1. Chapter 1 – Circuit Variables
Electrical Engineering Overview
International System of Units
Overview of Circuit Analysis
Voltage and Current
Ideal Elements
Power and Energy

2. Overview
Communication Systems – generate and transmit information (TVs, Radios)
Computer Systems – Process information (mathematical computations, simulation, software)

3. Control Systems – Regulate processes (PLC’s, Assembly line)

4. Power Systems – Generate and distribute electrical power (Turbines, Wind, Coal)
Signal Processing – Translates signals to a more suitable form (Xrays, CT Scans, Radar, Sonar)

5. Circuit Theory – A Mathematical Model that approximates the behavior of an actual electrical system.
3 basic Assumptions:
Electrical effects happen instantaneously throughout a system (lumped parameter system)
The net Charge on every component in the system is always zero
There is no magnetic coupling between components in a system

6. Problem Solving Identify what’s given and what’s to be found
Sketch a circuit diagram or model
Think of multiple solution method and decide on a way of choosing them
Calculate a solution
Does your answer make sense
Test your solution

7. International System of Units

8. International System of Units

9. International System of Units

10. Example : If a Signal can travel at 80% of the speed of light, what is the length of cable, in inches, represents 1 ns?
𝑐=3 × 𝑚 𝑠
1 𝑛𝑠= 10 −9 𝑠
0.8𝑐= × =2.4× 𝑚 𝑠
80% of the speed of light
Conversions
2.4× 𝑚𝑒𝑡𝑒𝑟𝑠 1 𝑠𝑒𝑐𝑜𝑛𝑑 ∙ 1 𝑠𝑒𝑐𝑜𝑛𝑑 𝑛𝑎𝑛𝑜𝑠𝑒𝑐𝑜𝑛𝑑𝑠 ∙ 100 𝑐𝑒𝑛𝑡𝑖𝑚𝑒𝑡𝑒𝑟𝑠 1 𝑚𝑒𝑡𝑒𝑟 ∙ 1 𝑖𝑛𝑐ℎ 2.54 𝑐𝑒𝑛𝑡𝑖𝑚𝑒𝑡𝑒𝑟𝑠
= (2.4× 10 8 )(100) ( 10 9 )(2.54) =9.45 𝑖𝑛𝑐ℎ𝑒𝑠 𝑛𝑎𝑛𝑜𝑠𝑒𝑐𝑜𝑛𝑑

11. Overview of Circuit Analysis

12. Voltage and Current
Electric Charge is the basis for describing all electrical Phenomena
Charge is bipolar (electrical effects are positive and negative)
Charge exists in discrete quantities (1.6022× 10 −19 𝐶)
Electrical effects are due to both separation of charge and charges in motion

13. Voltage and Current
Voltage – is the energy per unit charge created by the separation
𝑣= 𝑑𝑤 𝑑𝑞
v is the voltage in Volts
w is the energy in Joules
q is the charge in Coulombs

14. Voltage and Current Current – is the rate of charge flow 𝑖= 𝑑𝑞 𝑑𝑡
i is the current in Amperes
q is the charge in Coulombs
t is the time in Seconds

15. Ideal Basic Circuit Element
Does not exist as a realizable physical element, but is used to model devices and systems
It has two terminals
Described in terms of current and/or voltage
It cannot be subdivided into other elements

16. Ideal Basic Circuit Element
Passive Sign Convention:
Whenever the reference direction for the current in a element is in the direction of the reference voltage drop across the element as shown, use a positive sign in any expression that relates voltage and current. Else, us a negative sign.

17. Ideal Basic Circuit Element
Passive Sign Convention:

18. Example: No charge exists at the upper terminal of the element for t <0. At t = 0, a 5A current begins to flow into the upper terminal.
Derive the expression for the charge accumulating at the upper terminal of the element for t > 0.
If the current is stopped after 10 seconds, how much charge has accumulated at the upper terminal?
A) Using 𝑖= 𝑑𝑞 𝑑𝑡 𝑑𝑞=𝑖 𝑑𝑡 so we must integrate so 𝑞 𝑡 = 0 𝑡 𝑖 𝑥 𝑑𝑥
So 𝑞 𝑡 = 0 𝑡 5𝑑𝑥 =5𝑡−5 0 =5𝑡 𝐶 𝑓𝑜𝑟 𝑡>0
B) The total charge that accumulates at the upper terminal in 10 seconds due to 5A current is 𝑞 10 =5 10 =50𝐶

19. Ideal Basic Circuit Element
Assessment Problem 1.3 on page 14:
The current at the terminals is
𝑖=0, 𝑡<0
𝑖=20 𝑒 −5000𝑡 𝐴, 𝑡≥0
Calculate the total charge (in microcoulombs) entering the element at its upper terminal
𝑖= 𝑑𝑞 𝑑𝑡
𝑞 𝑡 = 0 ∞ 20 𝑒 −5000𝑡 𝑑𝑡
𝑑𝑞=𝑖𝑑𝑡
To solve for q(t) we must integrate
𝑙𝑒𝑡 𝑢=−5000𝑡
𝑑𝑢=−5000𝑑𝑡
20 0 ∞ 𝑒 𝑢 𝑑𝑢 − = 20 − ∞ 𝑒 𝑢 𝑑𝑢
𝑑𝑡= 𝑑𝑢 −5000 ∞
20 −5000 𝑒 𝑢
= 20 − 𝑒 ∞ − 𝑒 0
= 20 − −1 =0.004𝐶
=4000𝜇𝐶

20. Ideal Basic Circuit Element
Assessment problem 1.4 in book page 14
The expression for charge entering the upper terminal is
𝑞= 1 𝛼 2 − 𝑡 𝛼 + 1 𝛼 2 𝑒 −𝛼𝑡 𝐶
Find the maximum value of the current entering the terminal if 𝛼= 𝑠 −1
𝑖= 𝑑𝑞 𝑑𝑡
Re-write q so that 𝑞= 1 𝛼 2 − 𝑡 𝛼 𝑒 −𝛼𝑡 − 𝑒 −𝛼𝑡 𝛼 2
The maximum occurs when the derivative of i is equal to zero
𝑑𝑞 𝑑𝑡 =0− 𝑡 𝛼 𝑒 −𝛼𝑡 −𝛼 + 𝑒 −𝛼𝑡 1 𝛼 −(−𝛼) 𝑒 −𝛼𝑡 𝛼 2
0=−𝛼𝑡 𝑒 −𝛼𝑡 + 𝑒 −𝛼𝑡
0= −𝛼𝑡+1 𝑒 −𝛼𝑡
𝑑𝑞 𝑑𝑡 =0+𝑡 𝑒 −𝛼𝑡 − 𝑒 −𝛼𝑡 𝛼 + 𝑒 −𝛼𝑡 𝛼
0= −𝛼𝑡+1
𝑡= 1 𝛼
𝑑𝑞 𝑑𝑡 = 𝑡− 1 𝛼 + 1 𝛼 𝑒 −𝛼𝑡
Substitute
𝑖= 1 𝛼 𝑒 −𝛼 1 𝛼
𝑖= 𝑒 −1 𝛼
𝑖= 𝑒 − =9.99𝐴 𝑜𝑟 10𝐴
𝑑𝑞 𝑑𝑡 =𝑖=𝑡 𝑒 −𝛼𝑡

21. w is the energy in Joules
Power and Energy
Power – The time rate of expending or absorbing energy
𝑝= 𝑑𝑤 𝑑𝑡
p is the power in Watts
w is the energy in Joules
T is the time in seconds

22. Where v is voltage and i is current
Power and Energy
Power also relates to Voltage and Current
𝑝= 𝑑𝑤 𝑑𝑡 = 𝑑𝑤 𝑑𝑞 𝑑𝑞 𝑑𝑡
or alternatively
𝑝=𝑣𝑖
Where v is voltage and i is current

23. Power and Energy Use the passive sign relationship
If the power is positive (if p >0), power is being delivered to the circuit inside the box. If power is negative (p <0), power is being extracted from the circuit inside the box.
Ex: Using polarity reference (b) for I = 5A and V= -12V
P= -(-12)(5) = 60W

24. Calculate the power supplied to the element at 1 ms.
Example: Assume that the voltage at the terminals of the element shown, whose current was defined as:
𝑣=0 𝑖= 𝑓𝑜𝑟 𝑡<0
𝑣=10 𝑒 −5000𝑡 𝑘𝑉 𝑖=20 𝑒 −5000𝑡 𝑓𝑜𝑟 𝑡≥0
Calculate the power supplied to the element at 1 ms.
Calculate the total energy (in Joules) delivered to the circuit element.
A) 𝑝=𝑣𝑖= 𝑒 −5000𝑡 20𝑒 −5000𝑡 = 𝑒 −10000𝑡 𝑊
At 1 ms 𝑝 = 𝑒 −10000 (0.001) = 𝑒 −10 =9.079 𝑊
B) Using p= 𝑑𝑤 𝑑𝑡 𝑑𝑤=𝑝 𝑑𝑡 so we must integrate so w 𝑡 = 0 𝑡 𝑝 𝑥 𝑑𝑥
𝑤 𝑡𝑜𝑡𝑎𝑙 = 0 ∞ 𝑒 −10000𝑥 𝑑𝑥= 𝑒 −10000(∞) −10000 − 𝑒 −10000(0) −10000
=−20 𝑒 −∞ − −20 𝑒 −0 =0+20=20𝐽

25. Practical Perspective
Balancing Power – If power is balanced then, total power supplied + total power absorbed = 0.
For p>0 (positive), power is being absorbed
For p<0 (negative), power is being supplied

26. Practical Perspective
Circuit model for power distribution in a home.
Calculate the power at each location using the voltages and currents on the next slide and describe if it is supplying or absorbing. Keep in mind passive sign convention.
Is it balanced?

27. Practical Perspective
Is it balanced?

28. Practical Perspective
𝑝 𝑎 = 𝑣 𝑎 𝑖 𝑎 = 120 −10 =−1200𝑊
𝑝 𝑏 = −𝑣 𝑏 𝑖 𝑏 =− =−1080𝑊
𝑝 𝑐 = 𝑣 𝑐 𝑖 𝑐 = =100𝑊
𝑝 𝑑 =− 𝑣 𝑑 𝑖 𝑑 =− =−10𝑊
𝑝 𝑒 = 𝑣 𝑒 𝑖 𝑒 = −10 −9 =90𝑊
𝑝 𝑓 =− 𝑣 𝑓 𝑖 𝑓 =− =500𝑊
𝑝 𝑠𝑢𝑝𝑝𝑙𝑖𝑒𝑑 =−2290𝑊 𝑝 𝑎𝑏𝑠𝑜𝑟𝑏𝑒𝑑 =2270𝑊
𝑝 𝑔 = 𝑣 𝑔 𝑖 𝑔 = =480𝑊
𝑝 ℎ = 𝑣 ℎ 𝑖 ℎ = −220 −5 =1100𝑊
𝑝 𝑠𝑢𝑝𝑝𝑙𝑖𝑒𝑑 + 𝑝 𝑎𝑏𝑠𝑜𝑟𝑏𝑒𝑑 =−2290𝑊+2270𝑊=−20𝑊
Not Balanced – indicating that there might be an error with
current or voltage