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UNIT 8: ALTERNATINGCURRENT (A.C.)

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8.1 Alternating Current (a.c.) Definition is defined as an electric current whose magnitude and

direction change periodically.

Figures 8.1a, 8.1b and 8.1c show three forms of alternating current.

T

I

t0 T2

1 T2

0I

0I

T2

3 Fig. 8.1a : Sinusoidal a.c.

T

I

t0 T2

1 T2

0I

0I

T

2

3Fig. 8.1b : Saw-tooth a.c.

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When an a.c flows through a resistor, there will be a potentialdifference (voltage) across it and this voltage is alternating voltage as

shown in figure 8.1d.

Fig. 8.1c : Square a.c.T

I

t0 T2

1 T2

0I

0I

T2

3

T

V

t0 T2

1 T2

0V

0V

T2

3

Fig. 8.1d : Sinusoidal alternatingvoltage

gepeak volta:oVwhere

period:T

currentpeak:oI

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8.1.1 Terminology in a.c.

Frequency (f) Definition: Number of complete cycle in one second.

Unit: hertz (Hz) ors-1

Period (T)

Definition: Time taken for one complete cycle.

Unit: second (s)

Equation :

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) :

f

1T

tII 0 sin

where frequencangular:

currentpeak:0I

phase

tVV 0 sin

gepeak volta:0V

(8.1a)

(8.1b)

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8.2 Mean or Average Current (Iav)

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

Equation :

Note :

Iav

forone complete cycle is zero because the current flows in

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

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

resistanceRis

2

ooav

I

I2I

8.3 Root Mean Square Values

RIP 2

(8.2a)

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The average power, Pav over one cycle is given by

where is the average ofI2 over one cycle and can be written as

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

Since then eq. (8.3b) becomes

Using a double-angle formula for trigonometry and trigonometryidentity,

thus

2I

RIP 2av

2

rms II

2rms

2 II

RIP 2rmsav

tII 0 sin

(8.3a)

(8.3b)

(8.3c)

tII22

0rms sin

2 22 sincoscos 1 22 sincos and212 2sincos

212

12 cossin and t

t212I

I

2

0

rms cos

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The mean value ofcos2tfor one cycle is zero, finally eq. (8.3b) canbe written as

Root mean square current (Irms) is defined as the value of thesteady d.c. which produces the same power in a resistor as themean (average) power produced by the 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 voltagewhich when applied across a resistor, produces thesame power as the mean (average) power produced bythe alternating voltage across the same resistor.

Its formula is

Note :

Eq. (8.3d) and eq. (8.3e) are valid only for a sinusoidal alternatingcurrent and voltage (p.d.)

2

II 0rms (8.3d)

2VV 0rms (8.3e)

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Example 1 :

An a.c. source V=200 sin tis connected across a resistor of 100 .Calculate

a. the r.m.s. current in the resistor.

b. the peak current.c. the mean power.

Solution: R=100

then the peak voltage,V0 = 200 V

a. The r.m.s. current is

b. The peak current is

R

VI rmsrms

t200V sin

A411Irms .

tVV 0 sin

2R

VI 0rms

compare with

and2

VV 0rms

2

II 0rms

A991I0 .

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c. By applying the equation of mean power, thus

Example 2 :

The alternating potential difference shown above is connected across aresistor of 10 k. Calculate

a. the r.m.s. current,

b. the frequency,

c. the mean power dissipated in the resistor.

RIP 2rmsav

W199Pav

and

040.

)(VoltV

)second(t0 020. 080.

200

200

060.

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Solution: R=10x103

From the graph, V0 = 200 V and T=0.04 sa. The r.m.s. current is

b. The frequency of the a.c. is

c. By applying the equation of mean power, thus

R

VI rmsrms

A10x411I2

rms

.2R

VI 0rms

and2

VV 0rms

T

1f

Hz025f .

RIP 2rmsav

W991Pav .

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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 asalternating current (a.c.) and alternating voltage (a.v.).

It also being used to determine the phase difference between currentand voltage in a.c. circuit.

8.4.1 In Phase

Consider a graph represents sinusoidal a.c. and sinusoidal a.v. asshown in figure 8.4a.

t0

0V

0V

0I

0I

TT2

1 T2T

2

3

Fig. 8.4a

I V

Fig. 8.4b : Phasor diagram

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From the figure 8.4a :

Thus the phase difference is

Conclusion : The currentIis in phase with the voltage Vand constantwith time.

8.4.2 Lead

Consider a graph represents sinusoidal a.c. and sinusoidal a.v. asshown in figure 8.4c.

t0

00 IV &

00 IV &

TT2

1 T2T2

3

Fig. 8.4c

tII 0 sin

0

tVV 0 sin

tt

I

V

Fig. 8.4d : Phasor diagram

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From the figure 8.4c :

Thus the phase difference is

Conclusion : The voltage V leads the currentIby /2 radians andconstant with time.

8.4.3 Lags behind

Consider a graph represents sinusoidal a.c. and sinusoidal a.v. as

shown in figure 8.4e.

t0

00 IV &

00 IV &

TT2

1 T2T

2

3

Fig. 8.4e

tII 0 sin

rad2

tVV 0 cos

t2t

IV

Fig. 8.4f : Phasor diagram

2

tVV 0 sin

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From the figure 8.4e :

Thus the phase difference is

Conclusion : The voltage V lags behind the currentIby /2 radiansand constant with time.

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

It is defined by

It is a scalar quantity and its unit is ohm ( )

In a d.c. circuit, impedance is the resistanceR.

tII 0 sin

rad2

tVV 0 cos

t2t

2

tVV 0 sin

8.5 Impedance (Z)

rms

rms

I

VZ

2

V0

2

I0

or

0

0

I

VZ (8.5a)

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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.

Pure resistor means that no capacitance and self-inductance effectin the a.c. circuit.

8.6.1 Phase difference between Voltage Vand currentI

Figure 8.6b shows an a.c. source connected to a pure resistorR.

Fig. 8.6a

The current flows in the resistor isgiven by

The voltage across the resistorVR at

any instant is given by

tII 0 sin

IRVR tRIV 0R sin

VtVV 0R sin

and00 VRI

Fig. 8.6b

a.c. source

R

I

RV

V

where tageSupply vol:V

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Therefore the phase difference between VandIis

In pure resistor, the currentIis in phase with the voltage V.

Figure 8.6c shows the variation ofVandIwith time while Figure 8.6d

shows the phasor diagram forVandIin a pure resistor.

0tt

t0

0V

0V

0I

0I

TT2

1 T2T

2

3

Fig. 8.6c

I V

Fig. 8.6d : Phasor diagram

8 6 2 I d i i t

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8.6.2 Impedance in a pure resistor

From the definition of the impedance, hence

8.6.3 Power in a pure resistor

Since VandIare in phase, the instantaneous powerP is given by

Therefore the graph of variation of power with time is shown in figure8.6e.

tPP2

0 sin

IVP

RI

V

I

VZ

0

0

rms

rms

tVtIP 00 sinsin

tVIP2

00 sin and 000 PVI

(8.6a)

(8.6b)

tPP 20 sin

where powerum)peak(maxim:0P

t0

0P

TT21 T2T

23

Fig. 8.6e

)(PPower Power being absorbed

The a erage (mean) po er P in a resistor is gi en b

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The average (mean) powerPav in a resistor is given by

Pure capacitor means that no resistance and self-inductance effectin the a.c. circuit.

8.7.1 Phase difference between Voltage Vand currentI Figure 8.7a shows an a.c. source connected to a pure capacitorC.

tPP 20av sin

2

VIP 00

av

and

2

1t

2 sin

0av

P2

1P or (8.6c)

8.7 Pure Capacitor in the A.C. Circuit

The voltage across the capacitorVCatany instant is equal to the supply

voltage Vand is given by

The charge accumulates at the platesof the capacitor is

tCVQ 0 sin

CCVQ

C0 VtVV sin

Fig. 8.7a

a.c. source

CV

V

C

I

Charge and current are related by

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Charge and current are related by

Hence the equation of a.c. in capacitor is

Therefore the phase difference between VandIis

In pure capacitor, the currentIleads the voltage V by /2 radians.

dt

dQI

tCVI 0 cos

tCVdtd

I 0 sin

tdt

dCVI 0 sin

and

00

ICV

tII 0 cos or

2tII 0

sin

rad2

t2

t

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Figure 8.7b shows the variation ofVandIwith time while Figure 8.7c

shows the phasor diagram forVandIin a pure capacitor.

8.7.2 Impedance in a pure capacitor

From the definition of the impedance, hence

t0

0V

0V

0I

0I

TT

2

1 T2T

2

3

Fig. 8.7b

rad2

0

0

CV

VZ

0

0

I

VZ and 00 CVI

I V

Fig. 8.7c : Phasor diagram

1

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whereXCis known as capacitive (capacitative) reactance.

Capacitive reactance is the opposition of a capacitor to the currentflows and is defined by

The unit of capacitive reactance is ohm ( )

From eq. (8.7a), the relationship between capacitive reactanceXCand

frequencyfcan be shown by using graph in figure 8.7d.

CXC

1Z

(8.7a)fC2

1XC

and f2

0

0

rms

rmsC

I

V

I

VX (8.7b)

sourca.c.offrequency:fcapacitortheofecapacitanc:C

f0

Fig. 8.7d

CX

f

1XC

8 7 3 P i it

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8.7.3 Power in a pure capacitor

Since the currentIleads the voltage V by /2 radians, the

instantaneous powerP is given by

Therefore the graph of variation of power with time is shown in figure8.7e.

t2P2

1P 0 sin

IVP

tVtIP 00 sincosttVIP 00 cossin and 000 PVI

(8.7c)

t22

1PP 0 sin

t0

2

P0

2

P0

TT2

1 T2T

2

3Fig. 8.7e

)(PPower

t2P2

1P 0 sin

Power being absorbed

Power being returned to supply

Th ( ) P i it i

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The average (mean) powerPav in a capacitor is

Pure inductor means that no resistance and capacitance effect in thea.c. circuit.

8.8.1 Phase difference between Voltage Vand currentI

Figure 8.8a shows an a.c. source connected to a pure inductorL.

t2P2

1P 0av sin and 0t2 sin

0Pav

8.8 Pure Inductor in the A.C. Circuit

The current flows in the inductor isgiven by

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

tLI0B cos

tIdt

dL 0B sin

Fig. 8.8a

a.c. source

V

I

L

LV

tII 0 sin

dtdILB

(8.8a)

At h i t t th l lt V t b l t th b k f

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

B (voltage across the inductor) but the back e.m.f always oppose the

supply voltage Vrepresents by the negative sign in eq. (8.8a).

By comparing the magnitude ofVand B ,thus

Therefore the phase difference between VandIis

In pure inductor,

the voltage Vleads the currentIby /2 radians.

or

the currentIlags behind thevoltage V by /2 radians.

tLIV 0B cos

tVV 0 cos

and00 VLI

rad2

t2

t

or

2

tVV 0 sin

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Figure 8.8b shows the variation ofVandIwith time while Figure 8.8c

shows the phasor diagram forVandIin a pure inductor.

8.8.2 Impedance in a pure inductor

From the definition of the impedance, hence

0

0

I

LIZ

0

0

I

VZ and 00 LIV

t0

0V

0V

0I

0I

TT

2

1 T2T

2

3

Fig. 8.8b

rad2

I

V

Fig. 8.8c : Phasor diagram

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whereXL is known as inductive reactance.

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

The unit of inductive reactance is ohm ( )

From eq. (8.8b), the relationship between inductive reactanceXL and

frequencyfcan be shown by using graph in figure 8.8d.

LXLZ

(8.8b)fL2XL

and f2

0

0

rms

rms

L I

V

I

V

X (8.8c)

sourca.c.offrequency:finductortheofinductanceself:L

f0

Fig. 8.8d

LX

fXL

8 8 3 Power in a pure inductor

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8.8.3 Power in a pure inductor

Since the voltage Vleads currentI the by /2 radians, the

instantaneous powerP is given by

Therefore the graph of variation of power with time is shown in figure8.8e.

t2P2

1P 0 sin

IVP

tVtIP 00 cossinttVIP 00 cossin and 000 PVI

(8.8d)

t22

1PP 0 sin

t0

2

P0

2

P0

TT2

1 T2T

2

3Fig. 8.8e

)(PPower

t2P2

1P 0 sin

Power being absorbed

Power being returned to supply

The average (mean) power P in a inductor is

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The average (mean) powerPav in a inductor is

Note :

For both pure capacitor and inductor, the average (mean) power iszero because the power is positive for quarter of a cycle andnegative for the next quarter of a cycle in each half cycle.

The term resistance is not used in pure capacitor and inductorbecause 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: C=4700x10-12 F, Vrms=240 V, f=50 Hz.a. By applying the equation of capacitive reactance, thus

t2P2

1P 0av sin and 0t2 sin

0Pav

10x677X3

C

fC2

1XC

b From the definition of the capacitive reactance thus

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b. From the definition of the capacitive reactance, thus

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. Calculatea. the reactance of the inductor.

b. the inductance of the inductor.

Solution:Irms=3.0 A, Vrms=240 V, f=50 Hz.

a. From the definition of the inductive reactance, thus

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

rms

rmsC

I

VX and

2

II 0rms

A10x015I4

0 .

80XL

C

rms

0 X

2V

I

rms

rmsL

I

VX

H260L .

fL2XL

2

XL

L

8 9 RC RL d LRC i S i Ci it

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8.9 RC, RL and LRC in Series Circuit8.9.1 RC in series circuit

Consider an a.c. source with voltage Vis connected in series with a

resistorR and a capacitorCas shown in figure 8.9a. The voltage across the resistorVR andthe capacitorVC are given by

The phasor diagram of RC circuit isshown by figure 8.9b.

Based on the phasor diagram, the total

voltage V(supply voltage) across bothresistorR and capacitorCis

IRVR

CC IXV

2

C

2

R

2

VVV

where

anglephase:

Fig. 8.9a

a.c. source

R

I

RV

V

CV

C

CV

IRV

V

Fig. 8.9b : Phasor diagram

2C22

IXIRV

2C222 XRIV

22

2

C

1

RIV

and

C

1XC

(8.9a)

V

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Since , hence eq. (8.9a) can be written as

From the phasor diagram (fig. 8.9b), the currentIleads the supplyvoltage Vby radians where

From the eq. (8.9b) and (8.9c), the new phasor diagram in terms ofR,

XCandZcan be sketched (refer to figure 8.9c)

rms

rms

I

VZ

R

C

V

Vtan

2

C

2XRZ

IR

IXCtan

RXCtan

22

2

C

1RZ or (8.9b)

CR1tanor (8.9c)

CX

R

Z

Fig. 8.9c : Phasor diagram in

terms ofR,XCandZ

8 9 2 RL in series circuit

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8.9.2 RL in series circuit

Consider an a.c. source with voltage Vis connected in series with a

resistorR and an inductorL as shown in figure 8.9d.

The voltage across the resistorVR and

the inductorVL are given by

The phasor diagram of RL circuit isshown by figure 8.9e.

Based on the phasor diagram, the total

voltage V(supply voltage) across bothresistorR and inductorL is

IRVR

LL IXV

2

L

2

R

2VVV

LV

IRV

V

Fig. 8.9e : Phasor diagram

2

L

22 IXIRV

2L222 XRIV 222

LRIV

and LXL

(8.9d)

Fig. 8.9d

a.c. source

R

I

RV

V

L

LV

V

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Since , hence eq. (8.9d) can be written as

From the phasor diagram (fig. 8.9e), the supply voltage Vleads thecurrentIby radians where

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

XL andZcan be sketched (refer to figure 8.9f)

rms

rms

I

VZ

R

L

V

Vtan

2

L

2 XRZ

IR

IXLtan

RXLtan

222LRZ or (8.9e)

RLtanor (8.9f)

Fig. 8.9f : Phasor diagram in

terms ofR,XL andZ

LXZ

R

8 9 3 LRC in series circuit

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8.9.3 LRC in series circuit

Consider an a.c. source with voltage Vis connected in series with an

inductorL, a resistorR and a capacitorCas shown in figure 8.9g.

The voltage across the inductorVL ,

resistorVR and resistorVR are givenby

The phasor diagram of LRC circuit is

shown by figure 8.9h.

Based on the phasor diagram, the total

voltage V(supply voltage) acrossinductorL , resistorR and capacitorCis

IRVR

CC IXV

2

CL

2

R

2

VVVV

LV

IRV

V

Fig. 8.9h : Phasor diagram

2CL22

IXIXIRV

2CL222

XXRIV

2

CL

2

XXRIV

Fig. 8.9g

a.c. source

I

V

R

RV

L

LV CV

C

LL IXV

CV

CL VV

(8.9g)

V

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

From the phasor diagram (fig. 8.9h), the supply voltage Vleads the

currentIby radians where

From the eq. (8.9h) and (8.9i), the new phasor diagram in terms ofXL,

,R,XCandZcan be sketched (refer to figure 8.9i)

rms

rms

I

VZ

R

CL

V

VV tan

2

2

C

1LRZ

IR

XXI CL tan

R

XX CL tan

2CL2 XXRZ or

R

C

1L

tanor

Fig. 8.9i : Phasor diagram in terms of

XL,R,XCandZ

(8.9h)

(8.9i)

LX

R

Z

CX

CL XX

8 10 Resonance and Po er in A C Circ it

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From the graph, the value ofZis minimum when

where its value is

This phenomenon occurs at

frequency offr known asresonant frequency.

8.10 Resonance and Power in A.C. Circuit8.10.1 Resonance in a.c. circuit

Definition is defined as the phenomenon that occurs when thefrequency of the applied voltage is equal to the

frequency of the LRC series circuit. Figure 8.10a shows the variation ofXC,XL,R andZwith frequencyf

of the LRC series a.c circuit.

Z

fXL

R

f

1XC

Fig. 8.10a

0 f

ZRXX LC ,,,

rf

2CL2

XXRZ

0RZ 2 min

CL XX

RZ min

(8.10a)

When the resonance occurred in the LRC series circuit, the Z is

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When the resonance occurred in the LRC series circuit, the Z isminimum but the r.m.s. current flows in the circuit is maximum andgiven by

The resonant frequencyfrof the LRC series circuit is given by

At frequencies above or below the resonant frequencyfr, the r.m.s.current is less than the maximum current.

The series resonance circuit is used for tuning a radio receiver.

When resonanceoccurred

CL XX LXL

R

V

Z

VI rmsrmsrms

C

1XC where and

C

1L

LC1 rf2 and

hence

LC2

1fr

where

frequencyresonant:r

ffrequencangularresonant:

8.10.2 Power in a.c. circuit

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We get , then eq. (8.10b)can be written as

wherecos is called the power factor ofthe circuit andPr is the average realpower.

8.10.2 Power in a.c. circuit

No power is absorbed by either inductors or capacitors over a completecycle and therefore the average (mean) power absorbed by bothinductors and capacitors is zero.

In an a.c. circuit in which there is resistanceR, inductanceL and

capacitance C, the total average powerPav is equal to that dissipatedin the resistance i.e.

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

Rav IVP orRIP2

av

IRV

V

CL VV

Fig. 8.10b

(8.10b)

V

VRcos

cosIVPav

r

2

av PZIP cos

and IZV

(8.10c)

Note :

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Power factor can also be calculated by using formula below :

From the figure 8.10b,

From the figure 8.10b, Since VandIare not in phase, then the

average apparent powerPa is given by

By substituting Pa=I2Zinto eq. (8.10c), the power factor can be

related to Pa and Prby expression below :

where

IZ

IRcos

(8.10d)

V

VRcos IRVR

Z

Rcos

and IZV

Z

VZIIVP

22

a (8.10e)

a

r

P

Pcos (8.10f)

Example 5 :

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p

A 10 F capacitor, a 2.0 H inductor and a 20 resistor are connectedin 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=10x10-6F, L=2.0 H, R=20

Gives the peak voltage, V0 = 300 V and

the angular frequency = 300 rad s-1.

a. By using the relationship between andf, hence the frequency is

t300300V sin

t300300V sin tVV 0 sincompare with

f2

z. H847f

b By applying the formulae of capacitive reactance and inductive

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b. By applying the formulae of capacitive reactance and inductive

reactance, thus

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

d. From the definition of the impedance, thus

Capacitivereactance : 333XC

C

1XC

LXL 600XL Inductivereactance :

2

CL

2

XXRZ 267Z

0

0

rms

rms

I

V

I

VZ

A121I0 .Z

VI 00

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

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

f. The mean (average) power is

rad501785 .@.

R

XX CL tan

R

XX CL1tan

RIP2

rmsav

W512Pav .

and

2

II 0rms

RI2

1P

2

0av

OR cosZIP2

rmsav

W512Pav .

cosZI2

1P

2

0av

and

2

II 0rms

Example 6 :

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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 H, Vrms=110 V, f =25 Hza. The inductive reactance is

The impedance of the circuit is

Therefore the r.m.s. current is

LXL

A44Irms .

2

L

2XRZ

and f2

25Z

fL2XL 22XL

Z

VI rmsrms

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

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c. The power factor is

d. The average power loss is

Example 7 :

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

phase angle in this circuit.

R

XLtan

rad071461 .@.

W232Pav

R

XL1tan

cosfactorpower

cosrmsrmsav VIP

480factorpower .

CRL

V3 27 V141 V133

I

Calculate :

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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 Va. The phasor diagram of the circuit is

LV

I

RV

V

CV

CL VV

From the phasor diagram :

The supply voltage Vis

The phase angle is

2CL2

R VVVV V240V

R

CL

V

VV tan

rad942.0054 @.

R

CL1

V

VVtan

b. GivenR =68

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Since L, R and C are connected in series, hence the current flows in

each devices is the same. Therefore

c. Givenf =50 HzThe inductive reactance is

therefore the inductance of the inductor is

The capacitive reactance is

therefore the inductance of the inductor is

IRVR

A072I .R

VI R

H5030L .

LL IXV 158XL

LXL

f2

XL L

and f2

CCIXV 364X

C

.

F10x954C5.

C

1XC

CfX2

1C

and f2

d. The resonant frequency is

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Example 8 : (exercise)An a.c. current of angular frequency of 1.0 x 104 rad s-1 flows through a10 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. voltageacross 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 Care connected in series to an alternating source 250 V, 600 Hz.Calculate

a. the inductive reactance and capacitive reactance when resonance isoccurred.

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.

Ans. : 2.83 k, 93.8 nF, 200 , 1.25 A

CL XX

Hz032fr .

C

1L

LC21fr

and rf2

8 11 A C Rectification

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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 thediode connected to the positive terminal of the battery and viceversa, hence a current will be able to flow (figure 8.11a).

Diode is said to be reverse biased when positive terminal of thediode connected to the negative terminal of the battery and viceversa, hence no current flows (figure 8.11b).

Fig. 8.11a : Forward biased

+

+ -

-

I I

Fig. 8.11b : Reverse biased

+

+-

-

0I

Diode

Two types of rectification are

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t0 T T2

RV

0V

0V

Fig. 8.11f

t0T

T2

DV

0V

0V

Fig. 8.11e

half-wave

full-wave

8.11.1 Half-wave Rectification

Half-wave rectification means that only one half of a cycle passesthrough the rectifier (diode).

Figure 8.11c shows a half-wave rectification circuit.

Fig. 8.11d

0V

0V

t0T T2

VVoltageSupply

Fig. 8.11c

RV

DV

D

R

A

B

V

Supplyvoltage

Explanation:

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First half cycle (Fig. 8.11d)

When terminal A is positive, diode is forward biased and offerslow resistance such that a pulse of current flows through thecircuit.

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 flowsthrough it.

The voltage across the diode, VD is almost equal to the supplyvoltage (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 voltageacross the resistor. Current flows through the resistor in onedirection only and only half of each cycle passes through is shownin figure 8.11g.

t0T T2

I

0I

0IFig. 8.11g

R.m.s. value after half-wave rectification:

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In half-wave rectification, half the supply voltage is suppressed andtherefore the mean of the square of the voltage is given by

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

2

1Mean of the squareafter rectification

= XMean of the squarebefore rectification

rect.)wavehalfbefore(rect.)wavehalf( 22 V

2

1V

tV 220 sin and2

1t

2 sin

4

V

2

V

2

1V

2

0

2

02

rect.)wavehalf(

rect.)wavehalf( 2rms VV

4

V

V

2

0

rms 2

VV 0

rms(8.11a)

2

V2

0

In similar way as to find the r.m.s. voltage of half-wave rectification,h f h lf ifi i i i b

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0V

0V

t0T T2

VVoltageSupply

Fig. 8.11i

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

8.11.2 Full-wave Rectification

While half-wave rectification only allows half of each cycle to passthrough the diode, full-wave rectification allows both halves of eachcycle to pass through the diode.

To obtain full-wave rectification, four diode are used and are arranged

in a form known as the diode bridge. Figure 8.11h shows a full-wave rectification circuit.

2

II 0rms (8.11b)

Fig. 8.11h

A

RRVV

Supplyvoltage

F

B

CD

E

1

2 3

4

Fig. 8.11j

0V

t0T

T2

RV

Explanation:

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First half cycle (Fig. 8.11i)

When terminal A is positive, diodes 1 and 2 are forward biasedand conduct the current.

The current takes the path ABC, R and DEF.

Diodes 3 and 4 are reverse biased and hence, do not conductthe current.

The voltages across diodes 1 and 2 are negligible, the voltage

across the resistorVR 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 biasedand 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 conductthe current.

The voltage across the resistor is again almost equal to thesupply voltage (Fig. 8.11j).

Both halves of the alternating voltage are rectified. The currentflowing through the resistor is in one direction only i.e. a varyingd.c. is obtained (Fig. 8.11k).

Fig. 8.11k

0I

t0 T T2

I

R.m.s. value after full-wave rectification:

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Notice that the negative side of supply voltage is flipped over tobecome positive side without being suppressed, thus the r.m.s.voltage of full-wave rectification is the same as ther.m.s.voltage of supply voltage and given by

The output obtained from half-wave and full-wave rectifications areunidirectional but varying d.c.

Usually a steady (constant) d.c. is required for operating variouselectrical and electronic appliances. To change a varying d.c. into asteady (constant) d.c., smoothing is necessary.

A simple smoothing circuit consists of a capacitor ( with a large

capacitance >16 F) connected parallel to the resistorR shown in figure8.12a. The capacitor functions as a reservoir to store charges.

2VV 0rms (8.11b)

8.12 Smoothing using Capacitor

+R outputVVR C

- Fig. 8.12a

Rectified unsmoothed

voltage V

I

8.12.1 Smoothing of a half-wave rectified Voltage

Fi 8 12b h ff f hi h lf ifi d

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Figure 8.12b shows an effects of smoothing a half-wave rectifiedvoltage.

Initially, the half-wave rectified input voltage Vcauses the current toflow through the resistorR. At the same time, capacitorCbecomescharged to almost the peak value of the input voltage.

At A (Fig. 8.12b), input V(dash line) falls below output VR, the

capacitorCstarts to discharge through the resistorR. Hence thecurrent flow is maintained because of capacitors action.

Along AB (Fig. 8.12b), Voutput falls. At B, the rectified current again

flows to recharge capacitorCto the peak of the input voltage V.

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

resistorR will look like the variation shown in figure 8.12b.

Fig. 8.12b

A B

Rectified unsmooth input

voltage V

outputVVR

Smoothed voltage VR

Discharge

Charge

ttime,

8.12.2 Smoothing of a full-wave rectified Voltage

Fi 8 12 h ff t f thi f ll tifi d

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Figure 8.12c shows an effects of smoothing a full-wave rectifiedvoltage.

The explanation of the smoothing process likes for a half-wave rectifiedvoltage.

The fluctuations of the smoothed output voltage are must less compareto the half-wave rectified.

The smoothing action of the capacitor is due to the large time constant

, given byRCso the output voltage cannot fall as rapidly as therectified unsmoothed input voltage.

Therefore a large capacitor performs greater smoothing. However, aninitially uncharged capacitor may cause a sudden surge of currentthrough the circuit and damage the diode.

Fig. 8.12c

A B

Rectified unsmooth input

voltage V

outputVVR

Smoothed voltage VR

Discharge Charge

ttime,

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THE END

UNIT 9 :

ELECTROMAGNETIC WAVES