Basic Circuit Laws
Units
1
1Nm = 1Ws
2
SI Prefixes
3
Independent Sources
Ideal voltage source
Ideal current source
Ideal assumption: no resistive element in source
Realistic assumption:
Voltage source has an internal series
Current source has an
resistor
internal parallel resistor
R
The internal resistor of an ideal voltage source is zero !
The internal resistor of an ideal current source is infinity !
R
4

R=
A
Ohm’s Law
v(t ) = R ⋅ i(t ) Ohm's Law
v(t )
R=
[] Resistance
i(t )
1 i(t )
G= =
[S] Conductance
R v(t )
[Siemens]
{ [-1 , mho,  ]}
Instantaneous Power
2
v
(
t
)
p(t ) = v(t ) ⋅ i(t ) = R ⋅ i(t )2 =
= G ⋅ v(t )2
R
Note: - p(t) is a parabolic (non-linear) function that is always positive.
- p(t) is no indicator for the direction of power flow.
5
Examples:
p = v⋅i >0
v = R⋅i
Box absorbes power (resistor)
p = -v ⋅ i > 0
v = -R ⋅ i
Box provides power (source)
p = -v ⋅ i > 0
v = -R ⋅ i
Box provides power (source)
p = -v ⋅ (-i) > 0
- v = R ⋅ (-i), v = R ⋅ i
Box absorbes power (resistor)
6
Kirchhoff’s Current Law (KCL)
s
R1
i2
is - i1 - i2 = 0
R2
-is + i1 + i2 = 0
is = i1 + i2
- The algebraic sum of currents entering a node is zero.
- The algebraic sum of currents leaving a node is zero.
- The algebraic sum of currents entering a node equals the sum of
currents leaving the node.
N
In general:
åi
n
=0
n=0
The algebraic sum of all currents at any node in a circuit equals zero.
Example: For the above circuit, find the equivalent resistance.
7
Example - continued
Note: vs = v1 = v2
vs vs
is = i1 + i2 = +
R1 R2
is
1
1
1
= + =
vs R1 R2 Req
i2
s
R1
R2
R1 ⋅ R2
or Req =
= R1  R2
R1 + R2
is
= G1 + G2 = Geq
vs
In general:
n
1
1
=å
Req
j =1 R j
n
Geq = å G j
j =1
8
Kirchhoff’s Voltage Law (KVL)
-v0 + v1 + v2 = 0
or v0 - v1 - v2 = 0
N
In general:
åv
n
=0
n=0
The algebraic sum of all voltages around any
closed loop in a circuit equals zero.
Example: Find the equivalent resistance.
v0 - v1 - v2 = v0 - R1 ⋅ i - R2 ⋅ i = 0  v0 = ( R1 + R2 ) ⋅ i
v0
= R1 + R2 = Req
i
In general:
n
Req = å R j
j =1
9
Voltage Divider
R1
v1 =
vs
R1 + R2
R2
v2 =
vs
R1 + R2
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ïï
ïï
ý
ïï
ïï
ïþ
R1 + R2
v1 + v2 =
vs = vs
R1 + R2
Current Divider
R1 + R2
i1 + i2 =
is = is
R1 + R2
ü
ï
ï
ï
ï
ï
ý
ï
G2
1 R2
R1 R2
R1
i2 =
is =
is =
is =
is ï
ï
Gtot
1 R1 +1 R2
1 + R1 R2
R1 + R2 ï
ï
þ
G1
1 R1
R2 R1
R2
i1 =
is =
is =
is =
is
Gtot
1 R1 +1 R2
R2 R1 +1
R1 + R2
10
Superposition
The principle of superposition states that whenever a linear system is
driven by more than one independent source, the total response can be
found by summing the individual responses to each independent
source.
When applying this principle, short-circuit a voltage source, and open
a current source.
Example: Find the voltage across the 3 resistor.
Example: Find the voltage
+
across the 3
v resistor.
3
-
Step 1: Deactivate current source (voltage divider)
3  (2 + 4)
Example: Find the voltage
v3vs = 120V
+ resistor.
6 + 3  (2 + 4)
across thev3
3vs
-
= 120V
2
= 30V
6 + 2
11
Step 2: Deactivate voltage source (current divider)
1
i3''
2 + (3  6)
=
1
12A
4  [ 2 + (3  6)]
+
v3cs
-
=
i2''
i3''
=-
1
3
1
1
+
3 6
4  [ 2 + (3  6)]
2 + (3  6)
=
4  [ 2 + 2]
2+2
2
=
4
 i3'' = 6A
=-
2
2
2
= -  i2'' = - i3'' = -4A
2 +1
3
3
v3cs = 3 ⋅ i2'' = 3 ⋅ (-4A) = -12V
Step 3: Superposition
v3 = v3vs + v3cs = 30V -12V=18V
12
Mesh-Current Analysis
R = Resistance matrix
13
Example: Find the matrix system of this circuit using mesh analysis.
14
Node-Voltage Analysis
G = Conductance matrix
15
Example: Find the matrix system of this circuit using node analysis.
1
2
3
4
é1 2 +1 1
ù é v1 ù é6Aù
0
0
-1 1
ê
ú ê ú ê ú
ê -1 1 1 1 +1 1.2 +1 4
ú êv2 ú ê 0 ú
-1 1.2
-1 4
ê
ú⋅ê ú = ê ú
ê 0
ú êv ú ê 0 ú
-1 12
1 1.2 +1 8 +1 12
0
ê
ú ê 3ú ê ú
ê 0
0
1 4 +1 3 +1 6úúû êêëv4 úûú êêë 0 úúû
-1 4
êë
16
Nodal Versus Mesh Analysis
 Given a network to be analyzed, how do we know which method is better or
more efficient? The choice of the better method is dictated by two factors.
 Networks that contain many series-connected elements, voltage sources, or
supermeshes are more suitable for mesh analysis.
 Networks with parallel-connected elements, current sources, or supernodes are
more suitable for nodal analysis.
 Also, a circuit with fewer nodes than meshes is better analyzed using nodal
analysis, while a circuit with fewer meshes than nodes is better analyzed using
mesh analysis.
 The key is to select the method that results in the smaller number of equations.
 If node voltages are required, it may be expedient to apply nodal analysis.
 If branch or mesh currents are required, it may be better to use mesh analysis.
 Mesh analysis is the only method to use in analyzing transistor circuits, but
mesh analysis cannot easily be used to solve op amp circuits.
 For non-planar networks, nodal analysis is the only option, because mesh
analysis only applies to planar networks.
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Thevenin Equivalent Circuit
Thevenin’s theorem states that a linear two-terminal circuit can be
replaced by an equivalent circuit consisting of a voltage source VTh
and a resistor RTh, where VTh is the open-circuit voltage at the
terminals and RTh is the input or equivalent resistance at the terminals
when the independent sources are turned off.
18
Norton Equivalent Circuit
Norton’s’s theorem states that a linear two-terminal circuit can be
replaced by an equivalent circuit consisting of a current source IN and
a resistor RN, where IN is the short-circuit current at the terminals and
RN is the input or equivalent resistance at the terminals when the
independent sources are turned off.
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Example: Find the Thevenin voltage and Thevenin resistance.
Solution: Node equation
v1 - 25V
v
+ 1 - 3A = 0
5
20
 v1 = vab = 32V=VTh
Node equation for v2
v2 - 25V
v
v
+ 2 - 3A+ 2 = 0
5
20
4
 v2 = 16V
isc =
16V
= 4A
4
VTh 32V
RTh =
=
 RTh = 8
isc
4A
20
Source Transformations
vs = R ⋅ is
vs
is =
R
Maximum Power Transfer
2
æ
ö
V
÷÷ R
Th
pL = is2 RL = ççç
çè RTh + RL ÷÷ø L
pmax
VTh2
=
4 RTh
Maximum power is transferred to the load when the load resistance
equals the Thevenin (Norton) resistance.
21
Example: Find the value of RL for maximum power transfer in the
following circuit. Find the maximum power.
.
Solution:
RTh = 2 + 3 + (6  12) = 5 +
6 ⋅12
 = 9
6 +12
22
Example - cont’d
Mesh equation:
-12V +18 ⋅ i1 -12 ⋅ i2 = 0 , i2 = -2A  i1 = 2 3A
KVL around the outer loop:
-12V + 6 ⋅ i1 + 3 ⋅ i2 + 2 ⋅ (0) + VTh = 0  VTh = 22V
Maximum Power:
pmax
2
VTh
(22V) 2


=13.44 W
4 RTh
4  9
23
Example: Determine the value of RL that will draw the maximum
power from the rest of the circuit. Calculate the maximum power.
Note: In order to find the Thevenin resistance of circuits with
dependent sources, we need to excite the circuit at its terminals.
24
Example - cont’d
We need to find RTh and VTh. To find RTh, we consider the circuit in Fig.
(a).
Fig. (a)
Fig. (b)
25
Example: Is the 6V source absorbing
power and, if so, how much?
Solution:
1.) 40V 5 = 8A
2.) 5  20 = 4 , 4⋅ 8A = 32V
i
3.) (6 + 4 +10) = 20 ,
4.) 20  30 = 12 , 12⋅1.6A = 19.2V
4⋅ 8A = 32V 20 = 1.6A
19.2V - 6V
= 0.825A , p6V = v6V ⋅ i = 6V ⋅ 0.825A = 4.95W
i=
16
Note: Source transformation in this way for independent sources only26!
Delta-to-Wye (-to-T) Equivalent Circuits
A  circuit viewed as a  circuit.
A Y circuit viewed as a T circuit.
27
Transformation  <> Y
28
Capacitor
C
[F 
s

q
v
]
A typical capacitor
A capacitor with applied
voltage v.
29
KCL =>
The equivalent capacitance of n parallel-connected capacitors is
the sum of the individual capacitances.
30
KVL =>
The equivalent capacitance of series-connected capacitors is the
reciprocal of the sum of the reciprocals of the individual
capacitances.
31
Inductor
L

i
Figure 6.21 Typical form of an inductor.
[ H   s]
32
The equivalent inductance of series-connected inductors is the
sum of the individual inductances.
KVL =>
33
The equivalent inductance of parallel inductors is the reciprocal of
the sum of the reciprocals of the individual inductances.
KCL =>
34
Important characteristics of the basic elements
35