Circuit Elements
Ideal voltage source is a circuit element that maintains a prescribed voltage across its terminals
regardless of the current flowing in those terminals.
Ideal current source is a circuit element that maintains a prescribed current through its terminals
regardless of the voltage across those terminals.
An independent source establishes voltage or current in a circuit without relying on voltages or
currents elsewhere in the circuit. The value of the voltage or current supplied is specified by the
value of the independent source alone.
A dependent source establish a voltage or current whose value depend on the value of a
voltage or current elsewhere in the circuit.
Vs
Vs=µV
is
+
-
(a)
(b)
+
-
Vs=ρix
(c)
(d)
(a) An ideal independent voltage source
(b) An ideal independent current source
(c) An ideal dependent voltage-controlled voltage source
(d) An ideal dependent voltage-controlled current source
(e) An ideal dependent current-controlled voltage source
(f) An ideal dependent current-controlled current source
is=αVx
+
-
is=βix
(e)
(f)
Resistor constrains its voltage and current to be proportional to each other.
Ohm’s Law establishes the proportionality of voltage and current in a resistor.
v = iR
The power absorved by a resistor
v2
p = vi = i R =
R
2
A node is appoint where two or more circuit elements join.
A close path is a loop traced through connecting elements, starting and ending at the same node
and encountering intermediate nodes only once each
Kirchhoff’s current law (KCL) states that the algebraic sum of all the current at any node in a
circuit equal zero.
We must assign an algebraic sign corresponding to every current at node. Assigning a positive
sign to a current leaving a node requires assigning a negative sign to a current entering a node.
Kirchhoff’s voltage law (KVL) states that the algebraic sum of all the voltages around any close
path in a circuit equal zero.
We must assign an algebraic sign to each voltage in the loop. Assigning a positive sign to a
voltage rise requires assigning a negative sign to a voltage drop.
Example: 1
a) Find the current ig and ia in the following circuit
b) Find the voltage vg
c) Verify that the total power developed equals the total power dissipated.
R1
a
30
R3
ig
Ia
+
1.6 A
80
I
R2
90
vg
-
a)
i g = i a + 1.6
From node (a) :
from path (I)
80ia = 1.6(30 + 90) = 192
i g = 2.4 + 1.6 = 4 A
b)
c)
therefore
ia =
192
= 2.4 A
80
V g = 90(1.6) = 144 V
∑P
∑P
dis
= 2.4 2 (80) + 1.6 2 (120) = 768W
dev
= (4)(192) = 768W
Therefore
∑P
= ∑ Pdev
dis
Example 2 The current i0 in the following figure is 4A.
a) Find i1
b) Find the power dissipated in each resistor
c) Verify that the total dissipated in the circuit equals the power developed by the 180V
source
i0
25
+ v0 5
i4
180V
a)
ig
10
+
v1
-
i3
70
i1
v0 = (4)(25) = 100V
v 2 = 180 − 100 = 80 V
+
v2
_
8
i2
v 2 80
=
= 10 A
8
8
i3 + 4 = i 2 ,
i3 = 10 − 4 = 6 A
v1 = v 2 + v3 = 80 + 6(10) = 140 V
i2 =
v1 140
=
=2 A
70 70
i4 = i1 + i3 = 2 + 6 = 8 A
i g = 4 + i4 = 4 + 8 = 12 A
i1 =
b)
p5Ω = 8 2 (5) = 320 W
p 25 Ω = 4 2 (25) = 400 W
p70 Ω = 2 2 (70) = 280 W
p10Ω = 6 2 (10) = 360 W
p8 Ω = 10 2 (8) = 800 W
c)
∑P
dis
= 320 + 400 + 280 + 360 + 800 = 2160 W
Pdev = 180 i g = 180(12) = 2160 W
Circuit Analysis with dependent sources
Example:
For the following circuit
a. Find the current i1
b. Find the voltage v
1V
54 K
a
+
i1
5V
v
-
(f)
1.8 K
30 i1
I
i2
6K
8V
a. Using KCL at node a
i2 = i1 + 30i1 = 31i1
Using KVL in pole I
5 = 54i1 − 1 + 6(31i1 ) Therefore i1 =
b. Using KVL inpole II
6
= 25 µA
240
8 = 1800(30)(25 x10 −6 ) − v + 6000(25 x10 −6 )
Therefore v = 1.35 + 4.65 − 8 = −2 V
HW Problem:
2.14 (page 57)
2.27 (page 61)
Simple Resistive Circuits
Resistor in Serial
R1
R2
R3
R4
V1
R5
R9
R8
R7
V1
Req
R6
In general, if k resistors are connected in series, the equivalent single resistor value is the sum of
the k resistances,
k
Req = ∑ Ri = R1 + R2 + .... + R k
i =1
Resistor in Parallel:
V1
R1
R2
R3
R4
R5
V1
Req
In general form, if k resistor is connected in parallel, the equvalent single resistor value is given in
the following form:
k
1
1
1
1
1
=∑
=
+
+ .... +
Req i =1 Ri R1 R2
Rk
If two resistor connected parallel, in the special case:
R + R2
1
1
1
=
+
= 1
Req R1 R2
R1 R2
! Req
Voltage-divider circuit:
i
R1
Vs
R2
v s = iR1 + iR2
!
+
V1
+
V2
-
i=
vs
R1 + R2
=
R1 R
R1 + R
v1 = iR1 =
R1
v s and
R1 + R2
v 2 = iR2 =
R2
vs
R1 + R2
The current-divider circuit
+
i1
is
R1
i2
v
R2
-
v = i1 R1 = i2 R2 =
i1 =
R1 R2
is
R1 + R2
R2
i s and
R1 + R2
i2 =
R1
is
R1 + R2
The Wheatstone Bridge:
The Wheatstone bridge circuit is used to precisely measure resistances of medium values (the
range 1Ω to 1MΩ).
i1
R1
R2
i2
V
ig
i3
R3
Rx
ix
In the figure, R1,R2,and R3 are known resistors and Rx is the unknown resistor. To find the value
of Rx, we adjust the variable resistor R3 until there is no current in the galvanometer. Then Rx can
be calculated as
i1 = i3 and i2 = i x
i3 R3 = i x R x and i1 R1 = i2 R2 , and then
R3 R x
=
R1 R2
!
Rx =
R2
R3
R1
i1 R3 = i2 R x
Delta-to-Wye (Pi-to-Tee) Equivalent Circuits
Rc
a
b
a
b
R2
R1
Rb
Ra
R3
c
c
In the ∆-connected circuit, the equivalent resistance can be computed as :
Rc ( Ra + Rb )
= R1 + R2
Ra + Rb + Rc
R ( R + Rc )
Rbc = a b
= R2 + R3
Ra + Rb + Rc
R ( R + Ra )
= R1 + R3
Rcd = b c
Ra + Rb + Rc
Rab =
Y-connected resistors can be computed from ∆-connected circuit as:
Rb Rc
Ra + Rb + Rc
Rc R a
R2 =
Ra + Rb + Rc
Ra Rb
R3 =
Ra + Rb + Rc
R1 =
The ∆-connected resistors can be computed from Y-connected circuit as :
R1 R2 + R2 R3 + R3 R1
R1
R R + R2 R3 + R3 R1
Rb = 1 2
R2
R R + R2 R3 + R3 R1
Rc = 1 2
R3
Ra =
Techniques of Circuit Analysis
Thevenin’s Therorem
Any two-terminal bilateral dc network can be replaced by an equivalent circuit consisting of a
voltage source and a series resistor.
To find the Thevenin voltage VTh and the Thevenin resistance RTh:
1. Mark the terminals of the remaining two-terminal network.
2. Calculate RTh by first setting all sources to zero(voltage sources are replaced by short
circuits and current sources by open circuit) and then finding the result resistance
between the two marked terminals
3. Calculate VTh by first replacing the voltage and current sources and then finding the
open-circuit voltage between the marked terminals.
4. Draw the Thevenin equivalent circuit with the marked terminals
Example: Find the Thevenin circuit of a and b terminals for for the following network.
a
3
12
6A
RTh
6
36V
VTh
b
Step 2.
3
12
6
RTh = 3 +
RTh
(6)(12)
= 3+ 4 = 7Ω
6 + 12
Step 3:
Use superposition. For the 36 V source we can apply the voltage divider rule
3
12
6
36 V
'
ETh
=
(6)(36) 216
=
= 12V
6 + 12
18
For the 6A source we have can apply the current divider rule
3
12
6A
6
(12)(6) 72
=
= 4A
12 + 6 18
= I 6 Ω (6) = (4)(6) = 24 V
I 6Ω =
"
ETh
'
"
ETh = ETh
ETh
= (12)(24) = 36 V
Norton’s Theorem
Any two-terminal bilateral dc network can be replaced by an equivalent circuit consisting of a
current source and a parallel resistor.
The steps to find IN and RN
1. Mark the terminals of the remaining two-terminal network.
2. Calculate RTh by first setting all sources to zero(voltage sources are replaced by short
circuits and current sources by open circuit) and then finding the result resistance
between the two marked terminals
3. Calculate IN by first replacing the voltage and current sources and then finding the shortcircuit current between the marked terminals.
4. Draw the Norton equivalent circuit with the marked terminals
Example: Find the Norton equivalent circuit of a and b terminals for the following network.
a
4
4
6
7V
2
8A
b
Step 2:
4
4
6
RN
2
RN =
(6 / 2)(4) 12
=
= 1.714Ω
(6 / 2) + 4 7
Step 3: Using superposition for the 7V source
IN
4
4
4
6
7V
7V
2
I N' =
7
= 1.75 A
4
For 8A source
IN
4
4
6
4
2
8A
2
8A
I N" =
2(8)
= 2.667 A
2+4
I N = I N" − I N' = 2.667 − 1.75 = 0.917
IN
RN
Maximum Power Transfer
RTh
VTh
RL
For maximum Power transfer RTh should be equal to RL
RTh = RL
The maximum power transfer is
p=
VTH2
4 RL
Superposition:
•
•
•
Activate one source at a time to calculate the voltage and currents of the circuits
( replacing ideal current sources with an open circuit to deactivate, replacing voltage
sources with a short circuit to deactivate)
Sum the resulting voltage and currents to determine the voltage and current that exits
when all independent sources are active.
Depending sources are always active when applying superposition
Techniques of Circuit Analysis
The three fundamental laws of network analysis: Ohm’s law,KCL, and KVL;
V = RI Ohm’s law
The sum of the currents at a node must equal zero
N
∑i
n =1
n
= 0 Kirchhoff’s current law (KCL).
The net voltage around a closed circuit is zero.
N
∑v
n =1
n
= 0 Kirchhoff’s voltage law, or KVL.
Current flows from a higher potential to a lower potential in a resistor.
We can express this principle as
i=
v high − vlow
R
Solution of Linear System of Equations Using Cramer’s Rule
For two unknown variables
a1 x + b1 y = c1
a 2 x + b2 y = c 2
c1
c
x= 2
a1
a2
b1
b2
c b − c 2 b1
= 1 2
b1
a1b2 − a 2 b1
b2
a1
a
y= 2
a1
a2
c1
c2
a c − a 2 c1
= 1 2
b1
a1b2 − a 2 b1
b2
For three unknown variables:
a1 x + b1 y + c1 z = d1
a 2 x + b2 y + c 2 z = d 2
a 3 x + b3 y + c3 z = d 3
a1
∆ = a2
a3
b1
b2
b3
c1
c 2 = (a1b2 c3 + b1c 2 a 3 + c1 a 2 b3 ) − (a3 b2 c1 + b3 c 2 a1 + c3 a 2 b1 )
c3
a1
∆ = a2
a3
b1
b2
b3
c1
c 2 = (a1b2 c3 + b1c 2 a 3 + c1 a 2 b3 ) − (a3 b2 c1 + b3 c 2 a1 + c3 a 2 b1 ) ,
c3
d1
d2
d3
b1
b2
b3
c1
c2
c3
x=
1.
y=
,
∆
a1
a2
a3
d1
d2
d3
c1
c2
c3
∆
,
z=
a1
a2
a3
b1
b2
b3
,
d1
d2
d3
∆
The Node-Voltage Method
•
•
•
•
Check the essential node and select one of them is reference node(usually it is a
ground).
Define the remaining n- 1 node voltages as the independent variables.
Apply KCL to write node-voltage equations for the other nodes. A voltage nodes is
defined as the voltage rise from the reference node to a nonreference node.
Solve the linear system equations with voltage as a independent variable.
Example:
Use the node-voltage method to find v in the circuit shown.
2
8V
From node a
a
6
4
b
10
3
1V
−
8 − v a v a v a − vb
+ +
=0
2
4
6
!
11
1
v a − vb = 4 ! 11v a − 2vb = 48
12
6
From node b
−
v a − v b 1 − v b vb
−
+ =0 !
6
10
3
1
3
1
− v a + vb =
! − 5v a + 18v b = 3
6
5
10
va =
48(18) − 3(−2)
582
=
= 3.095 V
11(18) − (−5)(−2) 188
va =
11(3) − (−05)48
273
=
= 1.452 V
11(18) − (−5)(−2) 188
The Mesh-Current Method
•
•
•
Define each mesh current consistently. For convenience, define mesh currents
clockwise. (A mesh is a loop with no other loop inside it.)
Apply KVL around each mesh, expressing each voltage in terms of one or more mesh
currents.
Solve the linear system of equations with mesh currents as the independent variables.
Example:
30
II
90
5
26
80 V
I
31i1 − 5i2 − 26i3 = 30
− 51i1 + 125i2 − 90i3 = 0
− 26i1 − 90i 2 − 124i3 = 0
i1 = 5 A
i2 = 12 A
i3 = 2.5 A
8
III