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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

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

Delmar's Standard Textbook Of Electricity
Delmar's Standard Textbook Of Electricity
7th Edition
ISBN:9781337900348
Author:Stephen L. Herman
Publisher:Stephen L. Herman
Chapter21: Resistive-capacitive Series Circuits
Section: Chapter Questions
Problem 1PP
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Can you explain to me in detail how we come up with these final answers to 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
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