1. Chapter 2: Single-Phase Circuits
Alternating current circuits with each of the individual components, resistor, capacitor or inductor. AC circuits with resistor, inductor and capacitor in series and parallel….
Introduction
Resistive circuit
Inductive circuit
Capacitive circuit
Series circuit
Parallel circuit

2. 1. Introduction Circuit Theory 1 & Engineering Math
Ohm’s Law, Kirchoff’s Law, Voltage divider, current divider will also be used in AC circuit analysis
We’ve learnt that AC voltages and currents can be expressed as :
Time domain (Function of time)
Frequency domain (Phasor)
We also have learnt 3 types about relationship between two waveforms:
In-phase (Sefasa)
Lead (Mendulu)
Lag (Mengekor)

3. In the real world, electrical transmission line is in
3 phase  Red-Yellow-Blue (RYB) or DC
To most home, only single-phase…..
either Red or Yellow or Blue
Three phase – Chapter 3

4. Basic AC circuit comprises voltage supply, and load that is connected to the electric wire…
Resistor  “Perintang” (R)
Inductor  “Pearuh” (L)
Capacitor  “Pemuat” (C)
We’ll analyze circuits that consist of these components, in series and parallel
R,L & C each has different properties/attributes which gives different voltage-current relationships

5. 2. AC circuit with resistive load
Purely resistive circuit:
Resistor attribute:
Current is directly proportional to voltage (Ohm’s Law)
Apply Ohm’s Law: V = IR
Noticed that there are no phase difference. Voltage an current are in phase.

6. Waveform of Voltage & Current in purely resistive circuit: :
Phasor diagram of Voltage & Current in the purely resistive circuit: : VR

7. Example: Solution: VR = IR x R = 12 sin(ωt – 18o) A x 5Ω
Consider a purely resistive circuit. If R=5Ω, and iR = 12 sin(ωt – 18o) A, determine VR and draw the waveform
Solution:
VR = IR x R = 12 sin(ωt – 18o) A x 5Ω
= 60 sin(ωt – 18o) V

8. 3. AC circuit with inductive load
Consider a purely inductive circuit:
Ideal Inductor attribute:
VL is proportional to the rate of change of current
Mathematically: Take IL = Im sin ωt
But we know inductive reactance (Unit in Ω) of an inductor
XL = ωL, thus ω L Im = XLIm = Vm
Also, cos x = sin (x + 90o), substituting….
We have VL = Vm sin (ωt + 90o)…..when IL = Im sin ωt

9. Phasor diagram of Voltage & Current in the purely inductive circuit: :
Waveform of Voltage & Current in the purely inductive circuit: Current lag Voltage by 90o or Voltage lead current by 90o
90o
90o
Phasor diagram of Voltage & Current in the purely inductive circuit: :

10. Example: Solution: Therefore,
Consider a purely inductive circuit. The voltage across a 0.2H inductance is VL = 100 sin (400t + 70o)V. Determine iL and sketch it.
Solution:
Therefore,
The current lags voltage by 900, therefore
BEKP 1423

11. 4. AC circuit with capacitive load
Now consider a purely capacitive circuit:
Ideal Capacitor attribute:
iC is proportional to the rate of change of voltage
Mathematically: Take VC = Vm sin ωt
But we know capacitive reactance (Unit in Ω) of a capacitor
XC = 1/ωC, thus ω C Vm = Vm/XC = Im
Again, cos x = sin (x + 90o), substituting….
iC = Im sin (wt + 90o)…….when VC = Vm sin ωt

12. Waveform of Voltage & Current in the purely capacitive circuit: Voltage lag Current by 90o or Current lead voltage by 90o
Phasor diagram of Voltage & Current in the purely capacitive circuit: :

13. Example: Solution: Therefore,
Consider a purely capacitive circuit. The current through
a 0.1µF capacitor is IC = 5 sin (1000t + 120o)mA. Determine VC.
Solution:
Therefore,
The voltage lags current by 900, therefore
VC = 50sin(1000t+30o)V ωt

14. C I V I L Conclusion Pure Resistive Pure Inductive Pure Capacitive
V and I in-phase
V leads I by 900
I lags V by 900
I leads V by 900
V lags I by 900
How to memorize the relationships?
C I V I L

15. 5. AC circuit with series loads
What is series connection? Connect all the components to make them in series between point A & B A B
What is parallel connection? Connect all the components to make them in parallel between point A & B A B

16. Resistance, Reactance and Impedance
Resistance is essentially friction against the motion of electrons. It is present in all conductors to some extent (except superconductors!), most notably in resistors. Resistance is mathematically symbolized by the letter “R" and is measured in the unit of ohms (Ω).
Reactance is essentially inertia against the motion of electrons. It is present anywhere electric or magnetic fields are developed in proportion to applied voltage or current, respectively; but most notably in capacitors and inductors. Reactance is mathematically symbolized by the letter “X" and is measured in the unit of ohms (Ω).
Impedance is a comprehensive expression of any and all forms of opposition to electron flow, including both resistance and reactance. It is present in all circuits, and in all components. Impedance is mathematically symbolized by the letter “Z" and is measured in the unit of ohms (Ω), in complex form.

17. Resistance, Reactance and Impedance
From previous and with Ohms Law where V = IZ
we get :
Resistor 100Ω Inductor 100mH Capacitor 10uF
f = Hz f = Hz
R = 100Ω R = 0Ω R = 0Ω
X = 0Ω X = jωL = j100Ω X = 1/jωC = -j100Ω

18. R & L in series Applying KVL Phasor diagram: Impedance Triangle: VR V
I (Reference)
Impedance Triangle: Z XL R

19. Example: Find: Reactance and impedance The current
Phase between current and supplied voltage
12Ω
100V, 50Hz
100mH

20. R & C in series Applying KVL Phasor diagram: Impedance Triangle: VR VVC
I (Reference)
Phasor diagram: VR VR VC V
Impedance Triangle: V R XC Z

21. Example:
0.5 A
What is the value of R?
8µF
230V 86.5Hz
Answer: 398Ω

22. R, L & C in series Apply KVL Phasor diagram: I (Reference)
If VL>VC [XL>XC]
V [Z]
VC-VL [XC-XL]
V [Z]
VL-VC [XL-XC]
If VC>VL [XC>XL]
I (Reference)
VR [R]

23. Example:
A circuit having a resistance of 12, an inductance of 0.15H and a
capacitance of 100F in series, is connected across a 100V, 50 Hz
supply. Calculate
the impedance;
the current;
the voltages across R, L and C;
the phase different between the current and the supply voltage
Ans:
19.439Ω;
5.1443A;
VR= V, VL= V VC= V ;
51.880

24. 6. AC circuit with parallel loads
The reference is now voltage, instead of current as in series circuit
CIVIL concept still can be applied
V reference I IR IL IR
KCL will be used V IL I I IR IC I IC V IR
V reference

25. R & L in parallel Applying KCL Phasor diagram: V (Reference) IR IL I I

26. R & C in parallel Applying KCL Phasor diagram: IC I IR V (Reference) I

27. Example:
Three branches, possessing a resistance of 50, an inductance of 0.15H and a capacitance of 100F respectively, are connected in parallel across a 100V, 50 Hz supply. Calculate:
the current in each branch
the supply current
the phase angle between the supply current and the supply voltage
100V
50Hz
50Ω
100μF
0.15H

28. Problem in AC circuit with series-parallel load can be solved using 2 methods:
Phasor method
Complex number method

29. Solution: First, draw the circuit: Then calculate the reactance: 100V,
50 Hz
100F
50
0.15H
Then calculate the reactance:

30. Now calculate the current in each branch;
Since it is parallel, we know that Voltage same at each branch. V is used as the reference
Complex Number Method:
Phasor Method:
I & V in phase in R
I lags V by 900
I leas V by 900

31. Same Answer………….. b) the supply current; Apply KCL, Phasor Method:
Complex Number Method: I
IC-IL
= 3.14A -2.12A
= 1.02A
IR = 2A
V ref
Same Answer…………..

32. Example:
A circuit having a resistance of 12, an inductance of 0.15H and a
capacitance of 100F are connected in series across a 100V, 50 Hz
supply. Calculate
the impedance;
the current;
the voltages across R, L and C;
the phase different between the current and the supply voltage
Solution:
First, draw the circuit:
R = 12
100V, 50 Hz
L = 0.15H
C = 100F

33. a) the impedance;
Phasor Method:
Complex Number Method:
I (Reference)
VR [R]
V [Z]
VL-VC [XL-XC] Z
= 15.29Ω θ I
12Ω

34. b) the current;
Phasor Method:
Complex Number Method:
Apply Ohm’s Law
Since we know that I is taken as reference (Refer to the phasor diagram), phase angle for the I is actually 00

35. c) the voltages across R, L and C;
Phasor Method:
Complex Number Method:
VR= I x R = 5.14 x 12 = 61.68V
VR= I x Impedance across R only
VR= V
= x (12+j0)
= x
VL= I x XL = 5.14 x = V
= V
VL= V
VL= I x Impedance across L only
VC= I x XC = 5.14 x = V
= x (0+j47.12)
VL= V
= x
= V
VL= I x Impedance across C only
= x (0-j31.83)
= x
= V

36. d) the phase different between the current and the supply voltage
Phasor Method:
Complex Number Method:
Refer to the phasor diagram
V leads I by 51.87o or I lags V by 51.87o

37. CHAPTER 2
To be continued