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)
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)...
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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

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

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
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