ESC IV: Introduction to Electronics and

Electrical Engineering

Unit 4
Fundamentals of Electrical Engineering: Introduction to circuit laws, Network
theorems, Amplitude, Phase, Phase difference, RMS value and Average value
of a AC signal
 Electric Circuit and Charge
• An electric circuit is an interconnection of
electrical elements
• Charge
Charge is a basic SI unit, measured in Coulombs (C)
Charge of single electron is 1.602*10-19 C
One Coulomb is quite large, 6.24*1018 electrons
Charge cannot be created or destroyed, only transferred
Electric Current
The movement of charge is called a current
Historically the moving charges were thought to be positive
Thus we always note the direction of the equivalent positive charges, even if the moving
charges are negative. dq
Current, i, is measured as charge moved per unit time through an element. i  dt
DC vs. AC
• A current that remains constant with time is called Direct Current (DC)
• Such current is represented by the capital I, time varying current uses
the lowercase, i.
• A common source of DC is a battery.
• A current that varies sinusoidally with time is called Alternating Current
(AC)
• Mains power is an example of AC
Direction of current
The sign of the current indicates the direction in which the charge
is moving with reference to the direction of interest we define.
We need not use the direction that the charge moves in as our
reference, and often have no choice in the matter.
A positive current through a component is the same as a negative
current flowing in the opposite direction.
 Voltage
• Electrons move when there is a difference in charge between two locations.
• This difference is expressed at the potential difference, or voltage (V).
• It is always expressed with reference to two locations.
• It is equal to the energy needed to move a unit charge between the locations.
• Positive charge moving from a higher potential to a lower yields energy.
• Moving from negative to positive requires energy.
Power
• Power: time rate of expending or absorbing energy
dw
• Denoted by p p p  vi
dt
• Circuit Elements that absorb power have a positive value of p
• Circuit Elements that produce power have a negative value of p
Energy

Ideal dependent voltage and current source

Passive Sign ConventIon

Ideal Independent Voltage and current source

Resistance Short Circuit as Zero Resistance Open Circuit

• Resist the flow of current

Conductance
• Inverse of resistance
Branches
• A single two terminal elements in a circuit.
loops
• Examples: voltage source, resistor, current
source
Nodes
• The point of connection between two or more branches
• May Include a portion of the circuit (more than a single point)
• Essential node: the point of connection between three or more branches
loops
• Any closed path in a circuit
 Kirchhoff’s Laws
• It tells us how the voltage and currents in different branches are related
• It is two types: KCL (Kirchhoff’s current law) and KVL (Kirchhoff’s voltage law)
Kirchhoff’s Current Law
• The algebraic sum of currents entering
a node (or a closed boundary) is zero
• The sum of currents entering a node
is equal to sum of currents leaving a
node

Kirchhoff’s Current Law for Boundaries

Kirchhoff’s Voltage Law - KVL
• The algebraic sum of voltages around a closed path (i.e. loop) is zero
Consider the circuit shown in Fig. Let us start at point 1 and traverse the
circuit in the clockwise direction. Voltage at point 1 minus the voltage at
point 2 is the voltage drop from point 1 to 2, which is positive. Similarly, the
voltage drop from point 2 to 3 is also positive. However, the voltage drop
from point 3 to 1 is negative. Therefore, some of the voltage drops must be
negative and the rest positive so that the sum around the closed path
(loop) is zero.
Resistors in series Resistors in Parallel
Voltage Divider
Current Divider
Nodal Analysis I1  I 2  i1  i2
Steps to Determine Node Voltages:
I 2  i2  i3
• Select a node as the reference
node. Assign voltage v1, v2, …vn-1 vhigher  vlower
to the remaining n-1 nodes. The i
voltages are referenced with R
respect to the reference node.
• Apply KCL to each of the n-1
nonreference nodes. Use Ohm’s v1  0
law to express the branch i1  or i1  G1v1
currents in terms of node R1
voltages. v1  v2
• Solve the resulting simultaneous 2
i  or i2  G2 (v1  v2 )
R2
equations to obtain the unknown
node voltages v2  0
i3  or i3  G3v2
R3
 v1 v1  v2
 I1  I 2  
R1 R2
v1  v2 v2
I2  
R2 R3

 I1  I 2  G1v1  G2 (v1  v2 )
I 2  G2 (v1  v2 )  G3v2

 G1  G2  G2   v1   I1  I 2 
 
  G2 G2  G3  v2   I 2 
Find the node voltage in the circuit shown in Fig.

At node 1

i1  i2  i3
v1  v2 v1  0
5 
4 2
At node 2 In matrix form

i2  i4  i1  i5 1 1 1 
v2  v1 v2  0 2  4    v  5
4
5   1  
1
  
1 1 v2  5
4 6    
 4 6 4
Determine the voltage at the nodes At node 1

3  i1  ix
v1  v3 v1  v2
3 
4 2

In Matrix Form
At node 3
 3 1 1
At node 2
i1  i2  2ix  4   
2 4  v1  3
ix  i2  i3 v1  v3 v2  v3 2(v1  v2 )  1 7 1    
      v2   0 
v1  v2 v2  v3 v2  0 4 8 2
    2 8 8
2 8 4  3 9 3  v3  0

 4 8 8 
Nodal Analysis with Voltage Sources A circuit with a supernode

• Case 1: The voltage source is connected

between a nonreference node and the
reference node: The nonreference node
voltage is equal to the magnitude of
voltage source and the number of
unknown nonreference nodes is
reduced by one.
• Case 2: The voltage source is connected
between two nonreferenced nodes: a
generalized node (supernode) is formed.
• A supernode is formed by enclosing a (dependent or independent) voltage source
connected between two nonreference nodes and any elements connected in parallel
with it.
• The required two equations for regulating the two nonreference node voltages are
obtained by the KCL of the supernode and the relationship of node voltages due to
the voltage source.
node voltages

i1 i2

2  7  i1  i 2  0
At suopernode 1-2
v1 v2
27   0 v3  v2 v1  v4 v1
2 4
 10  
v1  v2  2 6 3 2
At suopernode 3-4 v1  v2  20
v1  v4 v3  v2 v4 v3
  
3 6 1 4
v3  v4  3(v1  v4 )
Mesh Analysis (analyzing circuits, applicable to Planar circuit.)
• A Mesh is a loop which does not contain any other loops within it
Steps to Determine Mesh Currents

1. Assign mesh currents i1,

i2, .., in to the n meshes.

2. Apply KVL to each of the

n meshes. Use Ohm’s law to
express the voltages in
terms of the mesh currents.

3. Solve the resulting n

(b) The same circuit redrawn with no crossing branches.
simultaneous equations to
get the mesh currents.
 Apply KVL to each mesh. For mesh 1,
 V1  R1i1  R3 (i1  i2 )  0
( R1  R3 )i1  R3i2  V1

For mesh 2,
R2i2  V2  R3 (i2  i1 )  0
A circuit with two meshes.  R3i1  ( R2  R3 )i2  V2
Solve for the mesh currents
 R1  R3  R3   i1   V1 
 Use i for a mesh current and I for a branch current
  R3 R2  R3  i2   V2 
I1  i1 , I 2  i2 , I 3  i1  i2
Find the branch current I1, I2, and I3 using mesh analysis
For mesh 1
 15  5i1  10(i1  i2 )  10  0
3i1  2i2  1
For mesh 2
6i2  4i2  10(i2  i1 )  10  0
i1  2i2  1

We can find i1 and i2 by substitution method or

Cramer’s rule
I1  i1 , I 2  i2 , I 3  i1  i2
 Use mesh analysis to find the current I0 in the circuit

Apply KVL to each mesh. For mesh 1,

 24  10(i1  i2 )  12(i1  i3 )  0
11i1  5i2  6i3  12
For mesh 2
24i2  4(i2  i3 )  10(i2  i1 )  0
 5i1  19i2  2i3  0
In matrix from
For mesh 3
4 I 0  12(i3  i1 )  4(i3  i2 )  0  11  5  6  i1  12
At node A, I 0  I1  i2 ,  5 19  2 i2    0 
  1  1 2  i   0 
4(i1  i2 )  12(i3  i1 )  4(i3  i2 )  0   3   
 i1  i2  2i3  0 we can calculus i1, i2 and i3 by Cramer’s rule, and find I0
Mesh Analysis with Current Sources
Current source exist only in one mesh

i1  2A
One mesh variable is reduced
Current source exists between two meshes, a super-mesh is obtained
• a supermesh results when two meshes have a (dependent , independent)
current source in common.

6i1  14i2  20
Properties of a Supermesh
1. The current is not completely ignored i1  i2  6
• provides the constraint equation necessary to solve for the mesh current.
2. A supermesh has no current of its own.
3. Several current sources in adjacency form a bigger supermesh.
For the circuit in Fig, find i1 to i4 using mesh analysis

2i1  4i3  8(i3  i4 )  6i2  0

i1  i2  5
i2  i3  i4
8(i3  i4 )  2i4  10  0

If a supermesh consists of two meshes, two equations are needed; one is obtained using
KVL and Ohm’s law to the supermesh and the other is obtained by relation regulated due
to the current source.
Similarly, a supermesh formed from three meshes needs three equations: one is from
the supermesh and the other two equations are obtained from the two current sources.
Nodal and Mesh Analysis by Inspection
(a)For circuits with only resistors and independent current sources

I1  I 2  G1v1  G2 (v1  v2 ) (3.7)

I 2  G2 (v1  v2 )  G3v2 (3.8)
 MATRIX
G1  G2  G2   v1   I1  I 2 

  G2 G2  G3  v2   I 2 
(b) For planar circuits with only resistors and independent voltage sources

 R1  R3  R3   i1   v1 

  R3 R2  R3  i2   v2 
In general, the node voltage equations in terms In general, if the circuit has N meshes, the mesh-
of the conductances is current equations as the resistances term is

G11 G12  G1N  v1  i1   R11 R12  R1N  i1  v1 
G G  G  v  i   R R  R  i  v 
 21 22 2N
 2    2   21 22 2N
 2    2 
                     
G G  G  v  i   R R  R  i  v 
 N1 N 2 NN   N   N  N1 N 2 NN   N   N
where G : the conductance matrix, where R : the resistance matrix,
v : the output vector, i : the input i : the output vector, v : the input vector
vector
Write the mesh current equations equations

The mesh-current equations are

 9 2 2 0 0  i1 
 2 10  4  1  1  i2  4
   6
 2 4 9 0 0  i3     6
 0 1  3  i4  0
   6
0 8
 
  
The input voltage vector v in volts
 0 1 0 3 4  i5
v1  4, v2  10  4  6,
v3  12  6  6, v4  0, v5  6