Description
(you must find the problems to solve them, I do not have the textbook)
- Solve the following problems from Chapter 9 – “Kirchhoff’s Laws” of Grob's Basic Electronics textbook.
- SECTION 9–1 Kirchhoff’s Current Laws
- Problems 9-2 and 9-4
- SECTION 9–2 Kirchhoff’s Voltage Laws
- 9-6 to 9-14 (Even problems only)
- SECTION 9–3 METHOD OF BRANCH CURRENTS
- Problem 9-16
- SECTION 9-4 Node Voltage Analysis
- Problems 9-18 and 9-20
- SECTION 9–1 Kirchhoff’s Current Laws
- Show all work for full credit in a word document or on a paper and scan your work. Save all your work as a word or a pdf file format.
- Save file as “HW7_StudentID” with your student id substituted into the file name and upload it.
Explanation & Answer
Attached.
Chapter 9 – “Kirchhoff’s Laws” of Grob's Basic Electronics
SECTION 9–1 Kirchhoff’s Current Laws - Problems 9-2 and 9-4
9-2
Applying Kirchhoff’s current law, write an equation for the currents directed into
and out of point X in Prob. 9–1.
Answer: 2mA+9mA+I 3 +3mA-20mA=0
9-4
In Fig. 9–11, solve for the following unknown currents: I 3 , I 5 , and I 8 .
6A+11A+I 3 -25A=0
I 3 =25A-6A-11A
Answer: I 3 = 8A
2A+7A+I 5 -25A=0
I 5 =25A-2A-7A
Answer: I 5 = 16A
5A+I 8 -16A=0
I 8 =16A-5A
Answer: I 8 = 9A
SECTION 9–2 Kirchhoff’s Voltage Laws - 9-6 to 9-14 (Even problems only)
9-6
Write a KVL equation for the loop CEFDC going clockwise from point C
.
Answer: 4.5V+5.4V+8.1V-18V=0
Write a KVL equation for the loop ACDBA going clockwise from point A.
Answer: 6V+18V+12V-36V=0
Write a KVL equation for the loop ACEFDBA going clockwise from point A.
Answer: 6V+4.5V+5.4V+8.1V+12V-36V=0
9-8
In Fig. 9–13, solve for the voltages V AG and V BG. Indicate the proper polarity for
each voltage.
VAG = -10V+15V
VAG = 5V
VBG = 10V – 20V
VBG = -10V
9-10 In Fig. 9–13, solve for the voltages V AG and V BG. Indicate the proper polarity for
each voltage.
VAG = -3V+10V
VAG = 7V
VBG = 5V – 15V
V BG = -10V
9-12 In Fig. 9–13, solve for the voltages V AG and V BG. Indicate the proper polarity for
each voltage.
VAG = -100V+50V
V AG = -50V
VBG = 50V+40V–50V
V BG = 40V
VCG = 40V-50V
V CG = -10V
9-14 Solve for the voltages V AG, V BG, and V CG .Indicate the proper polarity for
each voltage.
Answer: -10V-2.5V+12.5V=0
SECTION 9–3 METHOD OF BRANCH CURRENTS - Problem 9-16
9-16 using the method of branch currents, solve for the unknown values of voltage and
current in Fig. 9–19. To do this, complete steps a through m. The assumed direction
Figure 9–16 of all currents is shown in the figure.
a) Using Kirchhoff’s current law, write an equation for the currents I1 , I2 , and I3 at point C.
Answer: I 2 +I 3 -I 1=0
b) Specify the current I3 in terms of I1 and I2 .
Answer: I 3 =I 1 -I 2
c) Writ...