AC Circuit

CHAPTER18:
ALTERNATING CURRENT

CIRCUITS

1

Overview:

Alternating Current (AC)

AC waveform Phasor Power
diagram

Root Avera Instantaneou factor
mean ge s
square
value C L RC RL RCL

R

Curre Volta
nt ge

2

18.1 Alternating current through a resistor
Learning Outcome:

At the end of this chapter, students should be able to:
⚪ Define alternating current (AC).
⚪ Sketch and interpret sinusoidal AC waveform.
⚪ Write and use sinusoidal voltage and current equations.

3

18.1 Alternating current (AC)

⚪ is defined as an electric current whose magnitude and
direction change periodically.

⚪ Figures 18.1a, 18.1b and 18.1c show three forms of alternating
current.

Figure 18.1a: sinusoidal AC

4

Figure 19.1b: saw-tooth AC

Figure 19.1c: square AC 5

⚪ 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 21.1d.

Figure 21.1d: sinusoidal alternating voltage
where

6

21.1.1 Terminology in AC
Frequency (f)

⚪ is defined as a number of complete cycle in one second.

⚪ Its unit is hertz (Hz) OR s−1.
Period (T)

⚪ is defined as a time taken for one complete cycle.

⚪ Its unit is second (s).

⚪ Formulae,

(21.1)

Peak current (I0)

⚪ is defined as a magnitude of the maximum current.

⚪ Its unit is ampere (A).

7

21.1.2 Equations of alternating current and voltage
⚪ Equation for alternating current (I),

(21.2)

⚪ Equation for alternating voltage (V), phase

(21.3)

where

8

Learning Outcome:

21.2 Root mean square (rms) (1 hour)

At the end of this chapter, students should be able to:
⚪ Define root mean square (rms) current and voltage for

AC source.
⚪ Use the following formula,

and

9

21.2 Root mean square (rms)

21.2.1 Mean or Average Current (Iav)

⚪ is defined as the average or mean value of current in a
half-cycle flows of current in a certain direction.

⚪ Formulae:

(21.4)

Note:

Iav for one complete cycle is zero because the current

flows in one direction in one-half of the cycle and in the
opposite direction in the next half of the cycle.

10

21.2.2 Root mean square current (Irms)

⚪ In calculating average power dissipated by an AC, the mean
(average) current is not useful.

⚪ The instantaneous power, P delivered to a resistance R is

where

⚪ The average power, Pav over one cycle of AC is given by

where is the average value of I2 over one cycle and is
given by (21.5)

Therefore

(21.6)

11

⚪ Since thus the square value of current is given by

and the graph of I2 against time, t is shown in Figure 21.2.

⚪ From Figure 21.2, the shaFdiegdurreeg2io1n.2under the curve and

above the dashed line for I02/2 have the same are as the

shaded region above the curve and below the dashed line for

I02/2.

Thus (21.7)

12

⚪ By equating the eqs. (21.5) and (21.7), the rms current is

(21.8)

⚪ Root mean square current (Irms) is defined as the value of

the steady DC which produces the same power in a resistor
as the mean (average) power produced by the AC.
⚪ The root mean square (rms) current is the effective value of the
AC and can be illustrated as shown in Figure 21.3.

Figure 21.3 13

21.2.3 Root mean square voltage (Vrms)

⚪ is defined as the value of the steady direct voltage which
when applied across a resistor, produces the same power
as the mean (average) power produced by the alternating
voltage across the same resistor.

⚪ Its formula is

(21.9)

⚪ The unit of the rms voltage (potential difference) is volt (V).
Note:
Equations (21.8) and (21.9) are valid only for a sinusoidal
alternating current and voltage.

14

Example 21.1 :

An AC source V=500 sin ωt is connected across a resistor of

250 Ω. Calculate

a. the rms current in the resistor,

b. the peak current,

c. the mean power.

Solution :

By comparing to the
Thus the peak voltage is

a. By applying the formulae of rms current, thus

and

15

Solution :
b. The peak current of AC is given by

c. The mean (average) power of the resistor is

16

Example 21.2 :

an
d

Figure 21.4
Figure 21.4 shows a graph to represent alternating current passes
through a resistor of 10 kΩ. Calculate
a. the rms current,
b. the frequency of the AC,
c. the mean power dissipated from the resistor.

17

Solution :
From the graph,
a. By applying the formulae of rms current, thus

b. The frequency of the AC is

c. The mean power dissipated from the resistor is given by

18

Learning Outcome:

21.3 Resistance, reactance and impedance

(2 hours)

At the end of this chapter, students should be able to:

⚪ Sketch and use phasor diagram and sinusoidal waveform
to show the phase relationship between current and
voltage for a single component circuit consisting of

● pure resistor

● pure capacitor

● pure inductor.

⚪ Define and use capacitive reactance, inductive reactance
and impedance.

⚪ Use phasor diagram to analyse voltage, current and
impedance of series circuit of :

● RC

● RL

● RCL. 19

21.3 Resistance, reactance and impedance

21.3.1 Phasor diagram

⚪ Phasor is defined as a vector that rotate anticlockwise about
its axis with constant angular velocity.

⚪ A diagram containing phasor is called phasor diagram.
⚪ It is used to represent a sinusoidally varying quantity such as

alternating current (AC) and alternating voltage.
⚪ It also being used to determine the phase angle (is defined as

the phase difference between current and voltage in AC
circuit).
⚪ Consider a graph represents sinusoidal AC and sinusoidal
alternating voltage waveform as shown in Figure 21.5a.

Meanwhile Figure 21.5b shows the phasor diagram of V and I.

20

Figure 21.5b: phasor diagram Figure 21.5a

⚪ From the Figure 21.5a: and
Thus the phase difference is

⚪ Therefore the current I is in phase with the voltage V and

Note: constant with time.

Leads

Lags behind 21
In antiphase

21.3.2 Impedance (Z)

⚪ The quantity that measures the opposition of a circuit to the
AC flows.

⚪ It is defined by

(21.10
)

OR

(21.11
)

⚪ It is a scalar quantity and its unit is ohm (Ω).

⚪ In a DC circuit, impedance likes the resistance.

22

21.3.3 Pure resistor in an AC circuit

⚪ The symbol of an AC source in the electrical circuit is shown in
Figure 21.6.

Figure 21.6
⚪ Pure resistor means that no capacitance and self-inductance

effect in the AC circuit.

Phase difference between voltage V and current I
⚪ Figure 21.7 shows an AC source connected to a pure resistor R.

AC source

Figure 21.7 23

⚪ The alternating current passes through the resistor is given by

⚪ The alternating voltage across the resistor VR at any instant is

given by

and

where

⚪ Therefore the phase difference between V and I is

In pure resistor, the current I always in phase with the
voltage V and constant with time.

⚪ Figure 21.8a shows the variation of V and I with time while
Figure 21.8b shows the phasor diagram for V and I in a pure

resistor.

24

Figure 21.8b: phasor diagram Figure 21.8a

Impedance in a pure resistor
⚪ From the definition of the impedance, hence

(21.12
)

25

21.3.4 Pure capacitor in an AC circuit

⚪ Pure capacitor means that no resistance and self-inductance
effect in the AC circuit.

Phase difference between voltage V and current I

⚪ Figure 21.9 shows an AC source connected to a pure capacitor

C.

AC source

Figure 21.9

⚪ The alternating voltage across the capacitor VC at any instant is
equal to the supply voltage V and is given by

26

⚪ The charge accumulates at the plates of the capacitor is
⚪ The charge and current are related by

Hence the equation of AC in the capacitor is

and
OR

27

⚪ Therefore the phase difference between V and I is

In the pure capacitor,

the voltage V lags behind the current I by π/2 radians.

OR

the current I leads the voltage V by π/2 radians.
⚪ Figure 21.10a shows the variation of V and I with time while

Figure 21.10b shows the phasor diagram for V and I in a pure

capacitor.

28

Figure 21.10b: phasor diagram Figure 21.10a

Impedance in a pure capacitor

⚪ From the definition of the impedance, hence

and

29

and

(21.13
)

where XC is known as capacitive (capacitative) reactance.

⚪ Capacitive reactance is the opposition of a capacitor to the
alternating current flows and is defined by

(21.14
)
⚪ Capacitive reactance is a scalar quantity and its unit is ohm

(Ω) .

30

⚪ From the eq. (21.13), the relationship between capacitive

reactance XC and frequency f can be shown by using a graph

in Figure 21.11.

Figure 21.11

21.3.5 Pure inductor in an AC circuit

⚪ Pure inductor means that no resistance and capacitance
effect in the AC circuit.

Phase difference between voltage V and current I

⚪ Figure 21.12 shows an AC source connected to a pure inductor

L.

31

AC source
Figure 21.12
⚪ The alternating current passes through the inductor is given by
⚪ When the AC passes through the inductor, the back emf caused
by the self induction is produced and is given by

(21.15 32

⚪ At any instant, the supply voltage V equals to the back emf εB in

the inductor but the back emf always oppose the supply voltage

V represents by the negative sign in the eq. (21.15).Thus

and

OR

⚪ Therefore the phase difference between V and I is

In the pure inductor,

the voltage V leads the current I by π/2 radians.

OR

the current I lags behind the voltage V by π/2 radians.

33

⚪ Figure 21.13a shows the variation of V and I with time while
Figure 21.13b shows the phasor diagram for V and I in a pure

inductor.

Figure 21.13b: phasor diagram Figure 21.13a

34

Impedance in a pure inductor
⚪ From the definition of the impedance, hence

and

and

(21.16

where XL is known as inductive )
reactance.

35

⚪ Inductive reactance is the opposition of a inductor to the
alternating current flows and is defined by

(21.17
)

⚪ Inductive reactance is a scalar quantity and its unit is ohm (Ω).

⚪ From the eq. (21.16), the relationship between inductive

reactance XL and the frequency f can be shown by using a

graph in Figure 21.14.

Figure 21.14 36

Example 21.3 :

A capacitor has a rms current of 21 mA at a frequency of 60 Hz
when the rms voltage across it is 14 V.
a. What is the capacitance of the capacitor?
b. If the frequency is increased, will the current in the capacitor

increase, decrease or stay the same? Explain.
c. Calculate the rms current in the capacitor at a frequency of

410 Hz.
Solution :
a. The capacitive reactance of the capacitor is given by

Therefore the capacitance of the capacitor is

37

Solution :
b. The capacitive reactance is inversely proportional to the

frequency, so the capacitive reactance will decrease if the
frequency increases. Since the current in the capacitor is
inversely proportional to the capacitive reactance, therefore
the current will increase when the capacitive reactance
decreases.
c. Given
The capacitive reactance is

Hence the new rms current in the capacitor is given by

38

Example 21.4 :

A rms voltage of 12.2 V with a frequency of 1.00 kHz is applied to a
0.290 mH inductor.
a. What is the rms current in the circuit?
b. Determine the peak current for a frequency of 2.50 kHz.
Solution :
a. The inductive reactance of the inductor is given by

Thus the rms current in the circuit is

39

Solution :
b. Given

The inductive reactance of the inductor is given by

Thus the peak current in the circuit is
and

40

21.3.6 RC, RL and RCL series circuit

RC series circuit
⚪ Consider an AC source of rms voltage V is connected in series

to a resistor R and a capacitor C as shown in Figure 21.15a.

AC source

Figure 21.15a

⚪ The rms current I passes through the resistor and the

capacitor is equal because of the series connection between
both components.

41

⚪ 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 in Figure

21.15b.
where

is an angle between the rms

current I and rms supply (or
total) voltage V of AC circuit.

Figure 21.15b: phasor diagram

⚪ Based on the phasor diagram, the rms supply voltage V (or total

voltage) of the circuit is given by

(21.18
)42

⚪ Rearrange the eq. (8.18), thus the impedance of RC series

circuit is and

(21.19

⚪ From the phasor diagram in Figure 8.15b ,)the current I leads
the supply voltage V by φ radians where

(21.20

⚪ A phasor diagram in terms of R, XC and Z is ) in Figure
illustrated

21.15c.

Figure 21.15c 43

RL series circuit
⚪ Consider an AC source of rms voltage V is connected in series

to a resistor R and an inductor L as shown in Figure 21.16a.

AC source
Figure 21.16a

⚪ The rms voltages across the resistor VR and the inductor VL

are given by
and

44

⚪ The phasor diagram of the RL series circuit is shown in Figure

21.16b.

Figure 21.16b: phasor diagram

⚪ Based on the phasor diagram, the rms supply voltage V (or total

voltage) of the circuit is given by

(21.21 45
)

⚪ Rearrange the eq. (21.21), thus the impedance of RL series

circuit is and

(21.22

)
⚪ From the phasor diagram in Figure 21.16b , the supply voltage

V leads the current I the by φ radians where

(21.23

⚪ The phasor diagram in terms of R, XL and Z is ) in
illustrated

Figure 21.16c.

Figure 21.16c 46

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 as shown in

Figure 21.17a.

AC source

Figure 21.17a

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

and 47

⚪ The phasor diagram of the RL series circuit is shown in Figure

21.17b.

Figure 21.17b: phasor diagram

⚪ Based on the phasor diagram, the rms supply voltage V (or total

voltage) of the circuit is given by

(21.24
) 48

⚪ Rearrange the eq. (21.24), thus the impedance of RL series

circuit is

and

(21.25
)
⚪ From the phasor diagram in Figure 21.17b , the supply voltage

V leads the current I the by φ radians where

(21.26 49
)

⚪ The phasor diagram in terms of R, XC, XL and Z is illustrated in

Figure 21.17c.

Figure 21.17c

50