Review of
EE 220 Circuits I
1
Overview
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
Current, voltage, power, energy
Circuit elements
Ohm’s Law
Node, branches, andloops
Kirchhoff’s Laws.
Series resistors
Parallel resistors current
Wye-Delta transformations
Nodal Analysis
Mesh Analysis
Superposition
Source Transformation
Thevenin Equivalent
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
Norton Equivalent
Maximum Power Transfer
Capacitors
Inductors
RC Circuits– source free
RL Circuits – source free
Unit Step Function
Step Response of RC Circuits
Step Response of RL Circuits
Series RLC circuit - source free
Parallel RLC circuit – source free
Step Response of series RLC ckt
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,
vab  dw / dq
(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
1. Power


Power is the time rate of expending or absorbing energy,
measured in watts (W).
Mathematical expression:
dw dw dq
p


 vi
dt
dq dt
i
i
+
+
v
v
–
–
Passive sign convention
P = +vi
absorbing power
p = –vi
supplying power
5
1. Energy
•
Energy is the capacity to do work,
measured in joules (J).
•
Mathematical expression
t
t
t0
t0
w   pdt   vidt
6
2. Circuit Elements
Active Elements
Passive Elements
• A dependent source is an active
element in which the source quantity
is controlled by another voltage or
current.
Independent Dependant
sources
sources
• They have four different types: VCVS,
CCVS, VCCS, CCCS. Keep in minds the
signs of dependent sources.
7
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 :
v  iR


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
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).
1 i
G 
R v

The power dissipated by a resistor:
2
v
p  vi  i R 
R
2
9
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.
N
Mathematically,
i
n 1
n
0
11
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.
Mathematically,
M
v
m 1
n
0
12
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.
N
Req  R1  R2      RN   Rn
n 1

The voltage divider can be expressed as
Rn
vn 
v
R1  R2      RN
13
7. Parallel Resistors and Current Division

The equivalent resistance of a circuit with N resistors in
parallel is:
1
1
1
1


  
Req R1 R2
RN

The total current i is shared by the resistors in inverse
proportion to their resistances. The current divider can be
expressed as:
v iReq
in 

Rn
Rn
14
8. Wye-Delta Transformations
Delta → Wye
Wye → Delta
Rb Rc
( Ra  Rb  Rc )
Ra 
R1 R2  R2 R3  R3 R1
R1
Rc Ra
R2 
( Ra  Rb  Rc )
Rb 
R1 R2  R2 R3  R3 R1
R2
Ra Rb
R3 
( Ra  Rb  Rc )
Rc 
R1 R2  R2 R3  R3 R1
R3
R1 
15
9. Nodal Analysis
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.
16
9. Nodal Analysis
Example 1: Use nodal analysis to determine the
power consumed by each resistor in the circuit below.
Apply KCl at
node 1 and 2
v1
v2
3
answer: v1 = -2V, v2 = -14V, P2 =2W, P7= 28W, P6 = 24W
17
9. Nodal Analysis
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 = ???
18
10. Mesh Analysis
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.
2.
3.
Assign mesh currents i1, i2, …, all the n meshes.
Apply KCL to each of the n meshes. Use Ohm’s law
to express the voltages in terms of the mesh
currents.
Solve the resulting n simultaneous equations to
get the mesh currents.
19
10. Mesh Analysis
Example 1 – Use mesh analysis to find the voltage
across the dependent source.
Answer: 4Io = 6 V
20
10. Mesh Analysis
Example 2 – Use mesh analysis to find i1 and i2 (hint:
use supermesh)
Answer:
-20 +6i1+10i2+4i2 = 0
i2 – i1 = 6
21
11. Superposition Theorem
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.
22
11. Superposition Theorem
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
23
11. Superposition Theorem
Example 2: Use superposition to find vx in the circuit
below.
2A is discarded by
open-circuit
20 
10 V
10V is discarded
by open-circuit
20 
v1
+

4
(a)
Answer: Vx = 12.5V
0.1v1
Dependant source
keep unchanged
v2
2A
4
0.1v2
(b)
24
12. Source Transformation
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.
v s  is R
vs
is 
R
25
12. Source Transformation
Example 1: Find io in the circuit shown below using
source transformation.
Answer: io = 1.78A
26
13. Thevenin’s Theorem
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.
27
13. Thevenin’s Theorem
Example 1: Find the Thevenin’s equivalent circuit to the
left of the terminals in the circuit shown below.
6
6
4
RTh
(a)
6
2A
6
2A
4
+
VT
h

(b)
Answer: VTh= 6V, RTh = 3, i = 1.5A
28
13. Thevenin’s Theorem
Example 2: Find the Thevenin equivalent circuit of the
circuit shown below to the left of the terminals.
5
6V
+

Ix
3
a
i2
i1
4
i1
1.5Ix
i2
3  Ix
5
1.5I
h

o
0.5
Ix
+
VT
b
i
a
+

4
x
1
V
b
Answer: VTh = 5.33V, RTh = 3
29
14. Norton’s Theorem
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.
30
14. Norton’s Theorem
Example 1: Find the Norton equivalent of the circuit
shown below.
2vx
+
vx

i
+

6
2
ix
+
vx

1V
+

(a)
2vx
+

6
2
10
A
+
vx

Isc
(b)
Answer: RN = 1, IN = 10A.
31
15. Maximum Power Transfer
The power consumed by the load
resistance RL is given by
2
 VTh 
 RL
P  i RL  
 RTh  RL 
2
Maximum power is obtained
by dP/dRL = 0.
2
RL  RTh

Pmax
VTh

4 RL
Power transfer profile with different
values of RL
32
15. Maximum Power Transfer
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
33
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.
q
C
v
C
Units of capacitance: pF (10–12),
nF (10–9), μF (10-6), mF (10-3), F
35
 A
d
16. Capacitors
Current-voltage relationship of
capacitor according to above
convention:
dv
i C
dt
and
1
v
C
t
 i d t  v(t )
t0
0
Energy stored in a capacitor:
1
2
w Cv
2
– A capacitor acts as open
circuit under constant voltage.
– The voltage across a capacitor
cannot change abruptly
36
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.
Answer: v(2ms) = 100 mV, v(5ms) = 500 mV
37
16. Series and Parallel Capacitors
The equivalent capacitance of N parallel-connected
capacitors is the sum of the individual capacitances.
Ceq  C1  C2  ...  C N
The equivalent capacitance of N series-connected
capacitors is the reciprocal of the sum of the reciprocals of
the individual capacitances.
1
1
1
1


 ... 
Ceq C1 C2
CN
38
16. Series and Parallel Capacitors
Example 2: Find the equivalent capacitance seen at the
terminals of the circuit in the circuit shown below:
Answer: Ceq = 40F
Example 3: Find the voltage across each of the capacitors
in the circuit shown below:
Answer:
v1 = 30V
v2 = 30V
v3 = 10V
v4 = 20V
39
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
• 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.
L

i
N A
L
l
2
• Units of inductance: pH (10–12),
nH (10–9), μH (10-6), mH (10-3), H
41
17. Inductors
Voltage-current relation in an inductor:
di
vL
dt
and
1 t
i   v (t ) d t  i (t 0 )
L t0
Energy stored in an inductor.
1
w  L i2
2
– An inductor acts as short circuit under constant current.
– The current flow through an inductor cannot change
abruptly
42
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.
Answer: iL = 3A, vC = 3V,
WL = 1.125J, WC = 9J
43
17. Series and Parallel Inductors

The equivalent inductance of series-connected
inductors is the sum of the individual inductances.
Leq  L1  L2  ...  LN

The equivalent capacitance of parallel inductors is the
reciprocal of the sum of the reciprocals of the individual
inductances.
1
1
1
1


 ... 
Leq L1 L2
LN
44
Recap: voltage-current relation and power in
passive circuit elements
+
+
+
45
18. RC Circuit (source free)
• Applying Kirchhoff’s laws to purely resistive circuit
results in algebraic equations.
• Applying the laws to RC and RL circuits produces
differential equations.
By KCL
iR  iC  0
Current
flow in
resistor R
v
dv
C
0
R
dt
Current flow in
capacitor C
Vo is the initial voltage
τ = RC is the time constant
46
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.
Time constant
RC
Decays more slowly
Decays faster
• 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.
47
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.
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
48
19. RL Circuit (source free)
By KVL
vL  v R  0
di
L
 iR  0
dt
di R
 ( )i  0
dt
L
i(t )  I 0 e t /
Io is the initial current
τ = L/R is the time constant
49
19. RL Circuit (source free)
Example 1: Find i and vx in the circuit. Assume i(0) = 5A.
Answer: i(t) = 5e–53t A
Example 2: For the circuit, find i(t) for t > 0.
Answer: i(t) = 2e–2t A
50
20. Unit-Step Function
The unit step function u(t) is 0 for negative
values of t and 1 for positive values of t.
 0,
u(t )  
1,
 0,
u (t  to )  
1,
t0
t0
t  to
t  to
Represent an abrupt change in voltage and current:
51
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.
• Initial voltage (given):
v(0-) = v(0+) = V0
• Applying KCL,
or
dv v  Vs u (t )
c 
0
dt
R
Vs u (t )
dv
1
(
)v 
dt
RC
RC
• Where u(t) is the unit-step function
52
21. Step-Response of a RC Circuit
t0
V0
v(t )   t /
t / 
V
e

V
(
1

e
)
s
 0
t 0
Complete Response = Natural response + Forced Response
(stored energy) (independent source)
General Solution:
v (t )  v ()  [v (0)  v ()] e
53
t /
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.
Answer:
v(t )  15e 2t  5 and v(0.5) = 0.52 V
54
22. Step-response of a RL Circuit
•
Initial current (given)
i(0-) = i(0+) = Io
•
Final inductor current
i(∞) = Vs/R
• Apply KVL:
Vs u (t )
di
R
 ( )i 
dt
L
L
•
Time constant  = L/R
Vs
Vs
i(t )   ( I o  )e
R
R
i (t )  i ()  [i (0)  i ()] et /

t

55
t 0
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.
Answer:
i(t )  2  e 10t
56
23. Series RLC Circuit (source free)
• 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):
d 2i
di
2

2



0i 0
2
dt
dt
R

2L
1
and 0 
LC
57
23. Series RLC Circuit (source free)
There are three possible solutions for the following 2nd
order differential equation:
d 2i
di
2

2



0i 0
2
dt
dt
1. If  > o, over-damped case
i(t )  A1e s1t  A2e s2t
2
where s1, 2       0
2
2. If  = o, critical damped case
i(t )  ( A2  A1t )et
where
s1, 2   
3. If  < o, under-damped case
i(t )  et ( A1 cos d t  A2 sin d t ) where d  02   2
The constants A1 and A2 are obtained from the
initial conditions
58
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.
Answer: i(t) = e–2.5t[5 cos1.6583t – 7.538 sin1.6583t] A
59
24. Parallel RLC Circuit (source free)
0
i(0)  I 0 
Let
v(0) = V0
1
v(t )dt

L 
Apply KCL to the top node:
t
v 1
dv
  vdt  C  0
R L 
dt
Taking the derivative with
respect to t and dividing by C
d 2v
dv
2

2



0v  0
2
dt
dt
1
where  
2 RC
and 0 
60
1
LC
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.
Answer:
v(t) = 66.67(e–10t – e–2.5t) V
61
25. Step-Response of Series RLC Circuit
vs
d 2v
dv
2
 2
 0 v 
2
dt
dt
LC
R

2L
1
and 0 
LC
The solution of the above equation has two components:
transient response vt(t) & steady-state response vss(t):
v(t )  vt (t )  vss (t )
• 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.
62
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.
Answer: v(t) = 10 +e-2t (–2cos3.464t – 1.1547sin3.464t) V
63
26. Step-Response of Parallel RLC Circuit
d 2i 1 di
i
Is



2
dt
RC dt LC LC

1
2 RC
and 0 
1
LC
The solution of the above equation has two components:
transient response it(t) & steady-state response iss(t):
i(t )  it (t )  iss (t )
• 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.
64
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.
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
65