This document contains notes from a tutorial on basic circuit analysis. It introduces circuit modelling using components like resistors, capacitors and inductors. It defines key circuit variables like current, voltage and power. It also covers circuit analysis techniques like Kirchhoff's voltage and current laws, and applying the passive sign convention to analyze power flow in circuits. Several practice problems are worked through applying these concepts to calculate values like charge, current, voltage, power and energy in various circuits.

1. ELEC221: Tutorial 1
Mina Wahib
9/22/2015 Mina Wahib

2. Introduction
Text book
9/22/2015 Mina Wahib
Electric Circuits (Newest Edition) 10th Edition
by James W. Nilsson (Author), Susan Riedel (Author)

3. Introduction
Simple Circuit Modelling (1)
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 Electrical systems can be modelled by using circuit models
 Circuits are the interconnection of electrical components such as resistors, capacitors, inductors
and switches

4. Introduction
Simple Circuit Modelling (2)
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Sources Cables Electrical
Equipment

5. Introduction
Electrical Components
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Circuits Elements
Passive Active
+
-

6. Introduction
Circuit Variables
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 Current (I) : Time-rate of change of electric charge (1 Ampere = 1 Coulomb/ 1 Sec.)
𝐼 =
𝑑𝑞
𝑑𝑡
𝐴
 Voltage (V) : Electromotive force or potential (1 Volt = 1 Joule (N.m) / 1 Coulomb)
𝑉 =
𝑑𝑊
𝑑𝑞
𝑉
 Power (P) : Time-rate of change of energy (1 Watt = 1 Joule / 1 Sec.)
𝑃 =
𝑑𝑊
𝑑𝑡
= 𝐼 𝑉 𝑊𝑎𝑡𝑡

7. Introduction
Passive Sign Convention
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Circuit
Element
1 2+ -
𝐼
𝑉
𝑷 = 𝑰 𝑽
𝑃 > 0
Power Absorbed
𝑃 < 0
Power Supplied

8. Introduction
Power Balance + Passive Sign Convention
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-
𝑃𝑎 = 120 −10 = −1200 𝑊
𝑃𝑏 = − 120 9 = −1080 𝑊
𝑃𝑐 = 10 10 = 100 𝑊
𝑃𝑑 = − 10 −1 = 10 𝑊
𝑃𝑒 = −10 −9 = 90 𝑊
𝑃𝑓 = − −100 5 = 500 𝑊
𝑃𝑔 = 120 4 = 480 𝑊
𝑃ℎ = −220 −5 = 1100 𝑊
𝑃𝑎𝑏𝑠 = 2280 𝑊 𝑃𝑠𝑢𝑝𝑝 = −2280 𝑊 𝑃𝑛𝑒𝑡 = 0 𝑊+ = Power Balanced !

9. Introduction
KVL and KCL
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KVL: The sum of voltage drops around a closed path must equal zero
KCL: The sum of currents leaving a node must equal zero.
−𝑉𝑎 + 𝑉2 + 𝑉𝑏 = 0 𝑃𝑎𝑡ℎ 1
−𝑉𝑏 − 𝑉3 + 𝑉𝑐 = 0 (𝑃𝑎𝑡ℎ 2)
−𝑉𝑎 + 𝑉2 − 𝑉3 + 𝑉𝑐 = 0 (𝑃𝑎𝑡ℎ 3)

10. Introduction
KVL and KCL
9/22/2015 Mina Wahib
KVL: The sum of voltage drops around a closed path must equal zero
KCL: The sum of currents leaving a node must equal zero.
−4 + 𝑖 + 10 = 24
𝑖 = 18 𝐴

11. Problems
Problem 1
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Solution
𝑞 =
0
∞
𝑖 𝑑𝑡 =
0
∞
20𝑒−5000𝑡 𝑑𝑡 =
20𝒆−𝟓𝟎𝟎𝟎𝒕
−𝟓𝟎𝟎𝟎
∞
0
= 4000 𝜇𝐶
Remember that’s the integration of an exponential
It’s the same exact exponential divided by the
derivative of the power of the exponential

12. Problems
Problem 2
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Solution
𝑎 𝑃 = 𝑉 𝐼 = 10,000𝑒−5000𝑡
20𝑒−5000𝑡
𝑃 0.001 = 0.908𝑊
𝑏 𝑤 𝑡 =
0
∞
𝑝 𝑡 𝑑𝑡 = 20 𝐽
Remember that’s the integration of an exponential
It’s the same exact exponential divided by the
derivative of the power of the exponential

13. Problems
Problem 3
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Remember that this could be a non-constant
function (linear 1st order) as well, and in that case
multiplying it by another non-constant function
(linear 1st order) will result in a 2nd order function

14. Problems
Problem 3 (Cont’d)
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Solution
𝑤 1 =
1
2
1 100 = 50 𝐽
𝑤 6 =
1
2
1 100 −
1
2
1 100 +
1
2
1 100 −
1
2
1 100 = 0𝐽
𝑤 10 = 𝑤 6 +
1
2
1 100 = 50 𝐽
Area of the triangles under the curve

15. Problems
Problem 4
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Solution
𝑎 𝑝 = 0.25𝑒−3200𝑡
− 0.5𝑒−2000𝑡
+ 0.25𝑒−800𝑡
𝑝 625 𝜇𝑠 = 42.2 𝑚𝑊
𝑏 𝑤 =
0
625 𝜇𝑠
0.25𝑒−3200𝑡
− 0.5𝑒−2000𝑡
+ 0.25𝑒−800𝑡
𝑑𝑡
= 12.14 𝜇𝐽
𝑐 𝑤 =
0
∞
0.25𝑒−3200𝑡 − 0.5𝑒−2000𝑡 + 0.25𝑒−800𝑡 𝑑𝑡
= 140.625 𝜇𝐽

16. Problems
Problem 5
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Solution
𝑖 =
𝑑𝑞
𝑑𝑡
= 𝑡𝑒−𝛼𝑡
𝒅𝒊
𝒅𝒕
= 𝟎 𝒇𝒐𝒓 𝒎𝒂𝒙. 𝒄𝒖𝒓𝒓𝒆𝒏𝒕
1 − 𝛼𝑡 𝑒−𝛼𝑡 = 0
∴ 𝑡 =
1
𝛼
∴ 𝑖 =
1
𝛼
𝑒−1 ≅ 10 𝐴

17. Problems
Problem 6
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Solution
𝑎 𝑃 = 𝑉 𝐼 = 80,000𝑡𝑒−500𝑡
15𝑡𝑒−500𝑡
𝑑𝑃
𝑑𝑡
= 𝑒−1000𝑡 2𝑡 − 1000𝑡2 = 0
𝑡 = 2 𝑚𝑠𝑒𝑐
𝑏 𝑃 2 𝑚𝑠𝑒𝑐 = 649.6 𝑚𝑊
𝒄 𝑬 𝒕𝒐𝒕𝒂𝒍 =
𝟎
∞
𝑷 𝒕 𝒅𝒕

18. Problems
Problem 6
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Solution (Cont’d)
𝑐 𝐸𝑡𝑜𝑡𝑎𝑙
= 1200,000
0
∞
𝑡2 𝑒−1000𝑡 𝑑𝑡 = 1200,000 {(
𝒆−𝟏𝟎𝟎𝟎𝒕
−𝟏𝟎𝟎𝟎
𝒕 𝟐) −
0
∞
𝒆−𝟏𝟎𝟎𝟎𝒕
−𝟏𝟎𝟎𝟎
𝟐𝒕} = 1200,000{
𝑒−1000𝑡
−10002 2𝑡 −
0
∞
𝑒−1000𝑡
−10002 2}
= 1200,000
−2
1000
0 −
1
1000 2
= 2.4 𝑚𝐽
Remember that’s the integration by parts
(Integration of the simpler function X the second function as it it) – (Integration of the function I
already integrated X Derivative of the second function)

19. Problems
Problem 7
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𝑃1 = −205 𝑊
𝑃2 = 60 𝑊
𝑃3 = 45 𝑊
𝑃4 = 45 𝑊
𝑃5 = 30 𝑊
𝑃3 =? 𝑃3 = 70 𝑊 (𝑅𝑒𝑐𝑒𝑖𝑣𝑒𝑑 𝑃𝑜𝑤𝑒𝑟)

20. Problems
Problem 8
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𝑃1 = −300 𝑊
𝑃2 = 100 𝑊
𝑃3 = 280 𝑊
𝑃4 = −32 𝑊
𝑃5 = −48 𝑊

21. Problems
Problem 9
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𝑃𝑡𝑜𝑡𝑎𝑙 = −100 + 20𝐼0 + 36 + 24 = 0 𝑊
𝐼0 = 2 𝐴

22. Problems
Problem 10
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𝑃𝑡𝑜𝑡𝑎𝑙 = −180 + 72 + 56 + 28 + 3𝑉0 − 30 = 0 𝑊
𝑉0 = 18 𝑉

23. Problems
Problem 11
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A constant current of 3 A for 4 hours is required to
charge an automotive battery. If the terminal voltage is
(𝟏𝟎 + 𝒕/𝟐 ) 𝑽, where t is in hours,
(a) How much charge is transported as a result of the
charging?
(b) How much energy is expended?
(c) How much does the charging cost? Assume
electricity costs 9 cents/kWh.
Solution
𝑎 𝑞 =
0
𝑇
3 𝑑𝑡 = 3 𝑇 = 3 × 4 × 60 × 60 = 43.2 𝐾𝐶
𝑏 𝑊 =
0
𝑇
𝑝 𝑑𝑡 =
0
𝑇
(3)(10 +
𝑡
2
) 𝑑𝑡
= 3 10𝑡 +
𝑡2
4
= 132 𝑊ℎ
𝑐 𝐶𝑜𝑠𝑡 = 9
𝑐𝑒𝑛𝑡𝑠
𝑘𝑊ℎ
× 0.132 𝑘𝑊ℎ = 1.188 𝐶𝑒𝑛𝑡𝑠

24. Problems
Problem 12
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Solution
𝑎 𝑞 =
0
2
𝟑𝒆−𝟐𝒕 𝑑𝑡 = −
3
2
𝑒−4 − 1 = 1.472 𝐶
𝑏 𝑃 = 𝑉 𝐼 = −90𝑒−4𝑡 𝑊
𝑐 𝑊 =
0
3
−90𝑒−4𝑡 𝑑𝑡 =
90
4
𝑒−12 − 1 = −22.49 𝐽
The current entering the positive terminal of a device is
𝒊(𝒕) = 𝟑𝒆−𝟐𝒕
𝑨 and the voltage across the device is
𝒗(𝒕) = 𝟓 𝒅𝒊/𝒅𝒕 𝑽.
(a) Find the charge delivered to the device between t = 0
and t = 2 s.
(b) Calculate the power absorbed.
(c) Determine the energy absorbed in 3 s.

25. Problems
Problem 13
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Solution
𝑎 𝑞 =
0
1
10(1 − 𝑒−0.5𝑡 )𝑑𝑡 = 2.1306 𝐶
𝑏 𝑃 = 𝑣 𝑡 𝑖 𝑡 = 5𝑐𝑜𝑠𝑡 2𝑡 × 10 1 − 𝑒−0.5𝑡
= 5 cos 2 𝑟𝑎𝑑 × 10 1 − 𝑒−0.5
= −8.187 𝑊
The voltage v across a device and the current i through it
are 𝑣(𝑡) = 5 cos 2𝑡 𝑉, 𝑖(𝑡) = 10(1 − 𝑒−0.5𝑡 ) 𝐴
Calculate:
(a) The total charge in the device at t = 1 s
(b) The power consumed by the device at t = 1 s.

26. Problems
Problem 14
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Find 𝑉 using KVL and KCL
Solution:
KVL around the blue path:
−18 − 1 6 + 2 3 + 4 4 − 𝑉 = 0 ∴ 𝑉 = −2 𝑉

27. Problems
Problem 15
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Find 𝐼1, 𝐼2 using KVL and KCL
Solution:
KCL at a: 𝐼1 + 𝐼2 + 4 = 0
KCL at b: 𝐼2 + 𝐼3 + 4 = 6
KCL at c: 𝐼1 + 6 = 𝐼3 KVL around blue path: −12 + 3𝐼1 + 6𝐼1 + 9𝐼3 = 0 ∴ 𝐼1 = −
7
3
𝐴 , 𝐼2 = −
5
3
𝐴

28. Important Remarks
9/22/2015 Mina Wahib
 Don’t forget to revise the integration rules (integration by parts and the integration of other different
functions)
 Don’t forget to convert all units to the fundamental SI units (A, V, S, W….)
 Don’t forget to revise the differentiation rules (exponentials, sins, cosines…..)
 Email me whenever you have any questions at 13mmw@queensu.ca or Minawahib@Hotmail.com
Or directly come to my office at WLH506, Cubicle #9