Chapter 1. Basic concepts
ENGG 1008
Dr. K. K. Y. Wong
HKUEEE
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Motivation
• What is circuit analysis?
– To predict how a circuit behave without implementing the circuit
• For a simple circuit, that’s easy
– i=3/30=0.1A
30Ω
3V
• How about this?
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Systems of Units
e.g., 600,000,000mm=600,000m=600km
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Charge and Current
• Facts about charges:
–
–
–
–
–
–
The fundamental building block of all matters is atom
Each atom consists of electrons, protons, and neutrons
Electron carries -ve charge and proton carries +ve charge
The unit of charge is coulombs (C)
1C of –ve charge = 6.24×1018 electrons
Or equivalently one electron = 1.602 ×10-19 C
• What is current?
– Net movement of charges
– Due to historical reason, current flow in the same direction as +ve
charge movement
– Although we know that current in metallic conductors is due to –ve
charged electrons, we still follow the convention today
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• The amplitude of current is the rate of change of charge
(w.r.t. time across a fixed surface)
dq (t )
• Mathematically, it is given by i (t ) dt
• 1 ampere = 1 coulomb of charge moving across a fixed
surface in 1 second
• The charge transferred between time to and t is given by
t
the inverse relationship:
Q ∫t i (t )dt
o
• Example: The total charge entering a terminal is given by
q(t)=5tsin4πt mC. Calculate the current at t=0.5s
dq (t ) d
= (5t sin 4π t ) = (5sin 4π t + 20π t cos 4π t )mA
i (t ) =
dt
dt
i (0.5) = 5sin 2π + 10π cos 2π = 31.42mA
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• Example: Find the current at (a) t=1ms, (b)
t=6ms, (c) t=10ms, if the charge entering a
certain element is shown in the figure.
–
–
–
–
Current = slope of the curve
(a) i=80/2=40A,
(b) i=0,
(c) i=-80/4=-20A
• Note: negative current means opposite
direction to the reference
• Example: Determine the total charge
entering a terminal between t=1s and t=2s if
the current passing the terminal is i(t)=(3t2t)A.
2
2
Q = ∫1 i (t )dt = ∫1 (3t 2 − t )dt
(
3
2
= t −t /2
) 1 = (8 − 2) − (1 − 1/ 2) = 5.5C
2
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Voltage
• To move an electron from point a to point b, external
electromotive force (emf), typically a battery, is needed
• The voltage vab between two points a and b is the energy
needed to move a unit charge from a to b
dw
• Mathematically, vab dq
• 1 volt = 1 joule / coulomb
• Two equivalent representations:
– Point a is vab=+9V above point b
– Point b is vba=-9V above point a
– In general, vab= -vba
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Power and Energy
• Power is the rate (w.r.t. time) of expending or
absorbing energy, measured in watts (W)
• Mathematically, p(t ) dw(t )
dt
• Rewrite the above equation,
p (t ) =
dw(t ) dq(t )
= v(t )i (t )
dq dt
• Current direction and voltage polarity
play a major role in determining the
sign of power
–When the current enters through the +ve
terminal, p=vi, the element absorbs power
–When the current enters through the –ve
terminal, p=-vi, the element supply power
• The above rule is called the “passive
sign convention”
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• Example: Element
absorbing +12W
• Example: Element supplying
+12W, or absorbing –12W
• Note: +power absorbed = -Power supplied
• Energy absorbed or supplied by an element from time to to
t1:
t1
t1
w = ∫t p (t )dt = ∫t v(t )i (t )dt
o
o
• Note: If p(t) is a constant, w = p(t1-to)
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• Example: Find the power delivered to an element at t=3ms if
the current entering its positive terminal is i(t)=5cos60πtA
and the voltage is: a) v=3i, b) v=3di/dt
– a)
p(t) = v(t)i(t) = 3i(t)2 = 75cos260πt W
=> p(3ms) = 75cos20.18π = 53.48 W
– b)
v(t) = 3di/dt = -900πsin60πt
=> p(t) = v(t)i(t) = -4500πsin60πtcos60πt W
=> p(3ms) = -4500πsin0.18πcos0.18π = - 6.396kW
• Example: How much energy does a 100W electric bulb
consume in two hours?
– Since p is a constant =100, so w=pt=100×2×3600=720kJ
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Circuit Elements
•
Passive elements: not capable of generating energy
– e.g., resistors, capacitors, and inductors
•
Active elements: capable of generating energy
– e.g., generators, batteries, and operational amplifiers
• The most important active elements are voltage and
current sources (sources can be dependent or independent)
– The ideal voltage source will produce any current required
to ensure that the terminal voltage is as stated
– The ideal current source will produce the necessary voltage
to ensure the stated current flow
–Ideal independent sources: provides a specified voltage or
current that is completely independent of other circuit
elements
– Ideal dependent sources: source quantity is controlled by
another voltage or current
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• Example: Calculate the power supplied
or absorbed by each element
– For p1, the 5A current is out of the +ve
terminal; hence p1=-vi=-20(5)=-100W
– For p2 and p3, the current flows into the +ve
terminal, so p2=vi=12(5)=60W and
p3=vi=8(6)=48W
– For p4 , note that the voltage is 8V with +ve
terminal at the top. Since the current flow
out of the +ve terminal, we have
p4=-vi=-8(0.2I)= -8(0.2×5)=-8W
• The 20V independent voltage source and 0.2I dependent
current source are supplying power, while the two passive
elements are absorbing power
• Also note that p1+ p2 + p3 + p4=-100+60+48-8=0. This is the
law of conservation of energy
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