Description
- Solve the following problems from Chapter 4 – “Series Circuits” of Grob's Basic Electronics textbook.
- SECTION 4–1 WHY CURRENT (I) IS THE SAME IN ALL PARTS OF A SERIES CIRCUIT
- Problem 4-2
- SECTION 4–2 TOTAL RESISTANCE (R) EQUALS THE SUM OF ALL SERIES RESISTANCES
- 4-4 to 4-8 (Even problems only)
- SECTION 4–3 SERIES (IR) VOLTAGE DROPS
- 4-10 to 4-14 (Even problems only)
- SECTION 4-4 KIRCHHOFF’S VOLTAGE LAW (KVL)
- 4-16 to 4-20 (Even problems only)
- SECTION 4-5 POLARITY OF IR VOLTAGE DROPS
- Problem 4-22
- SECTION 4-6 TOTAL POWER IN A SERIES CIRCUIT
- 4-24 to 4-28 (Even problems only)
- SECTION 4-7 SERIES–AIDING AND SERIES–OPPOSING VOLTAGES
- Problems 4-30 and 4-32
- SECTION 4–10 TROUBLESHOOTING: OPENS AND SHORTS IN SERIES CIRCUITS
- Problems 4-50 and 4-52
- SECTION 4–1 WHY CURRENT (I) IS THE SAME IN ALL PARTS OF A SERIES CIRCUIT
- 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.
I DO NOT HAVE TEXTBOOK...YOU MUST LOCATE THE PROBLEMS YOURSELF AND COMPLETE THEM
Explanation & Answer
Attached.
4-6
Solution
RT = R1+R2+R3
Putting values from figure we get
RT = 1K + 1.2K +1.8K
RT= 4KΩ
I = V/ RT
I = 24/4K = 6mA
4-8
Solution
RT = R1+R2+R3 +R4 + R5
Putting values from figure we get
RT = 22K +68K +10K+ 1M +100K
RT= 1.2 MΩ
I = V/ RT
I = 240/1.2M = 200µA
4-10
Solution
I = VT / R1+R2
I = 9/300 = 30mA
VR1 =voltage drop across R1
VR1= I X R1 = 30m X120 = 3.6V
VR2 =voltage drop across R2
VR2= I X R2 = 30m X180 = 5.4...
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