This document discusses key concepts related to electrical circuits including: 1) It defines average/active power, reactive power, and apparent power as well as the relationship between power, voltage, and current using mathematical equations. 2) It explains the significance of power factor as the ratio of real power to apparent power and how power delivered is affected by the power factor. 3) It describes how to use phasor and impedance diagrams to represent voltages, currents, and phase relationships in AC circuits and how to use them to calculate unknown values.

1. ELECTRICAL CIRCUIT
Course Code: CSE132

2.  Definition : Average/active/real power,
reactive power,
apparent power, power factor.
 Relation between P & V & I
 Signification of Power Factor, cos θ
 Diagram : Phasor and Impedance Diagram
 Procedure to find voltage or current using
Out Line:

3. SUBMITTED TO
Mosharraf Hossain Khan
(MHK)
Senior Lecturer Dept. of
CSESUBMITTED BY
 Shahriar Alam »»» 161-15-7651
 Shohag Uddin »»» 163-15-8488
 Md. Musfiqur Rahman Foysal »»» 163-15-8489
 Amina Ahsani »»» 163-15-8533
 Abdullah Bin Kamal 163-15
 Tanvir Hossain »»» 163-15-8534
 Shahidullah Munse »»» 163-15-8535
GROUP-4

4. Definition :
Average power : is the power delivered by the source and diss
or consumed in the load. Sometimes this is also known as Act
Mathematically the equation is –
P= VI cos 
The unit of Average power (P) is Watt.
Reactive Power : The power required by a reactive load in an ac
called reactive power. Reactive power is represented by (Q)
Mathematically the equation is –
Q = VI sin 
The unit of reactive power (Q) is Volt-Ampere Reactive (VAR).

5. Definition :
Apparent power : The product of voltage and current if and on
phase angle differences between current and voltage are igno
Apparent power is represented by (S).
Mathematically the equation is –
P= VI
The unit of Apparent power (S) is volt-ampere.
Power Factor : the power factor of an AC electrical power syst
Defined as the ratio of the real power flowing to the load to
apparent power in the circuit.
P= VI cos  , here cos  is power factor.
In an AC circuit, the product of the r.m.s voltage and the
r.m.s current is called apparent power.

6. Relation between P & V & I
Let, v = Vm sin (t + v) and i = Im sin(t + i)
Then power is defined by
By simplifying this we get -
p = VI cos(v - i) (1 - cos 2t ) + VI sin(v - i) (sin 2t)
Let (v - i) = 
p = VI cos  - VI cos  (cos 2t) + VI sin  (sin 2t)
So average power, P = [(Vm Im/2) cos 
P = VI cos  , this is the required equation.
P = vi = Vmsin(𝜔t+𝜃v)Imsin(𝜔t+𝜃i)
= VmImsin(𝜔t+𝜃v)sin(𝜔t+𝜃i)
P =
𝑉 𝑚 𝐼 𝑚
2
cos(𝜃v − 𝜃i) -
𝑉 𝑚 𝐼 𝑚
2
cos(2𝜔𝑡 + 𝜃𝑣 + 𝜃𝑖)
Fixed Value Time-varying (function of f)

7. Signification of Power Factor :
Power factor = PF = cos  = P/VI…… (1)
The factor cos has a significant control on the delivered power
By simplifying eqn (1) we get the power equation which is,
P= VI cos …….(2)
Here, whatever the value of V and I become if cos  becomes 0
Then the power (P) becomes zero.
For fig-01 P = (100 V)(5 A) (Cos 0) = 250 W
For fig-02 P = (100 V)(5 A) (Cos 90) = 0 W
fig-01
fig-02

8. Impedance Diagram
Introduction :An impedance is defined when an angle is
associated with resistance, inductive reactance, and capacitive
reactance. each can be placed on a complex plane
diagram,
which is known as Impedance diagram .
For any network, the resistance will always appear on the
positive real axis, the inductive reactance on the positive
imaginary axis, and the capacitive reactance on the negative
imaginary axis

9. Phasor Diagram
Introduction : A phasor diagram is used to show the
phase relationships between two or more sine waves
having the same frequency.
Let, Voltage V= 20 sin ( ωt + 120 )
Current I = 5 sin ( ωt + 30 )
Here, Vm = 20 and Im = 30

10. Find voltage/current using Phasor Diagram
i = 5sin(𝜔𝑡 + 30°) => phasor form I = 3.535 A < 30°
V = IZL = (I < 𝜃)(XL<
90°) = (3.535A < 30°)(4Ω < +90°)
= 14.140V < 120°)
And v = 2(14.140)sin(𝜔t+120°) = 20 sin(𝜔𝑡 + 120°)

11. Find voltage/current using Phasor Diagram
i = 15sin(𝜔𝑡 + 30°) => phasor notaion V = 10.605 V < 0°
I =
𝑉
𝑍 𝑐
=
𝑉 <𝜃
𝑋 𝑐 < −90°
=
10.605𝑉 <0°
2Ω <−90°
= 5.303 A < 90°
And i = 2(5.303)sin(𝜔t+90°) = 7.5 sin(𝜔𝑡 + 90°)

12. Thankyou
!