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CHAPTER 2
Basic Circuit Elements and Analysis
Techniques
2.1 The Independent Voltage and
Current Sources
Figure 2.1 The independent voltage source.
A given function of time
Types of Voltage Waveforms
DC voltage source
AC voltage source
Combining Voltage Sources
In Series
In Parallel
Exercise Problem 2.1
Replace the combination of voltage sources
with an equivalent source at the two
terminals
+
b
V (t )
ab
V (t )
ab
+
-
4 - 3t V
a
-
KVL  -V 10 - 3t - 6  0
ab
 V
ab
 4 - 3t V
Independent Current Source
Independent Current Source
Is a function of time
Combining Current Sources
In Parallel
In Series
Exercise Problem 2.2
Replace the combination of current sources with a single
source that is equivalent at the two terminals
b
3+2 sin 3t -10= (-7+2sin 3t) A
a
Exercise Problem 2.3
Determine v s t  and i s t 
Such that the two circuits are equivalent at
terminals a and b
Vs(t) = (3 + 2t - 2cos 3t) v
is(t) = (2 - 6t - 5sin 2t)A
2.2 The Linear Resistor and Ohm’s
Law
G
Ohm’s Law:
1
R
v (t )  Ri (t )
Current is assumed to enter the positive voltage
terminal (The passive sign Convention)
Ohm’s Law
Conductance
1
G
R
Siemen (S )
Ohm ' s Law : i (t )  G v (t )
The Power Absorbed by a Resistor
p  t   v(t )i (t )
 i (t ) R
2
 G v (t )
2
2
v (t )

R
Example 2.3
Write Ohm’s law for the following four resistors
V = -3 i
V = -5 i
V=2i
V = -10 i
Example 2.4
Determine the unknown voltage or current
v  iR  12
v
10
i  -  -  -2
R
5
v  -iR  -6 v  iR  18
v  iR  12
2.3 Single-Loop and Single-NodePair Circuits
• Single-loop circuit have one current common to
all elements
Analysis of single loop circuits
• Assume an arbitrary direction for current I
• The resulting current is
 voltage sources 

I
  resistors in the loop 
• The Voltage across any resistor R is the product
of ii and R with signs according to Ohm’s law
Example 2.5
Determine voltages v1, v2, and v3 in the following circuit
3-58- 2
I
 0.267A
1 2  3  4  5
v1  1 0.267  0.267v
v 2  -5  0.267  -1.333v
v3  5  2  0.267  5.533v
Example 2.7
Determine voltages v1 and v2 in the following circuit
3 - 5 - 6 - 10
I
 -1.5A
23 43
v1  3   -1.5   -4.5v
v 2  -10 - 2  -1.5   -10  3  -7
Single-node-pair circuit
To analyze a single-node-pair circuit
1. Assume a direction for V across all elements
  current sources 
2. The resulting voltage is V    conduc tan ces in the loop 
3. The current through an individual resistor R is V/R
Example 2.8:
Determine currents i1 and i2 in the following circuit
First label the voltage V as shown, then write
-2  5 - 4
-1
V

 -0.632V
1
1
19
S  S  1S
S
4
3
12
-0.632
i1 
 -0.158A
4
-0.632
i2   0.211A
3
2.4 Resistors in Series
Equivalent of resistors in series
Ex 2.11 Determine the equivalent R at ab
Resistors in Parallel
1
1
1



R eq R1 R 2
1

Rn
Ex. Determine equivalent R at ab
1
1 1 1 1
     1.283S
R eq 3 4 2 5
R=0.779
Two resistors in parallel
R eq  R1 || R 2
R 1R 2
R eq 
R1  R 2
R
R || R 
2
Repeated use of two-resistors
2.4.1 Circuit Solution by Circuit Reduction
Determine voltages V and vx and currents I and ix
b
5A
2W
a
b
5A
a
+
2W
V
-
+
ix
V
-
1W
1W
v x  i x 1W  V
5
2
ix
1W
2W
c
c
I
+
Vx
-
vx
I 
 54 A
2W
2W
b
5A
+
Vx
a
+
2W
V
-
ix
2W
Single node pair
5A
V
5
 5V
ix 
 2A V  1
1
2W
2W 2W
2.5 Voltage Division
vS  t 
i
R1  R 2  R 3
R1
v1  R1i 
vS  t 
R1  R 2  R 3
R2
v 2  R 2i 
vS  t 
R1  R 2  R 3
R3
v3  R 3i 
vS  t 
R1  R 2  R 3
Ri
vi 
R1  R 2 
 Rn
vS  t 
Ex 2.19 Determine V
vi 
Ri
R1  R 2 
 Rn
vS  t 
2
v 5  -2V
23
EX: 2.21 Determine V
9
v a  (9W || 9W) *10A  *10  45V
2
2W
2
V
v a  * 45  10V
2W  3W  4W
9
Current Division
iS  t 
v
G1  G 2  G 3
G1
i1  G1v 
iS  t 
G1  G 2  G 3
G2
i2  G 2 v 
iS  t 
G1  G 2  G 3
G3
i3  G 3 v 
iS  t 
G1  G 2  G 3
Case of two resistors in parallel
Ex: 2.22 Determine I and ix by current division
2i x
ix
Two-resistor current division rule
1W
1
I2A  1W  3W
2
v x  -3W *I  3/ 2V
From original circuit
vx
ix 
 3 / 4A
2W
Ex: 2.22 Determine I and ix by current division
2i x
ix
Two-resistor current division rule
1W
1
I2A  1W  3W
2
v x  -3W *I  3/ 2V
From original circuit
vx
ix 
 3 / 4A
2W
Ex: 2.23 Determine vx
Single loop circuit
10V
30
I
 A
4
1W  W 7
3
Current division in the original circuit
2W
10
ix 
I A
2W  4W
7
10
v x  -1W *i x  - v
7
2.6 Solutions for Circuits Containing More than one Source
Previously we solve circuit containing one source . We apply circuit reduction
By combining resistors in parallel or series
We also can have more than one source however in a single loop or single node
however when we have more than one source and more than one loop or node
In this type of circuit we can not do circuit reduction by combining resistors
due to the source
Example we can not combine the 2 W resistor and the 1 W or the other 2 W
Direct Method:
• KCL to all nonsimple nodes,
• KVL to all nonsimple loops,
• Ohm’s law to every resistor.
Simple node
Simple loop
Non simple node
Non simple loop
Ex. 2.26 Determine
i
using the direct method
KCL at a
Ohm’s law
By KVL
10V = -2i + 3(5 - i)
=>
i = 1A
Ex. 2.27 Determine V using the direct method
KVL around the two nonsimple loops
Ohm’s law to determine currents
KCL at node a
v-3 v v-3
4A 


2W 2W 1W
17
v V
4
Ex: 2.28 Determine v and I using the direct method
KCL at a
Ohm’s law
KVL around the outside loop 3V=2(I-2)+2I+2I => I=7/6A
KVL around the right loop v = 2I +2I = 14/3 V
2.7 Source Transformations
A method called Source Transformations will allow the transformations of a voltage
source in series with a resistor to a current source in parallel with resistor.
R
a
a
vs
is

-
R
b
b
The double arrow indicate that the transformation is bilateral , that we can start with either
configuration and drive the other
R
a
a
vs

-
is
RL
iL
R
RL
iL
b
b
Voltage source with a series resistor
Current source with a parallel resistor
R
iL 
is
R  RL
vs
iL 
R  RL
Equating we have
,
vs
R

is
R  RL R  RL
 is 
vs
R
OR v s  Ri s
Polarity of Vs in one form must be such that it tends to
push in the direction of Is in the other form

-
Ex. 2.29: Determine i and ix using source transformation
Single loop
KVL
15V - 10V
i
 1A
2W  3W
ix should be found from the original circuit
i x  - (i - 5)
= 5 - i  4A
Ex 2.30: Determine I and V
Transform the 3V in series with the 2 Ohm resistor
We get a single-node-pair circuit
14
3

3
 4
V   A  2A  * 2W || 4W   A  2A  * W  V
3
2

2
 3
V 7
 A
Using Ohm’s law in the original circuit: I 
4W 6
Example (a) find the power associated with the 6 V source
(b) State whether the 6 V source is absorbing or
delivering power
We are going to use source transformation to reduce the circuit, however note that we
will not alter or transfer the 6 V source because it is the objective.
40
 8A
5
(20W || 5W)  4Ω
(8A )(4Ω)  32 V
(6  4)  10Ω
(10  10)  20Ω
32
 1.6A
20
(30W || 20W)  12Ω
(4  12)  16Ω
(1.6A )(12Ω)  19.2 V
i
i=
-
 0.825 A

 P6V  (0.825)(6)  4.95 W
It should be clear if we transfer the 6V during these steps you will not be able to find
the power associated with it