ECR101 & EEE 131

Lectures

Introduction to Electrical
Engineering (ECR 101)
&
Electrical Circuit – I (EEE 131)

Spring 2017

Faculty
Tahsin Ferdous Ara Nayna

Faculty:Tahsin Ferdous Ara Nayna 1

Review: Positive versus negative

2A -2 A
T1 T2
≡ T1 T2

T1 T1
+ –
2V ≡ -2 V

– +
T2 T2
 Circuits
• A circuit is composed of elements
(sources, resistors, capacitors, inductors)
and conductors (wires).
• Elements are lumped.
• Conductors are perfect.
 Resistors
• A resistor is a circuit element that
dissipates electrical energy (usually as
heat).
• Real-world devices that are modeled by
resistors: incandescent light bulb, heating
elements (stoves, heaters, etc.), long
wires
ρL
R=
A
 Ohm’s Law
• Linear circuit
• Constant Temperature
V=IR I=V.1/R Slope=1/R

V
 Series Resistance
Two elements are in series if the current that flows
through one must also flow through the other.

R1
R2 Req
R3

Req = R1 + R2 + R3
 Parallel Resistance
Two elements are in parallel if they are connected
between the same two nodes.

R1 R2 R3 Req

1/Req = 1/R1 + 1/R2 + 1/R3

R1 R2 R3
Req =
R1 R2 + R2 R3 + R3 R1
 Short Circuit
R=0

i(t)
The +
Rest of v(t)=0
the
Circuit -
 Open Circuit

• R=∞
i(t)=0
The +
Rest of v(t)
the
Circuit -
 Circuits with Series and Parallel
Combinations
• The combination of series and parallel
resistances can be used to find voltages and
currents in circuits
• Simplification
– Resistances are combined to create a simple
circuit (usually one source and one resistance),
from which a voltage or current can be found. Start
from the furthest branch from the source.
• Backtracking
– Once the voltage or current is found, KCL and
KVL, Ohm’s Law, Voltage and Current Dividers
are used to work back through the network to find
voltages and currents.
 Resistance

What is the net resistance of the circuit connected to the
battery? Each resistance has R = 3 kΩ
 1&2 in series ⇒ 6kΩ 2
3 in parallel with 1&2 ⇒ 2kΩ
 4 in series ⇒ 5kΩ
3
 5 in parallel ⇒ 15/8kΩ = 1.875kΩ
1 4
 6 in series ⇒ 4.875kΩ
5

6
R1
R3 R2
R4
 FIRST WE PRACTICE COMBINING RESISTORS

3k
SERIES
6k||3k

(10K,2K)SERIES

6k || 12k = 4k

5kΩ
12k
3k
EXAMPLES COMBINATION SERIES-
SERIES-PARALLEL
9k
AN EXAMPLE WITHOUT REDRAWING

6k || ( 4k + 2k )

12k || 12k = 6k
3k || 6k = 2k
18k || 9k = 6k

RESISTORS ARE IN SERIES IF THEY CARRY

EXACTLY THE SAME CURRENT

6k + 6k + 10k RESISTORS ARE IN PARALLEL IF THEY ARE

CONNECTED EXACTLY BETWEEN THE SAME TWO
NODES
 VOLTAGE DIVIDER
Consider two resistors in series with a voltage
v(t) across them:
IMPORTANT VOLTAGE
DIVIDER EQUATIONS
+ +
R1
R1 v1(t) v1 (t ) = v(t )
-
R1 + R2
v(t) +
R2
R2 v2(t) v2 (t ) = v(t )
- -
R1 + R2
 Y − ∆ TRANSFORMATIONS

THIS CIRCUIT HAS NO RESISTOR IN THEN THE CIRCUIT WOULD

SERIES OR PARALLEL BECOME LIKE THIS AND
BE AMENABLE TO SERIES
IF INSTEAD PARALLEL TRANSFORMATIONS
WE COULD
OF THIS HAVE THIS
 Rab = R2 || ( R1 + R3 )

Rab = Ra + Rb
∆ →Y

R2 ( R1 + R3 ) R R Ra R1
= ⇒ =
Rb R1 Rb R2
=
RR
⇒ R2 = b 1
Ra + Rb = Ra = 1 2 R3
R1 + R2 + R3 R1 + R2 + R3 Rb R3 Ra Rc R1 Rc
R2 R3 REPLACE IN THE THIRD AND SOLVE FOR R1
R ( R + R2 ) Rb =
Rb + Rc = 3 1 R1 + R2 + R3 =
Ra Rb + Rb Rc + Rc Ra
R1 + R2 + R3 R 1
Rb
R3 R1
Rc =
R1 + R2 + R3 R R + Rb Rc + Rc Ra
R1 ( R2 + R3 ) R2 = a b
Rc + Ra = ∆ →Y Rc
R1 + R2 + R3 R R + Rb Rc + Rc Ra
R3 = a b
SUBTRACT THE FIRST TWO THEN ADD Ra
TO THE THIRD TO GET Ra Y −∆
LEARNING EXAMPLE: APPLICATION OF WYE-
WYE-DELTA TRANSFORMATION
c
COMPUTE IS c DELTA CONNECTION

R1 12k × 6k
R3 =
12k + 6k + 18k
a R2 b

R1 R2 a b
Ra =
R1 + R2 + R3
R2 R3
Rb =
R1 + R2 + R3
REQ = 6k + (3k + 9k ) || (2k + 6k ) = 10k
R3 R1
Rc = 12V
R1 + R2 + R3 IS = = 1.2mA
12k
∆ →Y

ONE COULD ALSO USE A

WYE - DELTA TRANSFORMATION ...
 More Complicated Circuits

Parallel and series rules are not enough!

Use Kirchoff’s rules
 KCL and KVL
Kirchhoff’s Current Law (KCL) and
Kirchhoff’s Voltage Law (KVL) are
fundamental properties of circuits that make
analysis possible.
 Kirchhoff’s Rules

Junction rule (conservation of charge)
 Current into junction = sum of currents out of it
I1
I
I2 I = I1 + I 2 + I 3
I3

Loop rule (conservation of energy)
 Algebraic sum of voltages around a closed loop is 0

− I 3 R1 − I 2 R2 − E2 − I 3 R1 + E2 = 0
2 1
− I1 R1 − E1 − I1 R1 + E2 + I 2 R2 = 0
 PROBLEM SOLVING TIP: KVL IS USEFUL
TO DETERMINE A VOLTAGE - FIND A LOOP
INCLUDING THE UNKNOWN VOLTAGE
THE LOOP DOES NOT HAVE TO BE PHYSICAL

+
Vbe
−
− VS + VR + VR + VR = 0
1 2 3

VR = 12V EXAMPLE : VR1 , VR3 ARE KNOWN

2

DETERMINE THE VOLTAGE Vbe

VR = 18V
VR +Vbe +VR −30[V] = 0
1

1 3

LOOP abcdefa
 Kirchhoff’s rules

Determine the magnitudes and directions of the currents
through R1 and R2 in the figure below.
Use I1 = I2 + I3

+6 − 15 I 3 = 0
I2
2 −22 I 2 + 9 + 15 I 3 = 0
I3

1 I 3 = 6 /15 = 0.40
I 2 = 15/ 22 = 0.68
I1
I1 = I 2 + I 3 = 1.08
 LINEARITY
• Linearity is a mathematical property of circuits
that makes very powerful analysis techniques
possible.
• Linearity leads to many useful properties of
circuits:
– Superposition: the effect of each source can
be considered separately.
– Equivalent circuits: Any linear network can
be represented by an equivalent source and
resistance (Thevenin’s and Norton’s
theorems)
 SUPERPOSITION
• Superposition is a direct consequence of
linearity
• It states that “in any linear circuit containing
multiple independent sources, the current or
voltage at any point in the circuit may be
calculated as the algebraic sum of the
individual contributions of each source acting
alone.”
 How to Apply Superposition?
• To find the contribution due to an individual
independent source, zero out the other
independent sources in the circuit.
– Voltage source ⇒ short circuit.
– Current source ⇒ open circuit.
• Solve the resulting circuit using your favorite
technique(s).
 Example

12V
2kΩ 4mA
- +

2mA 1kΩ 2kΩ

I0
 2mA Source

2kΩ

2mA 1kΩ 2kΩ

I’0

I’0 = -4/3 mA
4mA Source

2kΩ 4mA

1kΩ 2kΩ
I’’0

I’’0 = 0
12V Source

12V
2kΩ
- +

1kΩ 2kΩ
I’’’0

I’’’0 = -4 mA
 example (cont’d)

12V
2kΩ 4mA
- +

2mA 1kΩ 2kΩ

I0

I0 = I’0 +I’’0+ I’’’0 = -16/3 mA

 Thevenin’s Theorem
• Any circuit with sources (dependent
and/or independent) and resistors can
be replaced by an equivalent circuit
containing a single source and a single
resistor.
• Thevenin’s theorem implies that we can
replace arbitrarily complicated networks
with simple networks for purposes of
analysis.
Thevenin’s Theorem
R2 E2 A

R1 R3
R4 R
E

RTh B
A

ETH R

B
Find the Thevenen’s Equivalent

120Ω 120.24Ω
+ - +
10V
-
R
120Ω 120Ω
 120Ω 120.24Ω
+ - +
10V V1 Voc V2
-
120Ω 120Ω RTh
A

ETH
120Ω R
V1 = 10V = 5V
120Ω + 120Ω B

120Ω
V2 = 10V = 4.995V
120Ω + 120.24Ω
VTH = V2 − V1 = −0.005V RTH = 120 120 + 120 120.24
 Norton’s Theorem
R2 A

R1 R3
R4 R
E

B
SC

ISC RTh
 Nodal Analysis
1. Choose a reference node.
2. Assign node voltages to the other
nodes.
3. Apply KCL to each node other than the
reference-express currents in terms of
node voltages.
4. Solve the resulting system of linear
equations.
 Nodes, Branches , Loops

NODE: point where two, or more, elements are joined (e.g., big
node 1)
LOOP: A closed path that never goes twice over a node (e.g., the
blue line)
The red path is NOT a loop

BRANCH: Component connected between two nodes (e.g., component R4)

R4)
 Reference Node

500Ω 500Ω
+
I1 V 1kΩ 500Ω I2
500Ω
-

The reference node is called the ground

node.
 Nodal Analysis (cont’d)
1. Choose a reference node.
2. Assign node voltages to the other nodes.
3. Apply KCL to each node other than the
reference-express currents in terms of
node voltages.
4. Solve the resulting system of linear
equations.
 Node Voltages

V1 500Ω V2 500Ω V3

1 2 3
I1 1kΩ 500Ω I2
500Ω

V1, V2, and V3 are unknowns for which we

solve using KCL.
 Nodal Analysis (cont’d)

1. Choose a reference node.

2. Assign node voltages to the other
nodes.
3. Apply KCL to each node other than the
reference-express currents in terms of
node voltages.
4. Solve the resulting system of linear
equations.
 KCL at Node 1
KCL: Algebra sum of current leaving the node is 0.

V1 500Ω V2

I1 V1 − V2 V1
500Ω − I1 + + =0
500Ω 500Ω
 KCL at Node 2

V1 500Ω V2 500Ω V3

1kΩ
V2 − V1 V2 V2 − V3
+ + =0
500Ω 1kΩ 500Ω
 KCL at Node 3

V2 500Ω V3

V3 − V2 V3
500Ω I2 + − I2 = 0
500Ω 500Ω
 Nodal Analysis (cont’d)
1. Choose a reference node.
2. Assign node voltages to the other
nodes.
3. Apply KCL to each node other than the
reference-express currents in terms of
node voltages.
4. Solve the resulting system of linear
equations.
 System of Equations
• Node 1:
 1 1  V2
V1  + − = I1
 500Ω 500Ω  500Ω

• Node 2:
V1  1 1 1  V3
− + V2  + + − =0
500Ω  500Ω 1kΩ 500Ω  500Ω

• Node 3:
V2  1 1 
− + V3 +  = I 2
500 Ω  500 Ω 500 Ω 
 Equations
• These equations can be written by
inspection-the left side:
– The node voltage is multiplied by the sum
of conductances of all resistors connected
to the node.
– Other node voltages are multiplied by the
conductance of the resistor(s) connecting
to the node and subtracted.
• The right side of the equation:
– The right side of the equation is the sum of
currents from sources entering the node.
 Matrix Notation
• The three equations can be combined
into a single matrix/vector equation.
V1 500Ω V2 500Ω V3

1 2 3
I1 1kΩ 500Ω I2
500Ω

 1 1 1 
+
 500Ω 500Ω − 0  V
500Ω  1   I1 
 
 V2  =  0 
1 1 1 1 1
 − + + −
 500Ω 500Ω 1kΩ 500Ω 500Ω     
 1 1 1  V3   I 2 
 0 − +
500Ω 500Ω 500Ω 
 Loop (Mesh) Analysis
1. Identify mesh (loops).
2. Assign a current to each mesh.
3. Apply KVL around each loop to get an
equation in terms of the loop currents.
4. Solve the resulting system of linear
equations.
 Identifying the Meshes

1kΩ 1kΩ

+ +
V1 V2
Mesh 1 Mesh 2
- -
1kΩ

Mesh: A special kind of loop that doesn’t contain any loops within it.
 Mesh Analysis (cont’d)

1. Identify mesh (loops).

2. Assign a current to each mesh.
3. Apply KVL around each loop to get an
equation in terms of the loop currents.
4. Solve the resulting system of linear
equations.
 Assigning Mesh Currents
1kΩ 1kΩ

+ 1kΩ +
V1 V2
I1 I2
- -

A MESH CURRENT IS the CURRENT THE DIRECTION OF THE LOOP

ASSOCIATED TO A MESH. I1, I2 ARE MESH CURRENTS IS SIGNIFICANT
CURRENTS

FOR EVERY CIRCUIT THERE IS A MINIMUM FOR A GIVEN CIRCUIT LET

NUMBER OF LOOP CURRENTS THAT ARE B NUMBER OF BRANCHES
NECESSARY TO COMPUTE EVERY CURRENT N NUMBER OF NODES
IN THE CIRCUIT.
THE MINIMUM REQUIRED NUMBER OF
SUCH A COLLECTION IS CALLED A MINIMAL
LOOP CURRENTS IS
SET (OF LOOP CURRENTS).
L = B − ( N − 1)

MESH CURRENTS ARE ALWAYS INDEPENDENT

AN EXAMPLE

CLAIM: IN A CIRCUIT, THE CURRENT THROUGH

ANY COMPONENT CAN BE EXPRESSED IN TERMS
OF THE LOOP CURRENTS

TWO LOOP CURRENTS ARE

B=7 REQUIRED.
THE CURRENTS SHOWN ARE
N =6 MESH CURRENTS. HENCE
L = 7 − (6 − 1) = 2 THEY ARE INDEPENDENT AND
FORM A MINIMAL SET
 Mesh Analysis (cont’d)
1. Identify mesh (loops).
2. Assign a current to each mesh.
3. Apply KVL around each loop to get an
equation in terms of the loop currents.
4. Solve the resulting system of linear
equations.
Voltages from Mesh Currents
+ VR -
+ VR - I2
R R

I1 I1

VR = I1 R VR = (I1 - I2 ) R
 KVL Around Mesh 1
1kΩ 1kΩ

+ 1kΩ +
V1 V2
I1 I2
- -

-V1 + I1 1kΩ + (I1 - I2) 1kΩ = 0

I1 1kΩ + (I1 - I2) 1kΩ = V1
 KVL Around Mesh 2
1kΩ 1kΩ

+ 1kΩ +
V1 V2
I1 I2
- -

(I2 - I1) 1kΩ + I2 1kΩ + V2 = 0

(I2 - I1) 1kΩ + I2 1kΩ = -V2
 Mesh Analysis (cont’d)

1. Identify mesh (loops).

2. Assign a current to each mesh.
3. Apply KVL around each loop to get an
equation in terms of the loop currents.
4. Solve the resulting system of linear
equations.
 Matrix Notation
• The two equations can be combined into a
single matrix/vector equation.

1kΩ + 1kΩ − 1kΩ   I 1   V1 

 − 1kΩ    = 
 1k Ω + 1k Ω I −
 2   2  V
 WRITE THE MESH EQUATIONS
V2 WHENEVER AN ELEMENT
R1 HAS MORE THAN ONE
+ - LOOP CURRENT FLOWING
THROUGH IT WE COMPUTE
I1
V1 + R2 R3 NET CURRENT IN THE
- I2 DIRECTION OF TRAVEL

R5 R4

DRAW THE MESH CURRENTS. ORIENTATION

CAN BE ARBITRARY. BUT BY CONVENTION
THEY ARE DEFINED CLOCKWISE

NOW WRITE KVL FOR EACH MESH AND APPLY

OHM’
OHM’S LAW TO EVERY RESISTOR.

AT EACH LOOP FOLLOW THE PASSIVE SIGN

CONVENTION USING LOOP CURRENT REFERENCE
DIRECTION

− V1 + I1 R1 + ( I1 − I 2 ) R2 + I1 R5 = 0
V2 + I 2 R3 + I 2 R4 + ( I 2 − I1 ) R2 = 0
 What if there are current
sources?
• The current sources in this circuit will have
whatever voltage is necessary to make the
current correct.
• We can’t use KVL around the loop because we
don’t know the voltage.
 Advantages of Nodal Analysis
• Solves directly for node voltages.
• Current sources are easy.
• Voltage sources are either very easy or
somewhat difficult.
• Works best for circuits with few nodes.
• Works for any circuit.
 Advantages of Loop Analysis
• Solves directly for some currents.
• Voltage sources are easy.
• Current sources are either very easy or
somewhat difficult.
• Works best for circuits with few loops.
 General Test

1Ω 1Ω
V2

V1

1V 1V

i) V1=V2 ii) V1>V2

iii) V1<V2 iv) V1>>V2
≈

Prob-2

1Ω 1Ω

A2

A1

1V 1V

i) A1=A2 ii) A1≈A2

iii) A1<A2 iv) A1>>A2

 Prob-3

a b
R B
V

If V is negative, what is the direction of current through the

resistor?

i) a to b ii) b to a iii) both i & ii iv) no meaning

Prob-4
Ω
1Ω

A
VAB = ?
5V

B
Ω
1Ω

i) 5V ii) 0V iii) 2.5V iv) -2.5V

 Prob-5

Ω
1Ω

A
VAB = ?
Ω
2Ω
5V
B
Ω
1Ω

i) 5V ii) 3.33V iii) 2.5 V iv) 0V

Prob-6
Ω
1Ω

I
Ω
1Ω Ω
1Ω
2V I=?

Ω
1Ω

i) 0.8A ii) 2A iii) 0.5A iv) 1A

 Prob-7

Ω
1Ω

I
Ω
1Ω Ω
1Ω
2V
5V
(a) VAB=?
Ω
1Ω

A B

i) 1.5V ii) 2.5V iii) 0V iv) 0.8V

Prob-8
R

I1
2V
5V

R I2

i) I1=I2 ii) I1>I2 iii) I1<I2 iv) Other

 Prob-9
Ω
1Ω

Ω
1Ω Ω
2Ω
2V Ω
1Ω
A B
I

i) 2.98A ii) 0A iii) 0.33A iv) 1.19A

Prob-10
Ω
1Ω

Ω
1Ω Ω
1Ω Ω
1Ω
RT =?
RT
0.5 Ω

0.5 Ω

i) 0.5 Ω ii) 5Ω iii) 2.5Ω iv) 1Ω

Prob-11
I R

R R
E

If E= 4 V and R= 1 Ω, find I
i) 4A ii) 1A iii) 8A iv) 16A
Prob-12

Superposition theorem is only applicable when the circuit is:

i) Linear ii) nonlinear iii) bilateral iv) linear bilateral

Prob-13
Linear circuit means:
i) I ∞V ii) I∞ R iii) I ∞1/ R iv) V ∞ IR
Prob-14

Ω
1Ω Ω
2Ω

Ω
2Ω I=?
3V
I

Ω
1Ω Ω
2Ω

i) 0.375A ii) 0.5A iii) 0.75A iv) 0A

 Ω
1Ω
Prob-15

Ω
2Ω
4V V

Ω
1Ω

For a certain condition the measured voltage across 2Ω

resistor is 4V, what is the measured current through the
resistor?

i) 1A ii) 2A iii) 0A iv) 4A