UNIT 8: ALTERNATING CURRENT (A.C.

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8.1 Alternating Current (a.c.)

Definition is defined as an electric current whose direction changes periodically. Figures 8.1a, 8.1b and 8.1c show three forms of alternating current.

I0
0

I0
I0

1 T 2

3 T 2

2T

Fig. 8.1a : Sinusoidal a.c. a.c.

0
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I0

1 T 2

3 T 2

2T

Fig. 8.1b : Saw-tooth a.c. Sawa.c.

2

I
I0
0 1 T 2

I0

3 T 2

2T

Fig. 8.1c : Square a.c. a.c.

When an a.c flows through a resistor, there will be a potential difference (voltage) across it and this voltage is alternating voltage as shown in figure 8.1d.

V0
0 1 T 2

Fig. 8.1d : Sinusoidal alternating voltage

V0
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3 T 2

2T

t
where

Vo : peak voltage I o : peak current

T : period

8.1.1 Terminology in a.c. Frequency ( f ) Definition: Number of complete cycle in one second. second Unit: hertz (Hz) or s-1 Period ( T ) Definition: Time taken for one complete cycle. Unit: second (s) Equation : 1

T=

Peak (maximum) current ( Io ) Definition: Magnitude of the maximum current. 8.1.2 Equation for alternating current and voltage. Equation for the current ( I ) : Equation for the voltage ( V ) : where
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I = I 0 sin t
phase

(8.1a) (8.1b)

: angular frequency I 0 : peak current V0 : peak voltage

V = V0 sin t

8.2 Mean or Average Current (Iav)

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

I av =
Note :

2Io Io = (2 )

(8.2a)

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.

8.3 Root Mean Square Values

8.3.1 Root mean square current (Irms) In calculating average power dissipated by an a.c., the mean (average) current is not useful. The instantaneous power, P delivered to a resistance R is

P = I 2R
5

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The average power, Pav over one cycle is given by where

is the average of I2 over one cycle and can be written as

Pav = I 2 R

(8.3a)
2

I 2 = (I rms )
I2
2

I rms =

(8.3b) (8.3c)

therefore the eq. (8.3a) can be written as

Pav = (I rms ) R

Since

I = I 0 sin t
I rms =

then eq. (8.3b) becomes

I 02 sin 2 t

Using a double-angle formula for trigonometry and trigonometry identity, cos 2 = cos 2 sin 2 and cos 2 = 1 sin 2

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thus

cos 2 = 1 2 sin 1 sin 2 = (1 cos 2 ) and = t 2 I 02 (1 cos 2t ) I rms = 2

The mean value of cos2t for one cycle is zero, finally eq. (8.3b) can be written as

I rms =

I0 2

(8.3d)

Root mean square current (Irms) is defined as the value of the steady d.c. which produces the same power in a resistor as the d.c. mean (average) power produced by the a.c. a.c. The root mean square (rms) current is the effective value of the a.c. 8.3.2 Root mean square voltage/p.d. (Vrms) Definition 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

Vrms =

Note : Eq. (8.3d) and eq. (8.3e) are valid only for a sinusoidal alternating current and voltage (p.d.)
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V0 2

(8.3e)

Example 1 : An a.c. source V=200 sin t is connected across a resistor of 100 . Calculate a. the r.m.s. current in the resistor. b. the peak current. c. the mean power.

R=100 V = 200 sin t compare with V = V0 sin t then the peak voltage, V0 = 200 V a. The r.m.s. current is V0 V I rms = rms and Vrms = 2 R V0 I rms = R 2 I rms = 1.41 A
Solution: b. The peak current is

I rms =
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I0 2 I 0 = 1.99 A

c. By applying the equation of mean power, thus

Pav = (I rms ) R
2

Example 2 :

Pav = 199 W

V (Volt )
200
0

0.02

0.04

0.06
and

0 .08

t (second)

200

The alternating potential difference shown above is connected across a resistor of 10 k. Calculate a. the r.m.s. current, b. the frequency, c. the mean power dissipated in the resistor.
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R=10x103 From the graph, V0 = 200 V and T=0.04 s V0 V a. The r.m.s. current is I rms = rms and Vrms = 2 R V0 I rms = R 2 I rms = 1.41x10 2 A
Solution: b. The frequency of the a.c. is

f =

1 T f = 25.0 Hz
2

c. By applying the equation of mean power, thus

Pav = (I rms ) R
Pav = 1.99 W

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10

8.4 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 (a.c.) and alternating voltage (a.v.). It also being used to determine the phase difference between current and voltage in a.c. circuit. 8.4.1 In Phase Consider a graph represents sinusoidal a.c. and sinusoidal a.v. as shown in figure 8.4a.

I
V

V0 I0
0 1 T 2

I0 V0
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3 T 2

2T

Fig. 8.4b : Phasor diagram

Fig. 8.4a

11

From the figure 8.4a :

Thus the phase difference is

I = I 0 sin t V = V0 sin t

= t t = 0 Conclusion : The current I is in phase with the voltage V and constant

with time. 8.4.2 Lead Consider a graph represents sinusoidal a.c. and sinusoidal a.v. as shown in figure 8.4c.

V0 & I 0

1 T 2

3 T 2

2T

V0 & I 0
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Fig. 8.4d : Phasor diagram

Fig. 8.4c

12

From the figure 8.4c :

constant with time. 8.4.3 Lags behind Consider a graph represents sinusoidal a.c. and sinusoidal a.v. as shown in figure 8.4e.

V = V0 sin t + Thus the phase difference is 2 = rad = t + t 2 2 Conclusion : The voltage V leads the current I by /2 radians and

I = I 0 sin t V = V0 cos t

V0 & I 0

V0 & I 0

1 T 2

3 T 2

2T

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Fig. 8.4f : Phasor diagram

Fig. 8.4e

13

From the figure 8.4e :I

V = V0 sin t Thus the phase difference is 2 = rad = t t 2 2 Conclusion : The voltage V lags behind the current I by /2 radians
and constant with time.

= I 0 sin t V = V0 cos t

8.5 Impedance (Z)

The quantity that measures the opposition of a circuit to the a.c. flows. V0 It is defined by

Z=

Vrms I rms

It is a scalar quantity and its unit is ohm () In a d.c. circuit, impedance is the resistance R.
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2 or I0 2

Z=

V0 I0

(8.5a)

8.6 Pure Resistor in the A.C. Circuit

The symbol of an a.c. source in the electrical circuit is shown in figure 8.6a.

Fig. 8.6a Pure resistor means that no capacitance and self-inductance effect selfin the a.c. circuit. 8.6.1 Phase difference between Voltage V and current I Figure 8.6b shows an a.c. source connected to a pure resistor R.

VR
I
a.c. source a.c.

The current flows in the resistor is given by The voltage across the resistor VR at any instant is given by

I = I 0 sin t

VR = IR VR = I 0 R sin t and I 0 R = V0
15

Fig. 8.6b
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VR = V0 sin t = V where V : Supply voltage

Therefore the phase difference between V and I is

= t t = 0 In pure resistor, the current I is in phase with the voltage V . Figure 8.6c shows the variation of V and I with time while Figure 8.6d shows the phasor diagram for V and I in a pure resistor.

I
V

V0 I0
0 1 T 2

I0 V0
Fig. 8.6d : Phasor diagram

3 T 2

2T

Fig. 8.6c

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16

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

Z=

Vrms V0 = =R I rms I 0

(8.6a)

8.6.3 Power in a pure resistor Since V and I are in phase, the instantaneous power P is given by

P = IV P = (I 0 sin t )(V0 sin t ) P = I 0V0 sin 2 t and I 0V0 = P0

(8.6b)

P = P0 sin 2 t

where P : peak(maximum) power 0 Therefore the graph of variation of power with time is shown in figure Power (P ) Power being absorbed 8.6e.

P0
Fig. 8.6e
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P = P0 sin 2 t

1 T 2

3 T 2

2T

17

The average (mean) power Pav in a resistor is given by

Pav = P0 sin 2 t
Pav = 1 P0 2
or

and

sin 2 t =
I 0V0 2

1 2
(8.6c)

Pav =

8.7 Pure Capacitor in the A.C. Circuit

Pure capacitor means that no resistance and self-inductance effect selfin the a.c. circuit. 8.7.1 Phase difference between Voltage V and current I Figure 8.7a shows an a.c. source connected to a pure capacitor C.

VC
I
a.c. source a.c.

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

V = V0 sin t = VC
Q = CVC Q = CV0 sin t

The charge accumulates at the plates of the capacitor is

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Fig. 8.7a

18

Charge and current are related by

I= I=

Hence the equation of a.c. in capacitor is

dQ dt

d (CV0 sin t ) dt d I = CV0 (sin t ) dt I = CV0 cos t and CV0 = I 0 I = I 0 cos t or I = I 0 sin t + 2 Therefore the phase difference between V and I is = t + t 2 = rad 2 In pure capacitor, the current I leads the voltage V by /2 radians.
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Figure 8.7b shows the variation of V and I with time while Figure 8.7c shows the phasor diagram for V and I in a pure capacitor.

rad

V0 I0
0 1 T 2

V
T

I0 V0
Fig. 8.7c : Phasor diagram

3 T 2

2T

Fig. 8.7b

8.7.2 Impedance in a pure capacitor From the definition of the impedance, hence

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V0 and I 0 = CV0 I0 V Z= 0 CV0 Z=

20

1 = XC C 1 XC = 2fC Z=

and

= 2f
(8.7a)

where XC is known as capacitive (capacitative) reactance. (capacitative) Capacitive reactance is the opposition of a capacitor to the current flows and is defined by

f : frequency of a.c. source C : capacitance of the capacitor

XC =

Vrms V0 = I rms I 0

(8.7b)

The unit of capacitive reactance is ohm () frequency f can be shown by using graph in figure 8.7d. From eq. (8.7a), the relationship between capacitive reactance XC and

XC
Fig. 8.7d
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X C

1 f

21

8.7.3 Power in a pure capacitor Since the current I leads the voltage V by instantaneous power P is given by

/2 radians, the

P = IV P = (I 0 cos t )(V0 sin t ) P = I 0V0 sin t cos t and I 0V0 = P0 1 P = P0 sin 2t 2 1 (8.7c) P = P0 sin 2t 2

Therefore the graph of variation of power with time is shown in figure 8.7e. Power (P ) Power being absorbed 1 P
0

2
Fig. 8.7e

P=

P0 sin 2t

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P0 2

1 T 2

3 T 2

2T

t
22

Power being returned to supply

The average (mean) power Pav in a capacitor is

1 P0 sin 2t 2 Pav = 0 Pav =

and

sin 2t = 0

8.8 Pure Inductor in the A.C. Circuit

Pure inductor means that no resistance and capacitance effect in the a.c. circuit. 8.8.1 Phase difference between Voltage V and current I Figure 8.8a shows an a.c. source connected to a pure inductor L. The current flows in the inductor is L given by I = I sin t

VL
I
a.c. source a.c.

When the current flows in the inductor, the back e.m.f. caused by the self induction is produced and given by

B = L

Fig. 8.8a
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dI dt d B = L (I 0 sin t ) dt B = LI 0 cos t

(8.8a) 23

B (voltage across the inductor) but the back e.m.f always oppose the supply voltage V represents by the negative sign in eq. (8.8a).
By comparing the magnitude of V and B ,thus

At each instant the supply voltage V must be equal to the back e.m.f

LI 0 = V0 V = V0 cos t or V = V0 sin t + 2 Therefore the phase difference between V and I is = t + t 2 = rad 2

V = B = LI 0 cos t
and In pure inductor, inductor the voltage V leads the current I by

/2 radians. /2 radians.
24

or
the current I lags behind the voltage V by
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Figure 8.8b shows the variation of V and I with time while Figure 8.8c shows the phasor diagram for V and I in a pure inductor.

rad

V0 I0
0 1 T 2

I0 V0

3 T 2

2T

Fig. 8.8b Fig. 8.8c : Phasor diagram 8.8.2 Impedance in a pure inductor From the definition of the impedance, hence

Z=

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V0 and V0 = LI 0 I0 LI Z= 0 I0

25

Z = L = X L X L = 2fL

and

= 2f
(8.8b)

where XL is known as inductive reactance.

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

f : frequency of a.c. source L : self inductance of the inductor

XL =

Vrms V0 = I rms I 0

(8.8c)

The unit of inductive reactance is ohm () From eq. (8.8b), the relationship between inductive reactance XL and frequency f can be shown by using graph in figure 8.8d.

XL
X L f
Fig. 8.8d
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26

8.8.3 Power in a pure inductor Since the voltage V leads current I the by instantaneous power P is given by

/2 radians, the

P = IV P = (I 0 sin t )(V0 cos t ) P = I 0V0 sin t cos t and I 0V0 = P0 1 P = P0 sin 2t 2 1 (8.8d) P = P0 sin 2t 2

Therefore the graph of variation of power with time is shown in figure 8.8e. Power (P ) Power being absorbed 1 P
0

2
Fig. 8.8e

P=

P0 sin 2t

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P0 2

1 T 2

3 T 2

2T

t
27

Power being returned to supply

The average (mean) power Pav in a inductor is

Note : For both pure capacitor and inductor, the average (mean) power is zero because the power is positive for quarter of a cycle and negative for the next quarter of a cycle in each half cycle. The term resistance is not used in pure capacitor and inductor because no heat is dissipated in both devices. Example 3 : A capacitor with C=4700 pF is connected to an a.c. supply with r.m.s. voltage of 240 V and frequency of 50 Hz. Calculate a. the capacitive reactance. b. the peak current in the circuit. Solution:

1 P0 sin 2t 2 Pav = 0 Pav =

and

sin 2t = 0

C=4700x10-12 F, Vrms=240 V, f=50 Hz.

a. By applying the equation of capacitive reactance, thus

XC =
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1 2fC

X C = 677 x10 3

28

b. From the definition of the capacitive reactance, thus

XC =

Example 4 : A 240 V r.m.s. supply with a frequency of 50 Hz causes an r.m.s. current of 3.0 A to flow through an inductor which can be taken to have zero resistance. Calculate a. the reactance of the inductor. b. the inductance of the inductor. Solution: Irms=3.0

I0 Vrms and I rms = 2 I rms V 2 I 0 = rms XC I 0 = 5.01x10 4 A

A, Vrms=240 V, f=50 Hz.

a. From the definition of the inductive reactance, thus

XL =

b. By using the equation of the inductive reactance, thus

Vrms I rms

X L = 80

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X L = 2fL X L= L 2f

L = 0.26 H

29

8.9 RC, RL and LRC in Series Circuit

8.9.1 RC in series circuit Consider an a.c. source with voltage V is connected in series with a resistor R and a capacitor C as shown in figure 8.9a. C R The voltage across the resistor VR and the capacitor VC are given by

VR

VC

I
a.c. source a.c.

VR = IR VC = IX C

The phasor diagram of RC circuit is shown by figure 8.9b. Based on the phasor diagram, the total voltage V(supply voltage) across both resistor R and capacitor C is

Fig. 8.9a

VR

VC
where

: phase angle
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Fig. 8.9b : Phasor diagram

V 2 = VR2 + VC2 2 2 V 2 = (IR ) + (IX C ) 1 2 V 2 = I 2 R 2 + X C and X C = C 1 2 (8.9a) V =I R + 2 2 30 C

Vrms , hence eq. (8.9a) can be written as I rms 1 2 (8.9b) Z = R 2 + 2 2 or Z = R 2 + X C C From the phasor diagram (fig. 8.9b), the current I leads the supply voltage V by radians where V IX tan = C tan = C VR IR 1 XC (8.9c) or tan = tan = CR R From the eq. (8.9b) and (8.9c), the new phasor diagram in terms of R, XC and Z can be sketched (refer to figure 8.9c) R Fig. 8.9c : Phasor diagram in XC terms of R, XC and Z Z
Since

Z=

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31

8.9.2 RL in series circuit Consider an a.c. source with voltage V is connected in series with a resistor R and an inductor L as shown in figure 8.9d.

VR
I

VL

The voltage across the resistor VR and the inductor VL are given by

VR = IR VL = IX L

a.c. source a.c.

The phasor diagram of RL circuit is shown by figure 8.9e. Based on the phasor diagram, the total voltage V(supply voltage) across both resistor R and inductor L is

Fig. 8.9d

VL

VR

V 2 = VR2 + VL2 2 2 V 2 = (IR ) + (IX L ) 2 V 2 = I 2 R 2 + X L and X L = L

Fig. 8.9e : Phasor diagram

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V = I R 2 + 2 L2

(8.9d)
32

Since

Z=

Vrms , hence eq. (8.9d) can be written as I rms 2 Z = R 2 + 2 L2 or Z = R 2 + X L

(8.9e)

From the phasor diagram (fig. 8.9e), the supply voltage V leads the current I by radians where

tan =

VL VR

tan =

From the eq. (8.9e) and (8.9f), the new phasor diagram in terms of R,

IX L IR XL tan = R

or

tan =

L R

(8.9f)

XL and Z can be sketched (refer to figure 8.9f)

XL
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Fig. 8.9f : Phasor diagram in terms of R, XL and Z

33

8.9.3 LRC in series circuit Consider an a.c. source with voltage V is connected in series with an inductor L, a resistor R and a capacitor C as shown in figure 8.9g.

VL
I

VR
a.c. source a.c.

VC

The voltage across the inductor VL , resistor VR and resistor VR are given by V = IX

The phasor diagram of LRC circuit is shown by figure 8.9h. Based on the phasor diagram, the total voltage V (supply voltage) across inductor L , resistor R and capacitor C is
2

VR = IR VC = IX C

VL

Fig. 8.9g

(VL VC )
VC
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V VR
I

V 2 = VR2 + (VL VC ) 2 2 V 2 = (IR ) + (IX L IX C ) V 2 = I 2 R2 + (X L X C )

]
34

Fig. 8.9h : Phasor diagram

V = I R2 + (X L X C )

(8.9g)

Since

Z=
2

Vrms I rms

, hence eq. (8.9g) can be written as

2

1 (8.9h) Z = R + ( X L X C ) or Z = R + L C From the phasor diagram (fig. 8.9h), the supply voltage V leads the current I by radians where VL VC I (X L X C ) tan = tan = VR IR 1 L X L XC C (8.9i) or tan = tan = R R From the eq. (8.9h) and (8.9i), the new phasor diagram in terms of XL , X L , R, X and Z can be sketched (refer to figure 8.9i) C
2
2

(X L X C )
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Z
R

Fig. 8.9i : Phasor diagram in terms of XL, R, XC and Z

XC

35

8.10 Resonance and Power in A.C. Circuit

8.10.1 Resonance in a.c. circuit Definition is defined as the phenomenon that occurs when the frequency of the applied voltage is equal to the frequency of the LRC series circuit. Figure 8.10a shows the variation of XC, XL,R and Z with frequency f of the LRC series a.c circuit. From the graph, the value of Z X C , X L , R, Z is minimum when

X L = XC

(8.10a)

XL f
R

where its value is

Z = R2 + (X L X C )

0
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fr
Fig. 8.10a

1 XC f f

Z min = R 2 + 0 Z min = R
This phenomenon occurs at frequency of fr known as resonant frequency. frequency
36

When the resonance occurred in the LRC series circuit, the Z is minimum but the r.m.s. current flows in the circuit is maximum and given by V V When resonance I rms = rms = rms occurred Z R The resonant frequency fr of the LRC series circuit is given by

1 X L = X C where X L = L and X C = C 1 L = C 1 and = 2f r = LC f r= 1 2 LC

where

hence

f r : resonant frequency : resonant angular frequency At frequencies above or below the resonant frequency fr, the r.m.s.
current is less than the maximum current. The series resonance circuit is used for tuning a radio receiver.
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37

8.10.2 Power in a.c. circuit No power is absorbed by either inductors or capacitors over a complete cycle and therefore the average (mean) power absorbed by both inductors and capacitors is zero. In an a.c. circuit in which there is resistance R, inductance L and capacitance C, the total average power Pav is equal to that dissipated in the resistance i.e.

Pav = I 2 R

or

Pav = IVR
VR V

(8.10b)

From the phasor diagram of the LRC series circuit in figure 8.10b,

(VL VC )

V VR
Fig. 8.10b

We get cos = can be written as

, then eq. (8.10b)

Pav = IV cos and V = IZ

Pav = I 2 Z cos = Pr
(8.10c) where cos is called the power factor of the circuit and Pr is the average real power. power
38

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Note : Power factor can also be calculated by using formula below : From the figure 8.10b,

VR V IR cos = IZ R cos = Z cos =

where VR

= IR and V = IZ

(8.10d)

From the figure 8.10b, Since V and I are not in phase, then the average apparent power Pa is given by

Pa = IV = I 2 Z =

V2 Z

(8.10e)

By substituting Pa=I2Z into eq. (8.10c), the power factor can be related to Pa and Pr by expression below :

cos =
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Pr Pa

(8.10f)
39

Example 5 : A 10 F capacitor, a 2.0 H inductor and a 20 resistor are connected in series with an alternating source given by the equation below : Calculate : a. the frequency of the source. b. the capacitive reactance and inductive reactance. c. the impedance of the circuit. d. the maximum (peak) current in the circuit. e. the phase angle. f. the mean power of the circuit. Solution: C=10x10-6

V = 300 sin 300t

= 300 rad s-1. a. By using the relationship between and f, hence the frequency is = 2f
the angular frequency

V = 300 sin 300t compare with V = V0 sin t Gives the peak voltage, V0 = 300 V and

F, L=2.0 H, R=20

f = 47.8 Hz
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b. By applying the formulae of capacitive reactance and inductive reactance, thus Capacitive reactance : Inductive reactance :

XC =

1 C

X C = 333

X L = L

X L = 600
2

c. By using the formula of impedance for LRC series circuit, thus

Z = R2 + (X L X C )
Z = 267
Vrms V0 = I rms I 0 V I0 = 0 Z I 0 = 1.12 A Z=
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d. From the definition of the impedance, thus

41

e. By using the formula of phase angle for LRC series circuit, thus

tan =

X L XC R X XC = tan 1 L R

= 85.7 o @ 1.50 rad

f. The mean (average) power is
2 Pav = I rms R and I rms =

Pav =

1 2 I0 R 2 Pav = 12.5 W

I0 I 2 OR Pav = I rms Z cos and I rms = 0 2 2 1 2 Pav = I 0 Z cos 2 Pav = 12.5 W

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42

Example 6 : A 0.14 H inductor and a 12 resistor are connected in series to the alternating source 110 V, 25 Hz. Calculate a. the r.m.s current flows in the inductor. b. the phase angle between the current and supply voltage. c. the power factor of the circuit. d. the average power loss to the surrounding. Solution: R=12 , L=0.14 a. The inductive reactance is

H, Vrms=110 V, f =25 Hz

X L = L and = 2f X L = 2fL X L = 22 2 The impedance of the circuit is Z = R 2 + X L Z = 25

Therefore the r.m.s. current is

I rms =

I rms
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Vrms Z = 4.4 A
43

b. By using the formula of phase angle for RL series circuit, thus

tan =

= tan 1
c. The power factor is

XL R = 61.4 o @ 1.07 rad

XL R

d. The average power loss is

power factor = cos power factor = 0.48

Pav = I rmsVrms cos Pav = 232 W L C R

Example 7 :

327 V

141V

133 V

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Based on the RCL series circuit above, the r.m.s. voltage across R, L and C is shown. a. By using the phasor diagram, calculate the supply voltage and the 44 phase angle in this circuit.

Calculate : b. the current flows in the circuit if the resistance of the resistor R is 68 . c. the inductance and capacitance if the frequency of the a.c. source is 50 Hz. d. the resonant frequency. Solution: VR=141 V, VL=327 V, VC=133 a. The phasor diagram of the circuit is

VL

From the phasor diagram :

(VL VC )
VC

V VR

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V = VR2 + (VL VC ) V = 240 V I The phase angle is V V tan = L C VR V V = tan 1 L C V R o = 54.0 @ 0.942 rad
2

The supply voltage V is

45

b. Given R =68 Since L, R and C are connected in series, hence the current flows in each devices is the same. Therefore

c. Given f =50 Hz The inductive reactance is VL = IX L XL therefore the inductance of the inductor is X L = L and = 2f

VR = IR V I= R R I = 2.07 A

= 158

XL L = 0.503 H 2f The capacitive reactance is VC = IX C X C = 64.3 L=

therefore the inductance of the inductor is

XC =
SF027

1 and = 2f C 1 C= C = 4.95 x10 5 F 2fX C

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d. The resonant frequency is

XL = XC 1 L = C
fr =

and

= 2f r
f r = 32.0 Hz

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Example 8 : (exercise) An a.c. current of angular frequency of 1.0 x 104 rad s-1 flows through a 10 k resistor and a 0.10 F capacitor which are connected in series. Calculate the r.m.s. voltage across the capacitor if the r.m.s. voltage across the resistor is 20 V. Ans. : 2.0 V Example 9 : (exercise) A 200 resistor, a 0.75 H inductor and a capacitor of capacitance C are connected in series to an alternating source 250 V, 600 Hz. Calculate a. the inductive reactance and capacitive reactance when resonance is occurred. b. the capacitance C. c. the impedance of the circuit at resonance. d. the current flows through the circuit at resonance. Sketch the phasor diagram. 47 Ans. : 2.83 k, 93.8 nF, 200 , 1.25 A

2 LC

8.11 A.C. Rectification

Definition is defined as the process of converting alternating current to direct current. Rectifier : is a device that allows current to flow in one direction only. diodes are usually used as rectifiers. Diode is said to be forward biased when positive terminal of the diode connected to the positive terminal of the battery and vice versa, hence a current will be able to flow (figure 8.11a). Diode is said to be reverse biased when positive terminal of the diode connected to the negative terminal of the battery and vice versa, hence no current flows (figure 8.11b). Diode

+
I

+
I =0

+ SF027

- +
Fig. 8.11b : Reverse biased
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Fig. 8.11a : Forward biased

Two types of rectification are half-wave full-wave 8.11.1 Half-wave Rectification Half-wave rectification means that only one half of a cycle passes through the rectifier (diode). Figure 8.11c shows a half-wave rectification circuit.

Supply Voltage V

Supply voltage V

B A

V0

V0
D

2T

Fig. 8.11d

Fig. 8.11c

VD

V0

VD

V0
VR
R

2T

Fig. 8.11e

V0

VR

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V0

2T

Fig. 8.11f
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Explanation: First half cycle (Fig. 8.11d) When terminal A is positive, diode is forward biased and offers low resistance such that a pulse of current flows through the circuit. There is negligible voltage across the diode, VD (Fig. 8.11e). Thus the voltage across the resistor, VR is almost equal to the supply voltage (Fig. 8.11f). Next half cycle (Fig. 8.11d) When terminal B is positive, diode is now reverse biased and has a very high resistance such that a very small current flows through it. The voltage across the diode, VD is almost equal to the supply voltage (Fig. 8.11e). The voltage across the resistor, VR is almost zero (Fig. 8.11f). An alternating voltage is thus rectified to give direct current voltage across the resistor. Current flows through the resistor in one direction only and only half of each cycle passes through is shown in figure 8.11g. I Fig. 8.11g
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I0

I0

2T

t
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R.m.s. value after half-wave rectification: In half-wave rectification, half the supply voltage is suppressed and therefore the mean of the square of the voltage is given by Mean of the square = after rectification

V2

( half wave rect.)

1 V2 2

1 2

Mean of the square X before rectification

( before half wave rect.)

V02 sin 2 t

and

sin 2 t =

1 2

2 ( half wave rect.)

1 V02 V02 = = 2 2 4

V02 2

Therefore the r.m.s voltage of half-wave rectification is given by half-

V rms=
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V2
V02 4

( half wave rect.)

V rms=

V rms=

V0 2

(8.11a)

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In similar way as to find the r.m.s. voltage of half-wave rectification, the r.m.s. current of half-wave rectification is given by r.m.s. half-

I rms=

I0 2

(8.11b)

8.11.2 Full-wave Rectification While half-wave rectification only allows half of each cycle to pass through the diode, full-wave rectification allows both halves of each cycle to pass through the diode. diode To obtain full-wave rectification, four diode are used and are arranged in a form known as the diode bridge. bridge Figure 8.11h shows a full-wave rectification circuit. Supply Voltage V B

V0
Supply A voltage V F Fig. 8.11h
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4 D 2

1 3 C

VR

V0

T
Fig. 8.11i

2T

R VR

V0
E

T
Fig. 8.11j

2T
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Explanation: First half cycle (Fig. 8.11i) When terminal A is positive, diodes 1 and 2 are forward biased and conduct the current. The current takes the path ABC, R and DEF. Diodes 3 and 4 are reverse biased and hence, do not conduct the current. The voltages across diodes 1 and 2 are negligible, the voltage across the resistor VR is almost equal to the supply voltage (Fig. 8.11j) Next half cycle (Fig. 8.11i) When terminal F is positive, diodes 3 and 4 are forward biased and conduct the current. The path taken by the current is FEC, R and DBA. Diodes 1 and 2 are reverse biased and hence, do not conduct the current. The voltage across the resistor is again almost equal to the supply voltage (Fig. 8.11j). Both halves of the alternating voltage are rectified. The current flowing through the resistor is in one direction only i.e. a varying d.c. is obtained (Fig. 8.11k). I
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Fig. 8.11k

I0

2T

t 53

R.m.s. value after full-wave rectification: Notice that the negative side of supply voltage is flipped over to become positive side without being suppressed, thus the r.m.s. r.m.s. voltage of full-wave rectification is the same as the r.m.s. fullr.m.s. voltage of supply voltage and given by

V rms=

V0 2

(8.11b)

8.12 Smoothing using Capacitor

The output obtained from half-wave and full-wave rectifications are unidirectional but varying d.c. Usually a steady (constant) d.c. is required for operating various electrical and electronic appliances. To change a varying d.c. into a steady (constant) d.c., smoothing is necessary. A simple smoothing circuit consists of a capacitor ( with a large capacitance >16 F) connected parallel to the resistor R shown in figure 8.12a. The capacitor functions as a reservoir to store charges.

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Rectified unsmoothed voltage V -

I
C

R V R = V output
Fig. 8.12a
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8.12.1 Smoothing of a half-wave rectified Voltage Figure 8.12b shows an effects of smoothing a half-wave rectified voltage. Charge VR Voutput Discharge Smoothed voltage VR

time, t Rectified unsmooth input Fig. 8.12b voltage V Initially, the half-wave rectified input voltage V causes the current to
flow through the resistor R. At the same time, capacitor C becomes charged to almost the peak value of the input voltage. capacitor C starts to discharge through the resistor R. Hence the current flow is maintained because of capacitors action. Along AB (Fig. 8.12b), V output falls. At B, the rectified current again flows to recharge capacitor C to the peak of the input voltage V.
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At A (Fig. 8.12b), input V (dash line) falls below output VR, the

resistor R will look like the variation shown in figure 8.12b.

This process is repeated and hence the output voltage VR across the
55

8.12.2 Smoothing of a full-wave rectified Voltage Figure 8.12c shows an effects of smoothing a full-wave rectified voltage. Discharge Charge

VR (Voutput )

A B

Smoothed voltage VR

time, t
Rectified unsmooth input Fig. 8.12c voltage V The explanation of the smoothing process likes for a half-wave rectified voltage. The fluctuations of the smoothed output voltage are must less compare to the half-wave rectified. The smoothing action of the capacitor is due to the large time constant , given by RC so the output voltage cannot fall as rapidly as the rectified unsmoothed input voltage. Therefore a large capacitor performs greater smoothing. However, an initially uncharged capacitor may cause a sudden surge of current through the circuit and damage the diode.
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THE END
Next Unit
UNIT 9 : ELECTROMAGNETIC WAVES

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