E E 1205 Circuit Analysis
Lecture 2 - Circuit Elements and
Essential Laws
Five Fundamental Elements
• Ideal Voltage Sources
– Independent
– Dependent
• Ideal Current Sources
– Independent
– Dependent
• Resistors
• Inductors (to be introduced later)
• Capacitors (to be introduced later)
Independent Voltage Source
• Voltage may be
constant or timedependent
• Delivers nominal
terminal voltage
under all conditions
Positive Terminal
Vg
Negative Terminal
Independent Current Source
• Current may be
constant or timedependent
• Delivers nominal
terminal current
under all
conditions
Negative Node
Ig
Positive Node
Voltage-Controlled Dependent
Voltage Source
• Terminal voltage is
a function of the
voltage drop of a
different branch
• Delivers nominal
terminal voltage
under all conditions
Positive Terminal
v
Negative Terminal
+
v
-
Current-Controlled Dependent
Voltage Source
• Terminal voltage is
a function of the
current flow in a
different branch
• Delivers nominal
terminal voltage
under all
conditions
Positive Terminal
i
i
Negative Terminal
Voltage-Controlled Dependent
Current Source
• Current is a function
of the voltage drop
of a different branch
• Delivers nominal
terminal current
under all conditions
Negative Node
v
Positive Node
+
v
-
Current-Controlled Dependent
Current Source
• Source current is a
function of the
current flow in a
different branch
• Delivers nominal
terminal current
under all conditions
Negative Node
i
Positive Node
i
Electrical Resistance
(Ohm’s Law)
• Electrical resistance is
the ratio of voltage drop
across a resistor to
current flow through the
resistor.
• Polarities are governed
by the passive sign
convention.
R
i
+
v
v
R
i
-
Power Consumed by Resistors
• Resistors consume
power.
• v and i are both
positive or both
negative.
R
i
+
v
-
p  v i
v  R i
v
i
R
p i R
2
2
v
p
R
Conductance Defined
• Conductance is the
reciprocal of
resistance.
• The units of
conductance are
called siemens (S)
• The circuit symbol
is G
1
G
R
i  v G
i
v
G
p  v2  G
2
i
p
G
Creating a Circuit Model
• A circuit model is usually two or more
circuit elements that are connected.
• A circuit model may have active elements
(sources) as well as passive elements (such
as resistors).
• By the assumption that electric signal
propagation is instantaneous in a circuit, our
circuit model has lumped parameters.
Example of a Circuit Model
1000 ft AWG 14
Copper Wire
100 W
Lamp
120 V Battery
0.25 
2.57 
144 
120 V
2.57 
Kirchhoff’s Voltage Law
• The sum of the voltage drops around a
closed path is zero.
• Example: -120 + V1 + V2 + V3 + V4 = 0
0.25 
+ V1 -
2.57 
+ V2 -
120 V
2.57 
- V4 +
+
V3
-
144 
Kirchhoff’s Current Law
• A node is a point where two or more circuit
elements are connected together.
• The sum of the currents leaving a node is
zero.
I1
I4
I1  I 2  I 3  I 4  0
I3
I2
Apply KCL to Example
I1
Is
0.25  I1
+ V1 -
I2
2.57  I2
120 V
2.57 
Is
I4
I3
+ V2 +
V3
-
- V4 + I4
I s  I1  I 2  I3  I 4
144 
I3
Combine KVL, KCL & Ohm’s Law
I1
Is
0.25  I1
+ V1 -
I2
2.57  I2
120 V
2.57 
Is
I4
I3
+ V2 +
V3
-
144 
I3
- V4 + I4
120  0.25  I s  2.57  I s  144  I s  2.57  I s
120V
Is 
 0.803 A
149
Lamp Voltage & Battery Voltage
I1
Is
0.25  I1
+ V1 -
I2
+
2.57  I2
Vb
120 V
-
Is
I4
I3
+ V2 -
2.57 
+
V3
-
144 
I3
- V4 + I4
V3  144  I s  115.67V
Vb  (2.57  2  144)  0.803  119.8V
Battery Power and Lamp Power
1000 ft AWG 14
Copper Wire
100 W
Lamp
120 V Battery
Pb  119.8V  0.8033 A  96.23W
Pl  115.67V  0.8033 A  92.91W
Loss: Ploss  Pb  Pl  3.32W
Pl 92.91

 96.55%
Efficiency:  
Pb 96.23
Using Loops to Write Equations
vb
R2
+ v2 va
a
+
v1
-
R1
b
+
v3
-
R3
c
va  v2  v1  0
KVL @ Loop b: vb  v3  v1  0
KVL @ Loop c: va  v2  vb  v3  0
KVL @Loop a:
Loop c equation same as a & b combined.
Using Nodes to Write Equations
i2
y
ia
va
R2
i2
x
+ v2 -
ib
vb
ib
z
i3
i1
+
v1
-
+
v3
-
R1
i1
ia
R3
i3
w
i2  i1  ib  0
KCL @ Node y: ia  i2  0
KCL @ Node z: ib  i3  0
KCL @ Node w: ia  i1  i3  0
KCL @ Node x:
<== Redundant
Combining the Equations
•
•
•
•
•
•
•
•
•
There are 5 circuit elements in the problem.
va and vb are known.
R1, R2 and R3 are known.
v1, v2 and v3 are unknowns.
ia, ib, i1, i2 and i3 are unknowns.
There are 2 loop (KVL) equations.
There are 3 node (KCL) equations.
There are 3 Ohm’s Law equations.
There are 8 unknowns and 8 equations.
Working with Dependent Sources
4
i
48 V
KVL @ left loop:
i
+
vo
-
3
i
48V  4  i  3  io
KCL @ top right node: io  4i
Substitute and solve: i  3 A
vo  36V
Example 1 (1/3)
20 A
Ie
50 V
By KCL:
30 A
Ic
Id
+ Va + + Vb - +
25  Vc
Vd 10 
- 50 V If
ie  20 A, id  30 A, i f  30 A, ic  10 A
By Ohm’s Law:
Vc  25I c  250V , Vd  10I d  300V
Example 1 (2/3)
20 A
Ie
50 V
30 A
Ic
Id
+ Va + + Vb - +
25  Vc
Vd 10 
- 50 V If
By KVL: Va  300V , Vb  600V
Power:
Pa  300V  20 A  6.0 kW
Pb  600V  30 A  18.0 kW
Example 1 (3/3)
20 A
Ie
50 V
30 A
Ic
Id
+ Va + + Vb - +
25  Vc
Vd 10 
- 50 V If
Pc  250V   10 A  2.5 kW
Pd  300V  30 A  9.0 kW
Pe    50V    20 A  1.0 kW
Pf  50V  30 A  1.5 kW
Example 2 (1/4)
1
3A
I
12 
84 V
R
4
8
8
12 
Find Source Current, I, and Resistance, R.
Example 2 (2/4)
1
3A
I
84 V
12 
6A
8
Ohm’s Law: 36 V
+
36 V
+4
48 V
12 
KVL: 48 V
R
8
Ohm’s Law: 6 A
Example 2 (3/4)
1
3A
I
84 V
12 
6A
8
KCL: 3 A
+
36 V
3 A -12 V+
+4 +
48 V 60 V
- 12 
Ohm’s Law: 12 V
R
8
KVL: 60 V
Example 2 (4/4)
1
3A
I
84 V
12 
6A
8
Ohm’s Law: 3 A
++
36 V 24 V
3 A -12 V+
+4 +
48 V 60 V
- 12 
KCL: 6 A
Ohm’s Law: R=3 
6A
R
3A
8
KVL: 24 V
KCL: I=9 A