Lecture 2
F d
Fundamentals
l off Electrical
El i l
Engineering
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Introduction
1. Define current, voltage, and power,
including their units.
2. Calculate p
power and energy,
gy, as well as
determine whether energy is supplied or
y a circuit element.
absorbed by
3. State and apply basic circuit laws.
4. Solve for currents, voltages, and powers in
simple circuits.
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
d (C/
(C/s).
)
Electrical Current
ddq (t )
i (t ) =
dt
t
q (t ) = ∫ i (t )dt + q (t0 )
t0
Direct Current
Alternating Current
When a current is constant with time, we
say that
th t we have
h
di t current,
direct
t
abbreviated as dc. On the other hand, a
currentt that
th t varies
i with
ith time,
ti
reversing
i
direction periodically, is called alternating
c rrent abbreviated
current,
bb i t d as ac.
.
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
i l t to
t joules
j l per coulomb
l b (J/C).
(J/C)
“uphill: battery”
“downhill: resistor”
Power and Energy
gy
p(t ) = v (t )i (t )
Watts
t2
w = ∫ p(t )dt
t1
Joules
Current is flowing in
the passive
configuration
If the current flows opposite to the passive
configuration, the power is given by p = -vi
Pa = iava = (2A)*(12V) = 24W, energy is being absorbed
Pb = -iibvb = -(1A)*(12V)
(1A)*(12V) = -12W,
12W energy is
i being
b i supplied
li d
Pc = icvc = (-3A)*(12V) = -36W, energy is being supplied
Power and Energy
gy
v(t) = 12 V
i(t) = 2e-t A
p(t) = v(t)i(t) = 24e-t
∞
w=
∞
∫ p(t )dt = ∫ 24e
0
0
−t
dt = −24e
−∞
− (−24e ) = 24 J
0
?
• An ideal voltage source has a voltage vx =12 V
independent of the load
• An ideal conductor requires that vx = 0
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
“ohms”
George
g 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
p
much time dancing,
g, ice skating
g and p
playing
y g billiards.
spent
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”
siemens
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)
(Ω-m)
Power dissipation in a resistor
2
v
p = vi = Ri =
R
2
KIRCHHOFF S CURRENT
KIRCHHOFF’S
LAW
• The net current entering
g a node is zero.
• Alternatively, the sum of the currents
entering a node equals the sum of the
currents leaving a node.
Gustav Kirchhoff (1824-1887)
Kirchhoff's Current Law
The principle of conservation of electric charge
implies that:
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
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
(a) Currents into the node = 1A+3A = 4A current out of the node ia
(b) C
Currents into
i
the
h node
d = 3A+1A+i
3A 1A ib = current out off the
h node
d =
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
Which elements are in series?
KIRCHHOFF S VOLTAGE
KIRCHHOFF’S
LAW
The algebraic
Th
l b i sum off the
h voltages
l
equals
l
zero for any closed path (loop) in an
electrical
l
i l circuit.
i i
Loop 1: -vva+vvb+vvc = 0
Loop 2: -vc-vd+ve = 0
Loop 3: va-vve+vd-vvb = 0
Parallel Connection
KVL through A and B: -va+vb = 0 → va = vb
KVL through A and C: -va - vc = 0 → va = -vc
-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:
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 flowingg through
g the resistor?
• What is the power for each element in the circuit?
• Which elements are absorbing power?
• 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
Th power dissipated
di i
d in
i eachh element:
l
pR = iR2R = (2A)2(5Ω) = 20W (absorbing)
pvs= iv = (2A)(10V) = 20W (absorbing)
pcs= -iv = -(2A)(20V) = 40W (supplying)
Example
Use Ohm’s law, KVL and KCL to find Vx
IT
R1
IR2
IR3
R2
R3
Oh ’ Law:
Ohm’s
L
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
)( ) = 12.5V
Ohm’s Law: VR1= ITR1 = ((2.5A)(5Ω)
KVL: Vx = VR1 + VR4 = 12.5V+5V = 17.5V
R4