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Lecture 2
Fundamentals of Electrical
Engineering
Introduction
1. Define current, voltage, and power,
including their units.
2. Calculate power and energy, as well as
determine whether energy is supplied or
absorbed by a circuit element.
3 State
3.
St t and
d apply
l b
basic
i circuit
i it llaws.
4. Solve for currents, voltages, and powers in
simple circuits.
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Electrical Current
Electrical current is the time rate of flow of
electrical charge through a conductor or
circuit element. The units are amperes (A),
which are equivalent to coulombs per
second (C/s).
Electrical Current
i (t ) =
dqq (t )
dt
t
q (t ) = ∫ i (t )dt + q (t0 )
t0
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Dissipates energy in
the form of heat
One example is a
battery that stores
i the
th form
f
off
energy in
an electrochemical
reaction
Stores energy in
a magnetic field
Stores energy
gy in
an electric field
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Direct Current
Alternating Current
When a current is constant with time, we
say that we have direct current,
abbreviated as dc. On the other hand, a
current that varies with time, reversing
direction periodically, is called alternating
current, abbreviated as ac.
.
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Voltages
The voltage associated with a circuit
element is the energy transferred per unit of
charge that flows through the element. The
units of voltage are volts (V), which are
equivalent to joules per coulomb (J/C).
“uphill: battery”
“downhill: resistor”
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Power and Energy
p(t ) = v (t )i (t )
Watts
t2
w = ∫ p(t )dt
Joules
t1
Current is flowing in the
passive configuration.
Energy is being
absorbed by the element
(e.g. a resistor)
If the current flows opposite to the passive
configuration, the power is given by p = -vi
Energy is being provided by the element (e.g. a battery)
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Pa = iava = (2A)*(12V) = 24W, energy is being absorbed
Pb = -ibvb = -(1A)*(12V) = -12W, energy is being supplied
Pc = icvc = (-3A)*(12V) = -36W, energy is being supplied
Power and Energy
v(t) = 12 V
i( ) = 2e
i(t)
2 -tt A
p(t) = v(t)i(t) = 24e-t
∞
w=
∫
0
∞
∫
p (t )dt = 24e −t dt = −24e −∞ − (−24e 0 ) = 24 J
0
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An example would
be a battery.
An example would
be a transformer.
?
You don’t
want to do
d
this!
• An ideal voltage source has a voltage vx =12 V
independent of the load
• An ideal conductor requires that vx = 0
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Resistors and Ohm’s Law
a
b
v = iR
vab = iab R
The units of resistance are Volts/Amp which are called
“ohms”. The symbol for ohms is omega: Ω
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George Simon Ohm, 1789-1854
V
I=
R
In 1805 Ohm entered the University of Erlangen but he became rather
carried away with student life. Rather than concentrate on his studies he
spent much time dancing, ice skating and playing billiards.
http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Ohm.html
Conductance
1
G=
R
i = Gv
The units of conductance are 1/Ω or Ω-1. The units
are called “siemens”
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Resistance Related to Physical
Parameters
R=
ρL
A
ρ is the resistivity of the material used to fabricate
the resistor. The units of resitivity are ohm-meters
(Ω-m)
Power dissipation in a resistor
2
v
p = vi = Ri 2 =
R
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KIRCHHOFF’S CURRENT
LAW
• The net current entering a node is zero.
• Alternatively, the sum of the currents
entering a node equals the sum of the
currents
t leaving
l i a node.
d
Gustav Kirchhoff (1824-1887)
Kirchhoff's Current Law
The principle of conservation of electric charge
i li that:
implies
h
At any point in an electrical circuit where charge
density is not changing in time, the sum of
currents flowing towards that point is equal to the
sum of currents flowing away from that point.
A charge density changing in time would mean
the accumulation of a net positive or negative
charge which typically cannot happen to any
charge,
significant degree because of the strength of
electrostatic forces: the charge buildup would
cause repulsive forces to disperse the charges.
http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Kirchhoff.html
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((a)) Currents into the node = 1A+3A = 4A current out of the node ia
(b) Currents into the node = 3A+1A+ib = current out of the node =
2A so ib=2A-4A=-2A
(c) Currents into the node = 1A+ic+3A+4A = current out of the
node= 0 amps so ic =-8A
Series Connection
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Which elements are in series?
KIRCHHOFF’S VOLTAGE
LAW
The algebraic sum of the voltages equals
zero for any closed path (loop) in an
electrical circuit.
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Loop 1: -va+vb+vc = 0
Loop 2: -vc-vd+ve = 0
Loop 3: va-ve+vd-vb = 0
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Parallel Connection
KVL through A and B: -va+vb = 0 → va = vb
KVL through
h
h A and
dC
C: -va - vc = 0 → va = -vc
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-3V - 5V + vc = 0 → vc = 8V
-8V - (-10V) + ve = 0 → ve = -2V
Which elements are in series? Which elements are in parallel?
Find the current, voltage and power for each element:
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Power
PR = iRvR = (2A)(10V) = 20W (power dissipated)
= iR2R
= VR2/R
PS = -vSiS = -(10V)(2A) = -20W (power supplied)
PR+ PS = 0
Example
• What is the current iR flowing through the resistor?
• What is the power for each element in the circuit?
• Which elements are absorbing power?
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• Since all of the elements are in series, the same current iR= 2A
runs through each of them
• The voltage drop across the resistor:
vR= iR = (2A)(5Ω) = 10V
• Apply KVL:
vc = vR + 10 = 20V
• The power dissipated in each element:
pR = iR2R = (2A)2(5Ω) = 20W (absorbing)
pvs= iv = (2A)(10V) = 20W (absorbing)
pcs= -iv = -(2A)(20V) = 40W (supplying)
Example
Use Ohm
Ohm’ss law
law, KVL and KCL to find Vx
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IT
R1
IR2
IR3
R2
R3
R4
Ohm’s Law:
VR4 = (1A)(5Ω) = 5V = VR2 = VR3
IR2 = 5V/10Ω = 0.5A
IR3 = 5V/5Ω = 1A
KCL: IT = IR2 + IR3 + IR4 = 0.5A + 1A + 1A = 2.5A
Ohm’s Law: VR1= ITR1 = (2.5A)(5Ω) = 12.5V
KVL: Vx = VR1 + VR4 = 12.5V+5V = 17.5V
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