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Topic --- Alternating Current

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Topic --- Alternating Current

Topic --- Alternating Current

21 ALTERNATING CURRENT

21.1

Alternating current

21.2

Root mean square (rms)

21.3

Resistance, reactance & impedance

21.4

Power and power factor

Topic --- Alternating Current

(a) Define alternating current

(AC)

(b) Sketch and analyse

sinusoidal AC waveform

(c) Use sinusoidal voltage and

current equations

tII sin0 tVV sin0

21.1 ALTERNATING CURRENT

Topic --- Alternating Current

Alternating current (AC) electricity is the type of electricity commonly used in

homes and businesses throughout the world. While direct current (DC)

electricity flows in one direction through a wire, AC electricity alternates its

direction in a back-and-forth motion.

The direction alternates between 50 and 60 times per second, depending on the

electrical system of the country.

AC created by an AC electric generator, which determines the frequency

the voltage can be readily changed, thus making it more suitable for long-distance transmission than DC electricity

can employ capacitors andinductors in electronic circuitry, allowing for a wide range of applications

21.1 ALTERNATING CURRENT

Topic --- Alternating Current

• is defined as an electric

current which magnitude

& direction change

periodically

• Symbol:

21.1 ALTERNATING CURRENT

Topic --- Alternating Current

CurrentVoltage The output of an ac

generator is sinusoidal and varies with time

locityangular ve ORfrequency angular :

currentpeak : 0I gepeak volta: 0V

time: t

)2( f In general,

21.1 ALTERNATING CURRENT

tII o sintVV o sin

Topic --- Alternating Current

fT

1

21.1 ALTERNATING CURRENT

Topic --- Alternating Current

21.2 ROOT MEANC SQUARE (rms)

(a) Define root mean square

(rms) current and voltage

for AC source

(b) Use the following formula,

and2

0rms

II

2

0rms

VV

Topic --- Alternating Current

rmsI

acdcpower averagepower

RIRI ave22

aveII 2

Mean or average current, Iave: the average or mean value of current in a half-cycle flows of current in a certain direction

0

2

0

ave

2III

avermsII 2

The r.m.s (root mean square) current means the square root of the average value of the current

21.2 ROOT MEAN SQUARE (rms)

Topic --- Alternating Current

• Since ac current

and

• So

and

• The root mean square (rms) current is the effective value of the AC

tII sin0

avermsII 2

avermstsinII

2

0

2

0rms

II

2

1sinsin 22

tt

21.2 ROOT MEAN SQUARE (rms)

Topic --- Alternating Current

• The equation:

• Most household electricity is 240 V AC which means that Vrms is 240 V

2

0rms

VV

21.2 ROOT MEAN SQUARE (rms)

Topic --- Alternating Current

A sinusoidal, 60.0 Hz, ac voltage is read to be 120 V by an ordinary voltmeter.

(a) What is the maximum valuethe voltage takes on during a cycle?

(b) b) What is the equation for the voltage?

Solution:

(a)

(b)

2o

rms

VV

)V(Vrmso

2

V 170

tsinVVo

tsin 120170

21.2 ROOT MEAN SQUARE (rms)

Topic --- Alternating Current

A voltage V= 60 sin 120πt is applied across a 20 Ω resistor.

(a)What will an ac ammeter in series with the resistor read?

(b)Calculate the peak current and mean power.

2o

rms

VV

R

VI rms

rms

)I(Irmso

2

20

II

rms

W90(20)(2.12) 2

2

RIP rmsav

Solution:(a)

(b)

21.2 ROOT MEAN SQUARE (rms)

Topic --- Alternating Current

1. An AC sourceV = 500 sin t

is connected across a resistor of 250 . Calculate

(a) the rms current in the resistor,(b) the peak current,(c) the mean power.

2. Figure below shows a graph to represent alternating current passes through a resistor of 10 k. Calculate

(a) the rms current,(b) the frequency of the AC,(c) the mean power dissipated from

the resistor.

21.2 ROOT MEAN SQUARE (rms)

Topic --- Alternating Current

(a) Sketch and use phasor

diagram and sinusoidal

waveform to show the

phase relationship

between current and

voltage for a circuit

consisting of (i) pure resistor(ii)pure capacitor (iii)pure inductor

(b) Use phasor diagram to

analyse voltage, current,

and impedance of series

circuit of:

(i) RL

(ii)RC

(iii)RLC

21.3 RESISTANCE, REACTANCE & IMPEDANCE

Topic --- Alternating Current

(c) Define and use:

(i) capacitive reactance,

(ii) inductive reactance,

(ii)impedance,

(iv)phase angle,

(d) Explain

graphically the

dependence of

R, XC, XL and Z

on f and relate it

to resonance

fCXC

2

1

fLX L 2

22

CL XXRZ

R

XX CL tan

Topic --- Alternating Current

A diagram containing phasor is called phasor diagram

Phasor a vector that rotate

anticlockwise about its axis with constant

angular velocity

Used to represent a sinusoidally varying quantitysuch as alternating current (AC) & alternating voltage

Also being used to determine the phase angle the phase difference between current and voltage in AC circuit

21.3 RESISTANCE, REACTANCE & IMPEDANCE

Topic --- Alternating Current

Phase difference is

22

tt

21.3 RESISTANCE, REACTANCE & IMPEDANCE

Note:

valuepositive

radian valuenegative

Leads

Lags behind

In antiphase

Topic --- Alternating Current

• The projection of OP on the y-axis is ON, represents the instantaneous value

• Ao is the

peak valueof the quantity

t0 TT2

1 T2T2

3

Ao

tAA o siny

ωN

O

P

y

21.3 RESISTANCE, REACTANCE & IMPEDANCE

Topic --- Alternating Current

• It is defined by

OR

• It is a scalar quantityand its unit is ohm ()

• In a DC circuit, impedance likes the resistance

rms

rms

I

VZ

2

0V

2

0I

0

0

I

VZ

21.3 RESISTANCE, REACTANCE & IMPEDANCE

V = IZ V = IR V = IX

V = IXC

V = IXL

Topic --- Alternating Current

Resistance, R: Opposition to current flow in purely resistive circuit

Reactance, X: Opposition to current flow resulting from inductance or capacitance in ac circuit

Capacitive reactance, XC: Opposition of a capacitor to ac

Inductive reactance, XL: Opposition of an inductor to ac

Impedance, Z: Total opposition to ac (Resistance and reactancecombine to form impedance)

21.3 RESISTANCE, REACTANCE & IMPEDANCE

Topic --- Alternating CurrentAC is defined as an electric

current which magnitude

& direction change

periodically

In general,

Root mean square current (Irms) is defined as the effective value of a.c. which produces the same power as the steady d.c. when the current passes through the same resistor

2

0rms

II

Root mean square voltage/ p.d (Vrms): the value of the steady direct voltage which when applied across a resistor, produces the same power as the mean (average) powerproduced by the alternating voltage across the same resistor

2

0rms

VV

A diagram containing phasor is called phasor diagram

Phasor a vector that rotate

anticlockwise about its axis with constant

angular velocity

Impedance, Z

Resistance, R

in resistor

Reactance, X

Capacitive reactance, XC

in capacitor, C

Inductive reactance, XL

in inductor, L

V = IZ V = IR V = IX

V = IXC

V = IXL

Topic --- Alternating Current

• The circuit

• The alternating current passes through

the resistor is given by

• The alternating voltage across the

resistor VR at any instant is given by

where, V = supply voltage

tII sin0

IRVR

00 VRI RtI sin0 and

VtVVR sin0

21.3 RESISTANCE, REACTANCE & IMPEDANCE

Pure Capacitor In AC Circuit

Topic --- Alternating Current

• From figure above: I = I0 sin tand V = V0 sin t

• Thus the phase difference is

Impedance in a pure resistor• From the

definition of the impedance, hence

t0

0I

0V

0I0V

TT2

1 T2T

2

3

ω

VI

0 tt

Therefore the current I is in phase with the voltage V and constant with time

RI

V

I

VZ

0

0

rms

rms

21.3 RESISTANCE, REACTANCE & IMPEDANCE

Pure Capacitor In AC Circuit

Topic --- Alternating Current

CIRCUIT

AC source

R

I

RV

V

PHASE DIFFERENCEtII sin0

VtVVR sin0

0

IMPEDANCE, ZPHASORDIAGRAM

VI

RZ

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

RI

V

I

VZ

0

0

rms

rms

21.3 RESISTANCE, REACTANCE & IMPEDANCE

Pure Capacitor In AC Circuit

Topic --- Alternating Current

• The circuit

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

• The charge accumulates at the plates of the capacitor is

21.3 RESISTANCE, REACTANCE & IMPEDANCE

AC source

CV

V

C

I

tVVVC sin0

CCVQ

tCVQ sin0

Pure Capacitor In AC Circuit

Topic --- Alternating Current

• The charge and current are related by

Hence the equation of AC in the capacitor is

and

OR

dt

dQI

tCVdt

dI sin0

tdt

dCV sin0

tCV cos0 00 ICV

tII cos0

2sin0

tII

21.3 RESISTANCE, REACTANCE & IMPEDANCE

Pure Capacitor In AC Circuit

Topic --- Alternating Current

• From figure above: V = V0 sin t and I = I0 sin (t + /2)

• Thus the phase difference is

the voltage V lags

behind the current Iby /2 radians

ORthe current I leads

the voltage V by /2radians

21.3 RESISTANCE, REACTANCE & IMPEDANCE

t0

0I

0V

0I0V

TT2

1 T2T

2

3

ωVI

rad 2

22

tt

Pure Capacitor In AC Circuit

Topic --- Alternating Current

Impedance in a pure capacitor• From the definition of the

impedance, hence

and

and

where XC:Capacitive

(capacitative)reactance

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

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

0

0

I

VZ 00 CVI

0

0

CV

V

CXC

Z

1f 2

fCXC

2

1

0

0

rms

rms

I

V

I

VXC

f0

CX

fX C

1

21.3 RESISTANCE, REACTANCE & IMPEDANCE

Pure Capacitor In AC Circuit

Topic --- Alternating Current

CIRCUITPHASEDIFFERENCE

IMPEDANCE, Z PHASOR DIAGRAM

21.3 RESISTANCE, REACTANCE & IMPEDANCE

AC source

CV

V

C

I

tVVVC sin0

2sin0

tII

rad 2

the voltage V lags behindthe current I by /2radians.

ORthe current I leads the voltage V by /2 radians

V

I

rad 2

fCXC

2

1

f0

CX

fX C

1

Pure Capacitor In AC Circuit

Topic --- Alternating Current

An 8.00 μF capacitor is connected to the terminals of an AC generator with an rmsvoltage of 150 V and a frequency of 60.0 Hz. Find the capacitive reactance rmscurrent and the peak currentin the circuit.

Solution:

21.2 ROOT MEAN SQUARE (rms)

fCXC

2

1

0

0

rms

rms

I

V

I

VXC

Pure Capacitor In AC Circuit

Topic --- Alternating Current

• The circuit

• 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

21.3 RESISTANCE, REACTANCE & IMPEDANCE

AC sourceV

I

L

LV

tII sin0

dt

dILB tI

dt

dL sin0

tLI cos0B

At any instant,

the supply

voltage V

equals to the

back emf B in

the inductor but

the back emf

always oppose

the supply

voltage V

represents by

the negative

sign

2sin 0

tLIVLB

Pure Inductor In AC Circuit

Topic --- Alternating Current

• From figure above: I = I0 sin tand V = V0 sin (t + /2)

• Thus the phase difference is

In the pure inductor, the voltage V leads

the current I by /2radians

ORthe current I lags

behind the voltage Vby /2 radians

21.3 RESISTANCE, REACTANCE & IMPEDANCE

22

tt

t0

0I

0V

0I0V

TT2

1 T2T

2

3

V

I

rad 2

ω

Pure Inductor In AC Circuit

Topic --- Alternating Current

Impedance in a pure inductor• From the definition of the

impedance, hence

where XL:inductive reactance

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

• Inductive reactance is a scalar quantity and its unit is ohm ()

21.3 RESISTANCE, REACTANCE & IMPEDANCE

0

0

I

VZ 00 LIV and

0

0

I

LI

LXLZ

fLX L 2

f 2and

0

0

rms

rms

I

V

I

VX L

f0

LX

fX L

Pure Inductor In AC Circuit

Topic --- Alternating Current

CIRCUITPHASEDIFFERENCE

IMPEDANCE, Z PHASOR DIAGRAM

21.3 RESISTANCE, REACTANCE & IMPEDANCE

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

ORthe current I lags behindthe voltage V by /2radians

AC sourceV

I

L

LV

tII sin0tVV cos0

OR

2sin0

tVV

rad 2

V

I

rad 2

f0

LX

fX L

fLX L 2

Pure Inductor In AC Circuit

Topic --- Alternating Current

A coil having an inductance of 0.5 H is connected to a 120 V, 60 Hz power source. If the resistance of the coil is neglected, what is the effective current through the coil.

A 240 V supply with a frequency of 50 Hz causes a current of 3.0 A to flow through an pure inductor. Calculate the inductance of the inductor.

0

0

rms

rms

I

V

I

VX L

21.3 RESISTANCE, REACTANCE & IMPEDANCE

Topic --- Alternating Current

Since the voltage on a capacitor to the charge on it, the current must lead the voltage in time & phase to conduct charge to the capacitor plates and raise the voltage

V

I

rad 2

When a voltage is applied to an

inductor, it resists the change in

current. The current builds up more slowly than the

voltage, lagging it in time and phase

V

I

rad 2

21.3 RESISTANCE, REACTANCE & IMPEDANCE

Topic --- Alternating CurrentPure Capacitor in AC Circuit Pure Inductor in AC Circuit

V

I

rad 2

the voltage V leads the current I by /2radians

ORthe current I lags behind the voltage Vby /2 radians

tII sin0

2sin0

tVV

rad 2

IV

tVVVC sin0

2sin0

tII

rad 2

IV

V

I

rad 2

the voltage V lags behind the current Iby /2 radians

ORthe current I leads the voltage V by /2 radians

tII sin0

VtVVR sin0

0

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

VI

Topic --- Alternating Current

Q1 A capacitor has a rms

current of 21 mA at a frequency of 60 Hz when the rms voltage across it is 14 V.(i) What is the capacitance of the capacitor?(ii) If the frequency is increased, will the current in the capacitor increase, decrease or stay the same? Explain.(iii) Calculate the rmscurrent in the capacitor at a frequency of 410 Hz.

Q2 A 2 F capacitor and a

1000 resistor are placed in series with an alternating voltage source of 12 V and frequency of 50 Hz. Calculate(i) the current flowing,(ii) the voltage across the capacitor,(iii) the phase angle of the circuit.

Q3 A rms voltage of 12.2

V with a frequency of 1.00 kHz is applied to a 0.290 mH inductor.(i) What is the rmscurrent in the circuit?(ii) Determine the peak current for a frequency of 2.50 kHz.

21.3 RESISTANCE, REACTANCE & IMPEDANCE

Topic --- Alternating Current

AC source

R

I

RV

V

CV

C

IRVR CC IXV

22

CR VVV

Phasor diagram: the current Ileads the supply voltage V by radians

R

C

V

Vtan

R

XCtan

CXZ

R

22

CXRZ

fCXC

2

1

22

2 1

CRZ

A phasor diagram in terms of R, XC and ZI

CV

RV

V

21.3 RESISTANCE, REACTANCE & IMPEDANCE

Topic --- Alternating Current

An alternating current of angular frequency of 1.0 x 104

rad s-1 flows through a 10 kresistor and a 0.10 F capacitor which are connected in series. Calculate the rmsvoltage across the capacitor if the rms voltage across the resistor is 20 V.

Solution:

fCXC

2

1

0

0

rms

rms

I

V

I

VXC

RC Series Circuit

21.3 RESISTANCE, REACTANCE & IMPEDANCE

Topic --- Alternating Current

IRVR

Phasor diagram: the supply voltage V leads the

current I the by radians

22

2 1

CRZ

A phasor diagram in terms of R, XL and Z

AC source

R

I

RV

V

L

LV

LL IXV

LV

V

IRV

22

LR VVV

R

L

V

Vtan

LX

Z

RR

X Ltan

22

LXRZ

fLX L 2

21.3 RESISTANCE, REACTANCE & IMPEDANCE

Topic --- Alternating Current

IRVR

LL IXV

CC IXV

AC source

I

V

R

RV CV

C L

LV

I

LV

RV

V

CV

CL VV

22

CL XXRIV

R

XX CL tan

LX

Z

CX

CL XX

R

22

CL XXRZ

21.3 RESISTANCE, REACTANCE & IMPEDANCE

Phasor diagram: the supply voltage V

leads the current I the

by radians

Topic --- Alternating Current

A series RLC circuit has a resistance of 25.0 Ω, a capacitance of 50.0 μF, and an inductance of 0.300 H. If the circuit is driven by a 120 V, 60 Hz source, calculate

(a) The total impedance of the circuit

(b) The rms current in the circuit

(c) The phase angle between the voltage and the current.

Suggested Answer:

RCL Series Circuit

21.3 RESISTANCE, REACTANCE & IMPEDANCE

• 64.9 Ω , 1.85 A, 67.3o

Topic --- Alternating Current

• is defined as the phenomenon that occurs when the frequency of the applied voltage is equal to the frequency of the RCL series circuit

• Since resonance in series RLC circuit occurs at particular frequency, so it is used for filtering and tuning purpose as it does not allow unwanted oscillations that would otherwise cause signal distortion, noise and damage to circuit to pass through it

• Figure below shows the variation of XC,

XL, R and Z with frequency f of the RCL

series circuit

Z

fX L

R

fXC

1

0 f

Z

rf

•at low frequency,

impedance Z is large

because 1/ωCis large

• at high frequency,

impedance Z is high

because ωL is large

Topic --- Alternating Current

• From the graph, Zmin at fr

• This will happen when

22

CL XXRZ

02min RZ

RZ min

fr : resonant frequency

R

V

Z

VI

min

max

• The resonant frequency, fr of the RCL series circuit is given by

CL XX

CL

1

LC

12

LC

f1

22

r

LCf

2

1r

• At resonance in the RCL series circuit,

the impedance is minimum Zmin

thus the rms current flows in the circuit is maximum Imax and is given by

21.3 RESISTANCE, REACTANCE & IMPEDANCE

The resistance in the circuit is only came from R

CL XX

Topic --- Alternating Current

A 200 resistor, a 0.75 H inductor and a capacitor of capacitance C are connected in series to an alternating source 250 V, fr = 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(e) Sketch the phasor diagram.

Suggested Answer:(a)

(b)

(c)

(d)

(e)

RCL Series Circuit

21.3 RESISTANCE, REACTANCE & IMPEDANCE

k 2.83

k 2.83

C

L

X

LX

nF 93.9 ,2

1 x102.83 3 C

fC

A1.25R

V

Z

VI rmsrms

rms

VL

VC

VR

I

200RZ

Topic --- Alternating Current

1. Based on the RCL series

circuit in Figure above, the

rms voltages across R, L and

C are shown.

(a)With the aid of the phasor

diagram, determine the

applied voltage and the

phase angle of the circuit.

(b) Calculate(i) the current flows in the circuit

if the resistance of the

resistor R is 26 ,

(ii) the inductance and

capacitance if the frequency

of the AC source is 50 Hz,

(iii) the resonant frequency.

21.2 ROOT MEAN SQUARE (rms)

Topic --- Alternating Current

Q2

• A 2 F capacitor and a 1000 resistor are placed in series with an alternating voltage source of 12 V and frequency of 50 Hz. Calculate

• (a) the current flowing,• (b) the voltage across the capacitor,• (c) the phase angle of the circuit.

Q3

• An AC current of angular frequency of 1.0 104 rad s1 flows through a 10 k resistor and a 0.10 F capacitor which are connected in series. Calculate the rms voltage across the capacitor if the rms voltage across the resistor is 20 V.

ANS: 2.0 V

21.3 RESISTANCE, REACTANCE & IMPEDANCE

Topic --- Alternating Current

Q4 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 of the circuit.

ANS: 2.83 k, 2.83 k; 93.8 nF; 200 ; 1.25 A

Q5 A capacitor of capacitance C, a coil

of inductance L, a resistor of resistance R and a lamp of negligible resistance are placed in series with alternating voltage V. Its frequency f is varied from a low to a high value while the magnitude of Vis kept constant.(a) Describe and explain how the brightness of the lamp varies.(b) If V=0.01 V, C =0.4 F, L =0.4 H, R = 10 and the circuit at resonance, calculate(i) the resonant frequency,(ii) the maximum rms current,(iii) the voltage across the capacitor.• (Advanced Level Physics,7th edition, Nelkon

& Parker, Q2, p.423)• ANS: 400 Hz; 0.001 A; 1 V

21.3 RESISTANCE, REACTANCE & IMPEDANCE

Topic --- Alternating Current

V

V V

22

LR VVV

22

LXRIV

22

LXRZ

IZV 21.3 RESISTANCE, REACTANCE & IMPEDANCE

Topic --- Alternating Current

A circuit is made up of a 3200 pF capacitor connected in series to a 30 H coil of resistance 4 . Calculate(i) impedance at frequency

30 kHz.(ii)resonant frequency.

Solution:C = 3200 10-12 F; L = 3010-6

H; R = 4

(i) Given f = 30103 Hz, The reactance of capacitor and inductor are

PSPM 2009/ 10: Q12(c):

fCXC

2

1

123 10320010302

1

1066.1 3

CX

21.3 RESISTANCE, REACTANCE & IMPEDANCE

Topic --- Alternating Current

(i) and

Therefore the impedance is givenby

(ii) Apply:

fLX L 2

63 103010302

22

CL XXRZ

2321066.166.54

1654Z

66.5LX

126 10320010302

1

Hz 1014.5 5rf

LCfr

2

1

21.3 RESISTANCE, REACTANCE & IMPEDANCE

Topic --- Alternating Current

Apply

(i) average power,

(ii) instantaneous power,

(iii)power factor,

in AC circuit consisting of R, RC, RL and RCL in series.

21.4 POWER & POWER FACTOR

• Power factor is a way of measuring how efficiently electrical power is being used within a facility's electrical system

cosVIPrmsrms

av

IVP

IV

P

P

P av

a

rcos

Topic --- Alternating Current

• In an ac circuit , the power is only dissipated by a resistance, none is dissipated by inductance or capacitance

• From the phasor diagram of the RCL series circuit

21.4 POWER & POWER FACTOR

RIIVP R2

av

ωLV

IRV

V

CV

CL VV

• We get

• Thenand

cosVVR V

VRcos

cosav IVP IZV

r2

av cos PZIP

where cos is called the

power factor of the AC circuit,

Pr is the average real

power and I2Z

is called the apparent

power

Topic --- Alternating Current

• Power factor is defined as

• From

• the power factor also can be calculated by using the equation below

• When = 0o (cos =+1) ,the circuit is completely resistive or when the circuit is in resonance (RCL)

• When = +90o (cos = 0), the circuit is completely inductive

• When = -90o (cos =0), the circuit is completely capacitive

a

r

2

rcosP

P

ZI

P

IZ

IR

V

VR cos

Z

Rcos

22

CL XXRIV

21.4 POWER & POWER FACTOR

Topic --- Alternating Current

An oscillator set for 500 Hz puts out a sinusoidal voltage of 100 V effective. A 24.0 Ω resistor, a 10.0μF capacitor, and a 50.0 mHinductor in series are wired across the terminals of the oscillator.(a)What will an ammeter in the circuit read?(b)What will a voltmeter read across each element?(c)What is the real power dissipated in the circuit?(d)Calculate the power supply(e)Find the power factor(f) What is the phase angle?

Z

Rcos 1

Z

VI rms

rms

CC

LL

R

IXV

IXV

IRV

cosVIPrmsave

rmsrmsplysupVIP

Z

Rcos

Z

Rcos 1

21.4 POWER & POWER FACTOR

Topic --- Alternating Current

A 100 F capacitor, a 4.0 H inductor and a 35 resistor are connected in series with an alternating source given by the equation. V = 520 sin 100t. Calculate:

(a)the frequency of the source,

(b)the capacitive reactance and inductive reactance,

(c)the impedance of the circuit,

(d)the peak current in the circuit,

(e)the phase angle,

(f) the power factor of the circuit.

21.4 POWER & POWER FACTOR

Topic --- Alternating Current

Suggested Answer:

By comparing

V = 520 sin 100t

to the, V = V0 sin t

Thus

V0 = 520V, = 100 rad s-1

(a) The frequency of AC source is

given by

(b) The capacitive reactance is

and the inductive reactance is

f 2

Hz 9.15ff2100

fCXC

2

1

100CX

6101009.152

1

CX

fLX L 2

400LX

0.49.152

21.4 POWER & POWER FACTOR

Topic --- Alternating Current

(c) The impedance of the circuit is

(d) The peak current in the circuit is

(e) The phase angle between the current and the supply voltage is

ORf. The power factor of the circuit is given by

22CL XXRZ

2210040035

302Z

ZIV 00

302520 0I

A 72.10 I

R

XX CL tan

35

100400tan 1

rad 45.1

R

XX CL1tan

3.83

cosfactorpower 383cos .

117.0factorpower

21.4 POWER & POWER FACTOR

Topic --- Alternating Current

Q1 A 22.5 mH inductor, a

105 resistor and a 32.3 F capacitor are connected in series to the alternating source 240 V, 50 Hz.(a) Sketch the phasor diagram for the circuit(b) Calculate the power factor of the circuit(c) Determine the average power consumed by the circuit.

ANS: 0.755, 313 W

Q2 A coil having inductance 0.14

H and resistance of 12 is connected to an alternating source 110 V, 25 Hz. Calculate(a) the rms current flows in the coil(b) the phase angle between the current and supply voltage(c) the power factor of the circuit(d) the average power loss in the coil.ANS: 4.4 A, 61.3o , 0.48, 0.23

kW

Q3 A series RCL circuit

contains a 5.10 μF capacitor and a generator whose voltage is 11.0 V. At a resonant frequency of 1.30 kHz the power dissipated in the circuit is 25.0 W. Calculate(a) the inductance(b) the resistance(c) the power factor when the generator frequency is 2.31 kHz.

ANS: 2.94 x 10-3 H , 4.84 Ω , 0.163

21.4 POWER & POWER FACTOR

Topic --- Alternating Current

Q4

• An RLC circuit has a resistance of 105 , an inductance of 85.0 mH and a capacitance of 13.2 F.

• What is the power factor of the circuit if it is connected to a 125 Hz AC generator?

• Will the power factor increase, decrease or stay the same if the resistance is increased? Explain.

• (Physics, 3rd edition, James S. Walker, Q47, p.834)• ANS: 0.962; U think

Q5

• A 1.15 k resistor and a 505 mH inductor are connected in series to a 14.2 V,1250 Hz AC generator.

• What is the rms current in the circuit?• What is the capacitance’s value must be inserted in series with the resistor and inductor to

reduce the rms current to half of the value in part (a)?• (Physics, 3rd edition, James S. Walker, Q69, p.835)

• ANS: 3.44 mA, 10.5 nF

21.4 POWER & POWER FACTOR

Topic --- Alternating Current

G E O M E T R I C A LO P T I C S