 v T. For loop 2: -E1 - IR2 - E3 + I3R3 = 0 (4) -10.00V + 12(3.002) – 12.00 + 13(4.002) = 0 %3D For loop 3: -E1 + IR1 – E3 + I3R3 = 0 (5) -5.00 V + I1(2.00N) – 12.00 V + I3(4.002) = 0 %3! From (1), I1 = I2 + I3 Substituting (6) to (3) (-5.00 V) + (12 + I3)(2.002) + I½(3.002) + 10.00V = 0 -5.00V + (2.00N) L2 + (2.00 N) 3+ (3.00N) I2+ 10.00V = 0 I3 = (-5.00 Q) I2 - 5.00 V (6) (7) 2.00 N Substituting (7) for I3 in (4) (-5.00Ω 12 )-(5.00 ν) (-10.00 V) – 12(3.00 N) – (12.00 V) + (4.00 2) = 0 2.00 N (-13.00 N) I2 = 32 V I2 = -2.46 A Using (7) I3 = (-5.00 N) ( -2.46A) – 5.00V 3.65 A %3D 2.00 N Using (6), I = 3.65 A - 2.46 A = 1.19 A
Sample Problem:
Determine I1, I2 and I3 given that Ɛ1 =5.00 V, E2= 10.00 V,
E3= 12.00 V, R1 =2.00 2, R2 =3.00 2, and R3 =4.002.
Solutions:
The current for each branch is labeled and a direction for each
current is assumed. There are only two junctions in the given
circuit, A and B. The junction equation for points A and B are
as follows:
For junction A: 1-l2-l3 = 0
For junction B: -I1+ I2 + I3 = 0
(1)
(2)
The three loops that may be considered for the given circuit are shown on the right
Starting with points A and going clockwise around each loop will yield the following
loop equation.
For loop 1:
-ɛ1 + IR1 + IR2 + E2 = 0
(3)
-5.00V + I1(2.002) + 12(3.000) + 10.00V = 0
Transcribed Image Text:Sample Problem: Determine I1, I2 and I3 given that Ɛ1 =5.00 V, E2= 10.00 V, E3= 12.00 V, R1 =2.00 2, R2 =3.00 2, and R3 =4.002. Solutions: The current for each branch is labeled and a direction for each current is assumed. There are only two junctions in the given circuit, A and B. The junction equation for points A and B are as follows: For junction A: 1-l2-l3 = 0 For junction B: -I1+ I2 + I3 = 0 (1) (2) The three loops that may be considered for the given circuit are shown on the right Starting with points A and going clockwise around each loop will yield the following loop equation. For loop 1: -ɛ1 + IR1 + IR2 + E2 = 0 (3) -5.00V + I1(2.002) + 12(3.000) + 10.00V = 0
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 3 steps with 3 images

Blurred answer
Knowledge Booster
Microscopic view of current
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.
Similar questions
Recommended textbooks for you
College Physics
College Physics
Physics
ISBN:
9781305952300
Author:
Raymond A. Serway, Chris Vuille
Publisher:
Cengage Learning
University Physics (14th Edition)
University Physics (14th Edition)
Physics
ISBN:
9780133969290
Author:
Hugh D. Young, Roger A. Freedman
Publisher:
PEARSON
Introduction To Quantum Mechanics
Introduction To Quantum Mechanics
Physics
ISBN:
9781107189638
Author:
Griffiths, David J., Schroeter, Darrell F.
Publisher:
Cambridge University Press
Physics for Scientists and Engineers
Physics for Scientists and Engineers
Physics
ISBN:
9781337553278
Author:
Raymond A. Serway, John W. Jewett
Publisher:
Cengage Learning
Lecture- Tutorials for Introductory Astronomy
Lecture- Tutorials for Introductory Astronomy
Physics
ISBN:
9780321820464
Author:
Edward E. Prather, Tim P. Slater, Jeff P. Adams, Gina Brissenden
Publisher:
Addison-Wesley
College Physics: A Strategic Approach (4th Editio…
College Physics: A Strategic Approach (4th Editio…
Physics
ISBN:
9780134609034
Author:
Randall D. Knight (Professor Emeritus), Brian Jones, Stuart Field
Publisher:
PEARSON