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Review of
EE 220 Circuits I
1
Overview
1. Current, voltage, power, energy
2. Circuit elements
3. Ohm’s Law
4. Node, branches, andloops
5. Kirchhoff’s Laws.
6. Series resistors
7. Parallel resistors current
8. Wye-Delta transformations
9. Nodal Analysis
10. Mesh Analysis
11. Superposition
12. Source Transformation
13. Thevenin Equivalent
14. Norton Equivalent
15. Maximum Power Transfer
16. Capacitors
17. Inductors
18. RC Circuits– source free
19. RL Circuits – source free
20. Unit Step Function
21. Step Response of RC Circuits
22. Step Response of RL Circuits
23. Series RLC circuit - source free
24. Parallel RLC circuit – source free
25. Step Response of series RLC ckt
26. Step Response of Parallel RLC ckt
2
1. Current
Electric current is defined as the rate of flow of
electric charge, i.e., i = dq/dt.
The unit of ampere can be derived as 1 A = 1C/s.
A direct current (DC) is a current that remains
constant with time.
An alternating current (AC) is a current that
varies sinusoidally with time
3
1. Voltage
Voltage (or potential difference) is the energy required to move a unit charge through an element.
Mathematically, (volt)
where w is energy in joules (J) and q is charge (C).
voltage, vab, is always defined across a circuit element or
between two points in a circuit.
◦ vab > 0 means the potential of a is higher than potential of b.
◦ vab < 0 means the potential of a is lower than potential of b.
4
dqdwvab /
1. Power
Power is the time rate of expending or absorbing energy, measured in watts (W).
Mathematical expression:
5
vidt
dq
dq
dw
dt
dwp
i
+
–
v
i
+
–
v
Passive sign convention P = +vi p = –vi
absorbing power supplying power
1. Energy
6
t
t
t
tvidtpdtw
0 0
• Energy is the capacity to do work,
measured in joules (J).
• Mathematical expression
2. Circuit Elements
7
Active Elements Passive Elements
Independent sources
Dependant sources
• A dependent source is an active element in which the source quantity is controlled by another voltage or current.
• They have four different types: VCVS,
CCVS, VCCS, CCCS. Keep in minds the signs of dependent sources.
3. Ohm’s Law
Ohm’s law states that the voltage across
a resistor is directly proportional to the
current I flowing through the resistor.
Mathematical expression for Ohm’s Law :
Note: the current enters the positive
side and leaves the negative side of v.
Two extreme possible values of R:
◦ R = 0 (zero) → v = 0 V ----- short circuit
◦ R = (infinity) → I = 0 A --- open circuit
8
iRv
3. Ohm’s Law
Conductance is the ability of an element to conduct electric current; it is the reciprocal of resistance R and is measured in Siemens (S).
The power dissipated by a resistor:
R
vRivip
22
9
v
i
RG
1
4. Nodes, Branches, Loops and Meshes
A branch represents a single element such as a voltage source
or a resistor (or a series connection of elements).
A node is the point of connection between two or more
branches.
A loop is any closed path in a circuit.
A mesh is a loop that contains no elements inside it.
Example: How many notes, branches , loops and meshed are
there in the circuit below?
10
5. Kirchhoff’s Current Law (KCL)
Kirchhoff’s current law (KCL) states that the algebraic sum of currents entering a node (or a closed boundary) is zero.
11
01
N
n
niMathematically,
5. Kirchhoff’s Voltage Law (KVL)
Kirchhoff’s voltage law (KVL) states that the algebraic sum of all voltages around a closed path (or loop) is
zero.
12
Mathematically, 01
M
m
nv
6. Series Resistors and voltage Division
The equivalent resistance of any number of resistors
connected in a series is the sum of the individual
resistances.
The voltage divider can be expressed as
13
N
n
nNeq RRRRR1
21
vRRR
Rv
N
n
n
21
7. Parallel Resistors and Current Division
The equivalent resistance of a circuit with N resistors in
parallel is:
The total current i is shared by the resistors in inverse
proportion to their resistances. The current divider can be
expressed as:
14
Neq RRRR
1111
21
n
eq
n
nR
iR
R
vi
8. Wye-Delta Transformations
15
)(1
cba
cb
RRR
RRR
)(2
cba
ac
RRR
RRR
)(3
cba
ba
RRR
RRR
1
133221
R
RRRRRRRa
2
133221
R
RRRRRRRb
3
133221
R
RRRRRRRc
Delta → Wye Wye → Delta
9. Nodal Analysis
16
Nodal analysis provides a general procedure for analyzing circuits using node voltages as the circuit variables.
• Steps
1.Select a node as the reference node.
2.Assign voltages v1,v2,…,vn-1 to the remaining n-1 nodes. The voltages are referenced with respect to the reference node.
3.Apply KCL to each of the n-1 non-reference nodes. Use Ohm’s law to express the branch currents in terms of node voltages.
4.Solve the resulting simultaneous equations to obtain the unknown node voltages.
9. Nodal Analysis
17
v1 v2
3
answer: v1 = -2V, v2 = -14V, P2 =2W, P7= 28W, P6 = 24W
Apply KCl at node 1 and 2
Example 1: Use nodal analysis to determine the power consumed by each resistor in the circuit below.
9. Nodal Analysis
18
Example 2 – Use nodal analysis to find the power supplied by the 2 V source (hint: use supernode)
Answer: 2= v1/2 +v2/4 +7
v2 – v1 = 2
P = ???
10. Mesh Analysis
19
Mesh analysis provides another general procedure for analyzing circuits using mesh currents as the circuit variables. Mesh analysis applies KVL to find unknown currents.
Steps
1. Assign mesh currents i1, i2, …, all the n meshes.
2. Apply KCL to each of the n meshes. Use Ohm’s law to express the voltages in terms of the mesh currents.
3. Solve the resulting n simultaneous equations to get the mesh currents.
10. Mesh Analysis
20
Example 1 – Use mesh analysis to find the voltage across the dependent source.
Answer: 4Io = 6 V
10. Mesh Analysis
21
Example 2 – Use mesh analysis to find i1 and i2 (hint: use supermesh)
Answer: -20 +6i1+10i2+4i2 = 0 i2 – i1 = 6
11. Superposition Theorem
22
The voltage across (or current through) an element in a linear circuit is the algebraic sum of the voltage across (or currents through) that element due to each independent source acting alone. Steps: 1. Turn off all independent sources except one source.
Independent voltage sources are replaced by 0 V (i.e.,
short circuit) Independent current sources are replaced by 0 A (i.e.,
open circuit) Dependent sources are left intact .
2. Find the output (voltage or current) due to that active source using nodal or mesh analysis.
3. Repeat step 1 for each of the other independent sources.
4. Find the total contribution by adding algebraically all the contributions due to the independent sources.
11. Superposition Theorem
23
Example 1: Use the superposition theorem to find v in the circuit shown below.
3A is discarded by open-circuit
6V is discarded by short-circuit
Answer: v = 10V
11. Superposition Theorem
24
Example 2: Use superposition to find vx in the circuit below.
Answer: Vx = 12.5V
2A is discarded by open-circuit
20 v1
4 10 V
+
(a)
0.1v1 4
2 A
(b)
20
0.1v2
v2
10V is discarded by open-circuit
Dependant source keep unchanged
12. Source Transformation
25
Source transformation is the process of replacing a voltage source vS in series with a resistor R by an equivalent circuit that consists of a current source iS in parallel with a resistor R, or vice versa.
R
vi
Riv
ss
ss
12. Source Transformation
26
Example 1: Find io in the circuit shown below using source transformation.
Answer: io = 1.78A
13. Thevenin’s Theorem
27
A linear two-terminal circuit can be replaced by an equivalent circuit consisting of a voltage source VTh in series with a resistor RTh, where • VTh is the open-circuit
voltage across terminals a-b.
• RTh is the equivalent resistance of the linear circuit with the independent sources turned off.
13. Thevenin’s Theorem
28
Example 1: Find the Thevenin’s equivalent circuit to the left of the terminals in the circuit shown below.
Answer: VTh= 6V, RTh = 3, i = 1.5A
6
4
(a)
RTh
6
2A
6
4
(b)
6 2A
+ VT
h
13. Thevenin’s Theorem
29
Example 2: Find the Thevenin equivalent circuit of the circuit shown below to the left of the terminals.
Answer: VTh = 5.33V, RTh = 3
6 V
5 Ix
4
+
1.5Ix
i1
i2
i1 i2
3
o
+ VT
h
b
a
1.5I
x 1 V
+
3 0.5Ix
5
a
b
4
Ix i
14. Norton’s Theorem
30
A linear two-terminal circuit can be replaced by an equivalent circuit of a current source IN in parallel with a resistor RN, Where • IN is the short circuit current
through the terminals. • RN is the equivalent resistance o
the circuit with the independent sources turned off.
The Thevenin and Norton equivalent circuits are related by a source transformation.
14. Norton’s Theorem
31
Example 1: Find the Norton equivalent of the circuit shown below.
Answer: RN = 1, IN = 10A.
2
(a)
6
2vx
+ +
vx
+ vx
1V
+ ix
i
2
(b)
6 10 A
2vx
+ +
vx
Isc
15. Maximum Power Transfer
L
ThThL
R
VPRR
4
2
max
L
LTh
ThL R
RR
VRiP
2
2
32
The power consumed by the load resistance RL is given by
Power transfer profile with different values of RL
Maximum power is obtained by dP/dRL = 0.
15. Maximum Power Transfer
33
Example 1: Determine the value of RL that will draw the maximum power from the rest of the circuit to the right. Then calculate the maximum power.
Answer: RL = 4.22 , Pmax = 2.9 W
16. Capacitors
A capacitor is a passive element designed to store energy
in its electric field. It consists of two conducting plates
separated by an insulator (or dielectric).
34
16. Capacitors
Capacitance C is the ratio of the charge q on one plate of a capacitor to the voltage difference v between the two plates, measured in farads (F).
• Capacitance can be calculated in
terms of - the permittivity of the
dielectric material between the plates, A - the surface area of each plate, and d - the distance between the plates.
• Units of capacitance: pF (10–12), nF (10–9), μF (10-6), mF (10-3), F
35
v
qC
d
AC
16. Capacitors
36
Current-voltage relationship of capacitor according to above convention:
td
vdCi )(
10
0
tvtdiC
vt
t and
Energy stored in a capacitor:
2
2
1vCw
– A capacitor acts as open circuit under constant voltage.
– The voltage across a capacitor cannot change abruptly
16. Capacitors
Example 1: The current flow into an initially discharged 1mF capacitor is shown below. Calculate the voltage across its terminals at t = 2 ms and t = 5 ms.
37
Answer: v(2ms) = 100 mV, v(5ms) = 500 mV
16. Series and Parallel Capacitors
The equivalent capacitance of N parallel-connected capacitors is the sum of the individual capacitances.
38
Neq CCCC ...21
The equivalent capacitance of N series-connected capacitors is the reciprocal of the sum of the reciprocals of the individual capacitances.
Neq CCCC
1...
111
21
16. Series and Parallel Capacitors Example 2: Find the equivalent capacitance seen at the terminals of the circuit in the circuit shown below:
39
Answer: Ceq = 40F
Example 3: Find the voltage across each of the capacitors in the circuit shown below:
Answer:
v1 = 30V
v2 = 30V
v3 = 10V
v4 = 20V
17. Inductors
An inductor is a passive element designed to store
energy in its magnetic field. It consists of a coil of
conducting wire.
40
17. Inductors
41
• Inductance L is the ratio of flux linkage λ within the coil to the current flow in the coil, measured in farads (F).
• Inductance L of a solenoid can be calculated in terms of μ - the
magnetic constant of the material inside the coil, A - the cross sectional surface of the coil, N – the number of turns, and l – the length of the coil.
• Units of inductance: pH (10–12), nH (10–9), μH (10-6), mH (10-3), H
iL
l
ANL
2
17. Inductors
Voltage-current relation in an inductor:
42
Energy stored in an inductor.
td
idLv and )()(
10
0
titdtvL
it
t
2
2
1iLw
– An inductor acts as short circuit under constant current.
– The current flow through an inductor cannot change abruptly
17. Inductors Example 1: Determine vc, iL, and the energy stored in the
capacitor and inductor in the circuit of circuit shown below
under steady-state conditions.
43
Answer: iL = 3A, vC = 3V, WL = 1.125J, WC = 9J
17. Series and Parallel Inductors The equivalent inductance of series-connected
inductors is the sum of the individual inductances.
44
Neq LLLL ...21
The equivalent capacitance of parallel inductors is the
reciprocal of the sum of the reciprocals of the individual
inductances.
Neq LLLL
1...
111
21
Recap: voltage-current relation and power in
passive circuit elements
45
+
+
+
18. RC Circuit (source free)
46
Current flow in resistor R
Current flow in capacitor C
0 dt
dvC
R
v0 CR ii
By KCL
• Applying Kirchhoff’s laws to purely resistive circuit results in algebraic equations.
• Applying the laws to RC and RL circuits produces differential equations.
Vo is the initial voltage τ = RC is the time constant
18. RC Circuit (source free)
The natural response of a circuit refers to the behavior
(in terms of voltages and currents) of the circuit itself, with
no external sources of excitation.
47
• The time constant of a circuit is the time required for the response to decay by a factor of 1/e or 36.8% of its initial value.
CRTime constant
Decays more slowly
Decays faster
18. RC Circuit (source free)
Example 1: Refer to the circuit below, determine vC, vx, and io
for t ≥ 0. Assume that vC(0) = 30 V.
48
Answer: vC = 30e–0.25t V ;
vx = 10e–0.25t ;
io = –2.5e–0.25t A
Example 2: The switch in circuit below was closed for a long
time, then it opened at t = 0, find v(t) for t ≥ 0.
Answer: V(t) = 8e–2t V
19. RL Circuit (source free)
49
0 RL vvBy KVL
0 iRdt
diL
0)( iL
R
dt
di /0 )( teIti
Io is the initial current τ = L/R is the time constant
19. RL Circuit (source free)
Example 1: Find i and vx in the circuit. Assume i(0) = 5A.
50
Answer: i(t) = 5e–53t A
Example 2: For the circuit, find i(t) for t > 0.
Answer: i(t) = 2e–2t A
20. Unit-Step Function The unit step function u(t) is 0 for negative values of t and 1 for positive values of t.
51
0,1
0,0)(
t
ttu
o
o
ott
ttttu
,1
,0)(
Represent an abrupt change in voltage and current:
21. Step-Response of an RC Circuit The step response of a circuit is its behavior when the excitation is the step function, which may be a voltage or a current source.
52
• Initial voltage (given): v(0-) = v(0+) = V0
• Applying KCL,
or
• Where u(t) is the unit-step function
0)(
R
tuVv
dt
dvc s
RC
tuVv
RCdt
dv s )()
1(
21. Step-Response of a RC Circuit
53
0)1(
0)(
//
0
0
teVeV
tVtv
t
s
t
Complete Response = Natural response + Forced Response (stored energy) (independent source)
/ )]( )0( [ )( )( tevvvtv General Solution:
21. Step-Response of a RC Circuit
Example 1: Find v(t) for t > 0 in the circuit in below. Assume the switch has been open for a long time and is closed at t = 0. Calculate v(t) at t = 0.5.
54
Answer: and v(0.5) = 0.52 V 515)( 2 tetv
22. Step-response of a RL Circuit
55
• Initial current (given) i(0-) = i(0+) = Io
• Final inductor current i(∞) = Vs/R
• Apply KVL:
• Time constant = L/R
0)()(
teR
VI
R
Vti
t
so
s
L
tuVi
L
R
dt
di s )()(
/ )]( )0( [ )( )( teiiiti
22. Step-Response of a RL Circuit
Example 1: The switch in the circuit shown below has been closed for a long time. It opens at t = 0. Find i(t) for t > 0.
56
Answer: teti 102)(
23. Series RLC Circuit (source free)
57
• The solution of the source-free series RLC circuit is called as the natural response of the circuit.
• The circuit is excited by the energy initially stored in the capacitor and inductor.
• Expression of current (using KVL):
02 2
02
2
idt
di
dt
id
LCand
L
R 1
20
02 2
02
2
idt
di
dt
id
58
There are three possible solutions for the following 2nd order differential equation:
1. If > o, over-damped case
tstseAeAti 21
21)( 2
0
2
2,1 swhere
2. If = o, critical damped case
tetAAti )()( 122,1swhere
3. If < o, under-damped case
)sincos()( 21 tAtAeti dd
t where
22
0 d
23. Series RLC Circuit (source free)
The constants A1 and A2 are obtained from the initial conditions
23. Series RLC Circuit (source free)
Example 1: The circuit shown below has reached steady state at t = 0-. If the make-before-break switch moves to position b at t = 0, calculate i(t) for t > 0.
59
Answer: i(t) = e–2.5t[5 cos1.6583t – 7.538 sin1.6583t] A
24. Parallel RLC Circuit (source free)
60
0
0 )(1
)0( dttvL
IiLet
v(0) = V0
Apply KCL to the top node:
t
dt
dvCvdt
LR
v0
1
Taking the derivative with
respect to t and dividing by C
LCRCv
dt
dv
dt
vd 1and
2
1 where0 2 0
2
02
2
24. Parallel RLC Circuit (source free)
Example 3: Refer to the circuit shown below where the switch has been closed for a long time. Find v(t) for t > 0.
61
Answer: v(t) = 66.67(e–10t – e–2.5t) V
25. Step-Response of Series RLC Circuit
62
LC
vv
dt
dv
dt
vd s 2
02
2
2
The solution of the above equation has two components: transient response vt(t) & steady-state response vss(t):
)()()( tvtvtv sst
• The steady-state response is the final value of v(t).
vss(t) = v(∞) = Vs.
• The transient response is the same as the source-free
response
Depends on the values of α and ωo.
LCand
L
R 1
20
25. Step-Response of Series RLC Circuit
Example 1: Having been in position for a long time, the switch in the circuit below is moved to position b at t = 0. Find v(t) for t > 0.
63
Answer: v(t) = 10 +e-2t (–2cos3.464t – 1.1547sin3.464t) V
26. Step-Response of Parallel RLC Circuit
64
LC
I
LC
i
dt
di
RCdt
id s1
2
2
LCand
RC
1
2
10
The solution of the above equation has two components: transient response it(t) & steady-state response iss(t):
)()()( tititi sst
• The steady-state response is the final value of i(t).
iss(t) = i(∞) = Is
• The transient response is the same as the source-free
response
Depends on the values of α and ωo.
26. Step-Response of Parallel RLC Circuit
Example 1: Find i(t) and v(t) for t > 0 in the circuit shown in circuit shown below. Assume that there was no energy stored in C and L.
65
Answer:iss(t) = Is =20 A, α =0, and ω0 = 1, oscillatory case i(0) = 0 A, di(0)/dt = 0 A/sec → A1 = -20, A2 = 0.
i(t) = -20 cost A
v(t) = Ldi/dt = 100 sint V
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