Basic Electric Circuits Tutorial 4
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This document contains tutorial problems about basic electric circuits, including resistors in series and parallel and current division. Problem 1 involves calculating the current through two resistors in series. Problem 2 involves calculating the voltage across resistors in parallel. Problem 3 involves finding the currents and voltage in a circuit with current division using multiple resistors in series and parallel. The document provides step-by-step workings and refers to examples from an electrical circuits textbook.
![Basic Electric Circuits – © 2020 Mouli Sankaran Email: mouli.sankaran@yahoo.com 6
T4_Problem_3: Current Division … contd.
Find i1 , i2 and v3 :
Note: This is Practice problem 3.15 on page 63, Figure 3.36 in the Ref 1 book (by Hayt)
100mA, 50 mA and 0.8 V
i1 = [125 /(125 + 25)] 120*10-3
i40
25 Ω
i1120 mA
i1 = [125 /(125 + 25)] 120*10-3
i1 = 100 mA
50 Ω
i2100 mA
i2 = [50 /(50 + 50)] 100*10-3
i2 = 50 mA
i40 = [240 /(240 + 60)] 50*10-3
i40 = 40 mA
v3 = i40* 20 = 40*10-3 * 20 = 0.8 V](https://image.slidesharecdn.com/becmoulisankarantutorial4-200502160342/85/Basic-Electric-Circuits-Tutorial-4-6-320.jpg)
Basic Electric Circuits Tutorial 4
- 1. Basic Electric Circuits – © 2020 Mouli Sankaran Email: mouli.sankaran@yahoo.com
- 2. Basic Electric Circuits – © 2020 Mouli Sankaran Email: mouli.sankaran@yahoo.com 2 Tutorial 4: Focus Resistors in Series and Parallel Current Division
- 3. Basic Electric Circuits – © 2020 Mouli Sankaran Email: mouli.sankaran@yahoo.com 3 T4_Problem_1: Resistors in Series Use source and resistor combinations to find i: Draw the simplified circuit too. Note: This is Practice problem 3.12 on page 57, Figure 3.28 in the Ref 1 book (by Hayt) -333 mA 15 = -i 45 i = -15/45 = -0.333 A + - 5 5V + - 5V 15 25 5 i 15 45Ω i
- 4. Basic Electric Circuits – © 2020 Mouli Sankaran Email: mouli.sankaran@yahoo.com 4 T4_Problem_2: Resistors in Parallel Find vx: Note: This is Practice problem 3.14 on page 60, Figure 3.32 in the Ref 1 book (by Hayt) 2.819 V Requ = 6 || 9 || R3equ Requ = 6 || 9 || 13 = 6*9/(6+9) || 13 Combine the resistors: R3equ = 3 * 3/(3+3) +10+ 3 * 3/(3+3) 1 A Requ = 2.819 Ω i + - vx R3equ = 13 Ω Requ = 3.6 || 13 = 3.6*13/(3.6+13) Requ =2.819 Ω
- 5. Basic Electric Circuits – © 2020 Mouli Sankaran Email: mouli.sankaran@yahoo.com 5 T4_Problem_3: Current Division Find i1 , i2 and v3 : Note: This is Practice problem 3.15 on page 63, Figure 3.36 in the Ref 1 book (by Hayt) RS2equ = 40 + 20 = 60 Ω Combine resistors connected to 50 Ω : RPequ = 240 || 60 = 48 Ω RS1equ = 2 + 48 = 50 Ω Requ = 50 || 50 = 25 Ω i40 25 Ω i1120 mA
- 6. Basic Electric Circuits – © 2020 Mouli Sankaran Email: mouli.sankaran@yahoo.com 6 T4_Problem_3: Current Division … contd. Find i1 , i2 and v3 : Note: This is Practice problem 3.15 on page 63, Figure 3.36 in the Ref 1 book (by Hayt) 100mA, 50 mA and 0.8 V i1 = [125 /(125 + 25)] 120*10-3 i40 25 Ω i1120 mA i1 = [125 /(125 + 25)] 120*10-3 i1 = 100 mA 50 Ω i2100 mA i2 = [50 /(50 + 50)] 100*10-3 i2 = 50 mA i40 = [240 /(240 + 60)] 50*10-3 i40 = 40 mA v3 = i40* 20 = 40*10-3 * 20 = 0.8 V
- 7. Basic Electric Circuits – © 2020 Mouli Sankaran Email: mouli.sankaran@yahoo.com 7 References Ref 1 Ref 2