Note Physics 2 SP025

Example 5.5.2 (i)

The current flowing in the first coil changes
from 2 A to 10 A in 0.4 sec.
(i) Find the mutual inductance between two

coils if an emf of 60 mV is induced in
the second coil. Also
(ii) determine the induced emf in the second
coil if the current in the first coil is
changed from 4 A to 16 A in 0.03 sec.
Consider only the magnitude of induced
emf.

47

Example 5.5.3

Consider two coplanar, co-axial circular coils A and B Let IA be the current flowing in coil A, then
as shown in figure. The radius of coil A is 20 cm while the magnetic field BA at the centre of the
that of coil B is 2 cm. The number of turns is 200 and circular coil A is
1000 for coils A and B respectively. Calculate the
mutual inductance of coil B with respect to coil A. If × T
the current in coil A changes from 2 A to 6 A in 0.04 .
sec, determine the induced emf in coil B.
The magnetic flux linkage of coil B is

=. ×



48

Wireless mobile charging : Energy transfer due to
mutual induction

49

NEXT TOPIC    Topic 6
Alternating current ,AC 50

Chapter 6
Alternating Current

6.1 Alternating Current
6.2 Root mean square (rms)
6.3 Resistance, Reactance and Impedance
6.4 Power and Power Factor

MODE Face to face Non Face to
SLT face
SLT
Lecture 2
Tutorial 6 2

6

1

6.1 Alternating Current

LEARNING OUTCOMES

At the end of this chapter, students should be able to:
a) Define alternating current (AC).

(Lecture : C2, CLO1, PLO1, MQF LOD1)

b) Sketch and interpret sinusoidal AC waveform.

(Tutorial : C3, CLO3, PLO4, CTPS3, MQF LOD6)

c) Use sinusoidal voltage and current equations.

(Tutorial : C3, CLO3, PLO4, CTPS3, MQF LOD6)

2

6.2 Root mean square (rms)

LEARNING OUTCOMES
At the end of this chapter, students should be able to:
a) Define root mean square (rms) current and voltage for AC source.

(Lecture : C1, CLO3, PLO1, MQF LOD1)

b) Use,

and

(Tutorial : C3, CLO3, PLO4, CTPS3, MQF LOD6)

3

6.3 Resistance, Reactance and Impedance

LEARNING OUTCOMES

At the end of this chapter, students should be able to:
a) Sketch and use phasor diagram and sinusoidal waveform to show the

phase relationship between current and voltage for a single
component circuit of:

i. Resistor, R
ii. Capacitor, C
iii. Inductor, L

(Tutorial : C3, CLO3, PLO4, CTPS3, MQF LOD6)

b) Use phasor diagram to analyse voltage, current, and impedance of
series circuit of RL, RC, RLC.

(Tutorial : C3, CLO3, PLO4, CTPS3, MQF LOD6)

4

b) Use phasor diagram to analyse voltage, current and impedance

of series circuit of RC, RL and RCL.

(Tutorial : C4, PLO4, CTPS3, MQF LOD6)

At the end of this chapter, students should be able to:
c) Define and use
i.capacitive reactance,

ii. inductive reactance,

iii. impedance,

iv. phase angle,

(Lecture : C2,PLO1, MQF LOD1)
(Tutorial : C4, PLO4, CTPS3, MQF LOD6)

d) Discuss and explain graphically the dependence of R,
XC, XL and Z on f and relate it to resonance.

(Tutorial : C4, PLO4, CTPS3, MQF LOD6)

5

6.4 Power and Power Factor

LEARNING OUTCOMES

At the end of this chapter, students should be able to:
Apply in AC circuit consisting of R, RC, RL and RCL in series:
i. average power,

*Also known as power loss that only occurs in resistor.
ii. instantaneous power,
iii. power factor,

(Tutorial : C3, CLO3, PLO4, CTPS3, MQF LOD6)

6

Lecture: 1st Hour

6.1 Alternating Current (AC)

− is defined as the flow of electric current whose its
magnitude and direction change periodically.

− symbol in a circuit :
− There are many forms of AC but the most common is a

sinusoidal AC.

Sinusoidal AC

saw-tooth AC square AC

- Another waveform 7
of alternating current

− The value of an AC is given either positive or negative
sign to indicate the flow direction in a circuit.

ex: positive : current flows in clockwise direction
negative : current flows in anticlockwise direction

− When an AC flows through a resistor, there will be a
potential difference (voltage) across it and this voltage is
alternating as shown in Figure below.

sinusoidal alternating voltage Despite AC current flowing
back and forth many times a
second, the energy still
essentially flows continuously
from the power plant to the
electronic devices.

where

8

Graphical Representation of Sinusoidal Current / Voltage

AC waveform takes the shape of a sine curve is known as
sinusoidal current or voltage

Peak (max) Peak Peak to
current (max) peak
voltage voltage

1 complete 1 complete
cycle, T cycle, T

9

Equation representing Sinusoidal Current & Voltage

Note:

phase

* I and V vary with time, t

where : 10

Io = maximum current or peak current
I = instantaneous current at time t

Vo = maximum voltage or peak voltage
V = instantaneous voltage at time t
ω = angular frequency

f – frequency of the ac voltage and current
T – time taken to complete 1 cycle.

EXAMPLE QUESTION 1
The equation for an alternating current is given by I = 77 sin 314t
Find the
(i) peak value
(ii) frequency
(iii) time period and
(iv) instantaneous current, I at t = 2ms
ANSWER

(Answer: (i) 77 A (ii) 50 Hz (iii) 0.02 s (iv) 45.24 A) 11

EXAMPLE QUESTION 2

Write down the equation for a sinusoidal voltage of 50 Hz and its peak
value is 20 V. Draw the corresponding voltage versus time graph.

ANSWER

(Answer: refer to the video Chapter 6-6.1)12

6.2 Root mean square (rms)

root means square (rms) root means square (rms)
current, voltage,
o DEFINITION: is defined o DEFINITION: is defined

as the value of the steady as the value of the steady
direct current (DC) direct voltage (DC) which
which produces the same produces the same power
power dissipation in a dissipation in a resistor as
resistor as the mean the mean (average) power
(average) power produced by the AC.
produced by the AC.
o FORMULA : =

o FORMULA : =

13

The root mean square (rms) value of current and voltage is
known as the effective dc value of current (or voltage) of an ac
current (or voltage) .

The bulbs light with the same
brightness

(that is, they are working at
the same power)

14

Example : If given Io = 10 A
I rms = 0.707(10) = 7.07 A

15

EXAMPLE QUESTION 3

An AC voltage source has an output of V = 2.00 × 102 sin 2πft . The source
is connected to a 1.00 × 102 Ω resistor. Find the rms voltage and rms current
in the resistor.

ANSWER

(Answer: 141 V; 1.41 A) 16

EXAMPLE QUESTION 4

An AC generator with a maximum voltage of 24.0 V and a frequency of
60.0 Hz is connected to a resistor with a resistance R= 265 Ω.
Find the rms voltage and the rms current in the circuit

ANSWER

(Answer: 17 V; 0.0642 A) 17

6.3 Resistance, Reactance and Impedance

18

19

20

(1) Pure Resistor in AC circuit

AC source

o The AC source delivers a sinusoidal voltage to a circuit
consisting of a single resistor.

o The current in the resistor is given by

o The value of the voltage across the resistor at any instant
is given by

where

21

Phasor Diagram Graph variation of V and I with time

lead
lag

o Both V and I are in step ( in phase ) and reach their zero

values, minimum & maximum at the same time.
o the opposition for the AC current in circuit is known as

resistance, R.

o Apply Ohm’s law:

22

o Instantaneous Power, P delivered to the resistor:

and

o Average value of sin2 ω t = ½, thus average Power (Pav)
received by the resistor :

where

o Graph of instantaneous Power being absorbed
power P being absorbed by
the resistor against time t.

o Knowing that :

Using :

23

(2) Pure Capacitor in the AC circuit AC source

o A pure capacitor with
capacitance C is connected to
a sinusoidal voltage supply.

o At a particular instant, charge, Q accumulated on the capacitor
plates is given by :

o The current, I flowing in the circuit is given by :

Let : , we have :

24

o In mathematic, cos ωt can be written as sin ( ωt + ),
hence :

Phasor Diagram Graph variation of V and I with time

lead
lag

CIV

o Current leads the voltage by 90° ( or π/2 radians)

o This means that when the current is maximum the voltage is zero.

25

o Capacitive reactance is the opposition of a capacitor to
the alternating current flows and is given by

The reactance is very large at
low frequencies.

or

Unit : ohm ( Ω ) The reactance is very small at
high frequencies.

o The maximum voltage across a capacitor is

o Also can be express in term of rms value :

o In general, voltage across a capacitor : 26

(* voltage & current can be in form of rms @ maximum value. )

o Instantaneous Power, P is given by :

Identity Trigonometry :

and Graph of instantaneous power P of the
pure capacitor against time t:
where
Power being absorbed

On the first quarter of the cycle,
energy is stored (+) in the
electrostatic filed of the capacitor
through charging process.

o Average Power, Pav = 0 Power being returned to the supply

o A pure capacitor in an AC On the next quarter cycle, capacitor
circuit does not dissipate
energy.

discharges and the energy is returned

(−) to the supply and so on. 27

(3) Pure Inductor in the AC circuit

o AC voltage produces a time varying
current,

AC source

o Since the current is changing with time, self induction occurs in
the inductor induces a voltage (back e.m.f.)

o According to Faraday’s Law, back emf produced is given by :

o At any instant, the supply voltage V equals to the back emf in the
inductor but the back emf always oppose the supply voltage V
represents by the negative sign :

Let : 28

In mathematic, cos ωt can be written as sin ( ωt + ),
hence :

Graph variation of V and I with time

Phasor Diagram

lead
lag

VIL

o The voltage leads current by 90° (or π/2 radians)

o This means that when the voltage is a maximum, the current

is zero. 29

o Inductive reactance is the opposition of a
inductor to the alternating current flows and is
given by:

or The reactance increase with
increasing frequency.
Unit : ohm ( Ω )
o The maximum voltage across a inductor :
o express in term of rms value :

o In general, voltage across a inductor : 30

o Instantaneous Power, P is given by :

Identity Trigonometry : Graph of instantaneous power P of
the pure inductor against time t:
o Average Power, Pav = 0
Power being absorbed
o A pure inductor in an
AC circuit does not On the first quarter of the current cycle,
dissipate energy. power is absorbed (+) in the magnetic
field of the coil (inductor).

Power being returned to the supply 31

On the next quarter cycle, the power is
returned (−) to the supply and so on.

1. Resistive - Pav= ½ Po

V and I are in I VR

phase I P Pav= 0 t

2.Capacitive - VC ++
VL ––
I leads V 9 0 º
I P Pav= 0
3. Inductive -
++ t
V leads I 9 0 º ––
32

EXAMPLE QUESTION 5

A 400 mH coil of negligible resistance is connected to an AC circuit in which
an effective current of 6 mA is flowing. Find out the voltage across the coil if
the frequency is 1000 Hz. Write down the equations of voltage and current.

ANSWER

(Answer: refer to the video Chapter 6-6.3 Part I) 33

EXAMPLE QUESTION 6

A capacitor of capacitance 102 µF is connected across a 220 V, 50 Hz A.C.

mains. Calculate the capacitive reactance, RMS value of current and draw

the phasor diagram. What is the average power for this circuit?

ANSWER

(Answer: 100 Ω; 2.2 A; 0 W) 34

Lecture: 2nd Hour
Combination of RC, RL & RCL series circuit

Inductor(L) : V leads I

CIVILRemember:

Capacitor(C): I leads V

35

(4) RC series circuit

Consider an AC source of rms voltage V is connected in
series to a resistor R and a capacitor C

AC source

The rms current I passes through the resistor and the
capacitor is equal because of the series connection
between both components.

36

The rms voltages across the resistor VR and the
capacitor VC are given by

and

The phasor diagram of the RC series circuit is shown as

below :
where

is an angle between the rms
Lead current I and rms supply (or

total) voltage V of AC circuit.

Lag

Based on the phasor diagram, the rms supply voltage V (or
total voltage) of the circuit is given by

37

and

Thus the impedance of RC series circuit is

From the phasor diagram, the current I leads the supply
voltage V by φ radians where

phasor
diagram in
terms of R,
XC and Z

Lead
Lag 38

(5) RL series circuit

Consider an AC source of rms voltage V is connected in
series to a resistor R and a inductor L

AC source

The rms voltages across the resistor VR and the
inductor VL are given by

and

39

The phasor diagram of the RL series circuit is shown as

below :

Lead

Lag

Based on the phasor diagram, the rms supply voltage V (or
total voltage) of the circuit is given by

and

40

Thus the impedance of RC series circuit is

From the phasor diagram, the supply voltage V leads the
current I the by φ radians where

phasor diagram in terms 41
of R, XL and Z

(6) RCL series circuit

Consider an AC source of rms voltage V is connected in
series to a resistor R, a capacitor C and an inductor L

AC source

The rms voltages across the resistor VR, the capacitor VC
and the inductor VL are given by

; and

42

The phasor diagram of the RCL series circuit is shown as

below :

Lead
Lag

Lead
Lag

Based on the phasor diagram, the rms supply voltage V (or
total voltage) of the circuit is given by

and 43

Thus the impedance of RCL series circuit is given by :

Impedance, Z is the effective resistance (opposition) of an
electric circuit or component to alternating current arising
from the combined effects of ohmic resistance and
reactance.

44

From the phasor diagram, the supply voltage V leads the
current I the by φ radians where

: Phase angle
Phase angle is defined as the an angle between the rms
current I and rms supply (or total) voltage V of AC
circuit.

45

phasor diagram in

terms of R, XC, XL
and Z

If XL > XC , the phase angle is positive ( ) , and the
circuit is said to be inductive.

If XC > XL , the phase angle is negative ( ), and the
circuit is said to be capacitive because the capacitor

dominates the reactance.

46