chapter 6: alternating current - ysl...

Chapter 6: Alternating

Current

An alternating current is an current that reverses its

direction at regular intervals.

Overview

Alternating

Current

A.C. Through a

Capacitor

A.C. Through a

Inductor

A.C. Through a

Resistor

RCL Circuit in

Series

Phasor Diagram Sinusoidal Waveform

6.1 Alternating Current

Define alternating current (AC)

Sketch and interpret sinusoidal AC waveform

Use sinusoidal voltage and current equations:

Learning Objectives

tVV sin0

tII sin0

Alternating Current

An alternating current (AC) is the electrical

current which varies periodically with time in

direction and magnitude.

The usual circuit-diagram symbol for an AC

source is

Sinusoidal AC Waveform

The output of an AC generator is sinusoidal and

varies with time.

T

VI or

t0

T2

1

00 or VI

00 or VI

Positive half cycle

Negative half cycle

1 Revolution

Peak-to-peak value

Peak value

Sinusoidal AC Waveform

Equation for alternating current (I):

Equation for alternating voltage (V):

where

tII o sin

tVV o sin

Phase

)2(

locityangular ve ORfrequency angular :

f

currentpeak : 0I

gepeak volta: 0V

Phase Angle

Phase denotes the particular point in the cycle

of a waveform, measured as an angle in degrees/

radian

Revision A

(s) t0

0A

0A

A

(s) t0

0A

0A

A

(s) t0

0A

0A

tAA sin0

2sin0

tAA

2sin0

tAA

Terminology in AC

Peak (maximum) current ( I0 )

◦ Definition: Magnitude of the maximum current.

Peak (maximum) voltage ( V0 )

◦ Definition: Magnitude of the maximum voltage.

Angular frequency ( )

Unit: radian per second (rad s-1)

Equation: or

f 2T

2

Terminology in AC

Frequency ( f )

◦ Definition: Number of complete cycle in one

second.

◦ Unit: Hertz (Hz) or s-1

Period ( T )

◦ Definition: Time taken for one complete cycle.

◦ Unit: second (s)

fT

1

Example 1

Figure shows the variation of voltage with time for a sinusoidal AC. Determine

a) the frequency

b) the phase angle at t = 15 s

c) the peak-to-peak voltage and

d) write the expression for the graph

Example 1 – Solution

Example 2

The current in an AC circuit is given by the

expression:

Sketch a I-t graph for the AC circuit. Determine

the current when t = 50 s.

tI 50sinmA 25

6.2 Root Mean Square (rms)

Define root mean square (rms) current and

voltage for AC source.

Use the following formula,

**Equations Irms and Vrms are valid only for a

sinusoidal alternating current and voltage**

Learning Objectives

2

0II rms and

2

0VVrms

Average Current (Iave) Average current is defined as the average or mean

value of current in a half-cycle flows of current in a

certain direction.

Iave 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.

T

I

t0T

2

1 T2

0I

0I

Iav

Iav

0)( aveave II

2

2π

ooave

I

π

II

Root Mean Square Current (Irms)

Root mean square current (Irms) is defined as the

effective value of AC which produces the same power

(mean/average power) as the steady d.c. when the

current passes through the same resistor.

acdc power average power

rms

2

ave

2

ave

2

III

RIRI

2

0rms

II

Root Mean Square Voltage (Vrms)

Root mean square voltage/ p.d (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

2

0

rms

VV

rms

2

ave

2

ave

2

VVV

R

V

R

V

acdc power average power

Example 3

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

ordinary voltmeter. Determine

a) the maximum value the voltage takes on during a

cycle

b) the equation for the voltage

Example 4

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

a) the rms current,

b) the frequency,

c) the mean power dissipated in the resistor.

Example 5

An AC source V = 200 sin t is connected across a resistor

of 100 . Calculate

a) the rms current in the resistor.

b) the peak current.

c) the mean power.

6.3 Resistance, Reactance and

Impedance 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 and

pure inductor.

Use phasor diagram to analyze voltage, current, and impedance of

series circuit of RL, RC and RLC.

Define and use capacitive reactance, inductive reactance,

impedance and phase angle.

Explain graphically the dependence of R, XC , XL and Z on f and

relate it to resonance.

Learning Objectives

Phasor Diagram

Phasor is defined as a vector that rotates 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).

Phasor Diagram

t0 TT2

1 T2

ωN

O

P

Ao tAA o sin

A

Phasor diagram

(not including the circle)

Sinusoidal Waveform

Instantaneous value

P

Resistance, Reactance and

Impedance

Key Term/Ω Meaning

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 reactance combine to

form impedance)

Pure Resistor in the AC Circuit


The current flows in the resistor is

ωtII sin0

IRV R

ωtRIV sin0R

VωtVV sin0R

00 VRI

tageSupply vol:V

The voltage across the resistor VR at any instant is

and


ωtII sin0

ωtVV sin0

t0

0I

0V

0I0V

TT2

1 T2T

2

3

ω

VI

Phasor diagram


The phase difference between V and I is

In pure resistor, the voltage V is in phase with the current I

and constant with time (the current and the voltage reach

their maximum values at the same time).

The resistance in a pure resistor is

IV ωtωt

0

0

0

rms

rms

I

V

I

VR


The instantaneous power The average power

tVIP

tVtIP

R

VRIIVP

2

00

00

22

sin

sinsin

0

00

2

0

2

rmsave

2

1

2

1

2

1

P

IV

RI

RIP

A resistor in AC circuit dissipates energy in the form of heat

Pure Capacitor in the AC Circuit


The voltage on a capacitor depends on the amount of

charge you store on its plates. The charge accumulates on

the plates of the capacitor is

The voltage across the capacitor VC

ωtVV sin0C

CCVQ

tCVQ sin0


The current flows in the ac circuit is

dt

dQI

tCVdt

dI sin0

tdt

dCVI sin0

)cos(0 ωtCVI 00 ICV

2sin0

ωtII

and

ωtII cos0

Compare with

general equation


t0

0I

0V

0I0V

TT2

1 T2T

2

3

ω

VI

rad 2

Phasor diagram ωtVV sin0

2sin0

ωtII



In pure capacitor, the voltage V lags behind the current I

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

radians.

2

Δ

2

ωttΔ


The capacitive reactance in a pure capacitor is

0

0

0

0

rms

rmsC

CV

V

I

V

I

VX

Definition

circuitcapacitor purein CXZ

fCCX

2

11C



tPP

tVIP

tVtIP

R

VRIIVP

2sin2

1

2sin2

1

sincos

0

00

00

22 0ave P

For the first half cycle where the power is positive, the capacitor is

saving the power (in electric field). For the second half of the cycle

where the power is negative, the power is returned to the circuit.

Pure Inductor in the AC Circuit


When the current flows in the inductor, the back emf

caused by the self-induction is produced and given by

The current flows in the resistor is

ωtII sin0

dt

dILε B

ωtIdt

dL sin0B

ωtωLI cos0B


At each instant the supply voltage V must be equal to the

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

e.m.f always oppose the supply voltage V. Hence, the

magnitude of V and B:

ωtωLIV cos0B

2sin0

ωtωLIV

00 LIV

ωtVV cos0

Compare with

general equation

2sin0

ωtVV


t0

0I

0V

0I0V

TT2

1 T2T

2

3

ω

V

I

rad 2

ωtII sin0

2sin0

ωtVV

Phasor diagram

2

Δ



In pure inductor, the voltage V leads the current I by /2

radians or the current I lags behind the voltage V by /2

radians.

ωtωtΔ

2


The inductive reactance in a pure inductor is

0

0

0

0

rms

rmsL

I

LI

I

V

I

VX

Definition fLLX 2L

circuitinductor purein LXZ



tPP

tVIP

tVtIP

R

VRIIVP

2sin2

1

2sin2

1

cossin

0

00

00

22 0ave P

For the first half of the cycle where the power is positive, the

inductor is saving the power. For the second half cycle where the

power is negative, the power is returned to the circuit.

Summary 1

Circuit Relationship

between V and I

Impedance, Z

(V = IZ)

Pinstantaneous Pave

Pure

resistor

The voltage V is

in phase with the

current I

Resistor, R

Pure

capacitor

The voltage V

lags behind the

current I by /2

radians

Capacitive

reactance, XC

Pure

inductor

The voltage V

leads the current

I by /2 radians

Inductive

reactance, XL

tVIP 2

00 sin

tPP 2sin

2

10

tPP 2sin

2

10

0ave2

1PP

0ave P

0ave P

fCCX

2

11C

fLLX 2L

Example 6

A capacitor with C = 4700 pF is connected to an AC

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.

Example 7

A 240 V rms supply with a frequency of 50 Hz causes an

rms 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.

Example 7 –Solution

RC Circuit

RC Circuit

IRVR CC IXV

The rms voltage across the

resistor VR and the capacitor VC

are given by:

I

ω

CV

RV

V

Based on the phasor diagram, the

rms supply voltage V (or total

voltage) of the circuit is given by:

22

CR VVV

Phasor diagram

and

I leads V by ϕ

RC Circuit

22

22

22

C

C

CR

XRIV

IXIRV

VVV

2

C

2 XRI

V

ω

CXZ

R

2

C

2 XRZ

I

VZ and

Phasor diagram

ωCX C

1and

RC Circuit

From the phasor diagrams,

Graph of Z against f

R

CtanV

V

R

X Ctan or

RL Circuit

RL Circuit

IRVR


resistor VR and the inductor VL

are given by:

and LL IXV

ω

LVV

IRV




Phasor diagram

22

LR VVV

V leads I by ϕ

RL Circuit

22

22

22

L

L

LR

XRIV

IXIRV

VVV

2

L

2 XRI

V

2

L

2 XRZ

I

VZ and

and ωLX L

Phasor diagram

ω

LXZ

R

RL Circuit



R

LtanV

V

R

X Ltan or

RCL Circuit

RCL Circuit

IRVR


resistor VR and the capacitor

VC are given by:

LL IXV CC IXV , ,

I

ωLV

RV

V

CV

CL VV

Phasor diagram




22)( cLR VVVV

V leads I by ϕ

RCL Circuit

22

22

22

)(

)(

CL

CL

cLR

XXRIV

IXIXIRV

VVVV

VL > VC

2

CL

2 XXRZ

ωLX

Z

CX

CL XX

R

Phasor diagram

RCL Circuit



R

CLtanV

VV

R

XX CLtan

or

VL < VC ?

VL = VC ?

Example 8

An alternating current of angular frequency of 1.0 × 104

rad s‒1 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.

Resonance in AC circuit

Resonance is defined as the phenomenon that occurs when

the frequency of the applied voltage is equal to the

frequency of the LRC series circuit.

The graph shows that:

i. at low frequency, impedance Z

is large because 1/ωC is large.

ii. at high frequency, impedance Z

is high because ωL is large.

iii. at resonance frequency,

impedance Z is minimum (Z =

R; XC = XL) and I is

maximum


At resonance frequency:

CL

1

LCf

CfLf

2

1

2

12

r

r

r

Resonance

frequency

CL XX


At resonance frequency:

2

CL

2 XXRZ

02

min RZ

RZ min

max min, , 1

rmsrms

rms

rms

IZZ

I

Z

VI

Example 9

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.

Calculate:

b. the current flows in the circuit if the resistance of the resistor R is

26 ,

c. the inductance and capacitance if the frequency of the AC source

is 50 Hz,

d. the resonant frequency.

6.4 Power and Power Factor

Apply:

i. average power,

ii. instantaneous power,

i. power factor,

in AC circuit consisting of R, RC, RL and RLC in

series.

Learning Objectives

cosrmsrmsave VIP

IVP

rmsrms

av

a

r

VI

P

P

Pcos

Power and Power Factor

In an AC circuit, the power is only dissipated by a

resistance; none is dissipated by inductance or

capacitance.

Therefore, the real power (Pr) that is used or gone is

equal to that dissipated from the resistor and given by

the average power (Pave)

rmsrmsave RVIP

r

2

ave rms PRIP


For RCL circuit:

From the diagrams above

Z

R

V

V cos and cos R


cos

or

cos

rmsrmsave

2

ave rms

VIP

ZIP

aapparent PP

a

avecosP

PRearranging

Example 10

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.

tV 300sin300

Example 11

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 mH inductor 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?

Summary

Alternating

Current

A.C. Through a

Capacitor

A.C. Through a

Inductor

A.C. Through a

Resistor

RCL Circuit in

Series

Phasor Diagram Sinusoidal Waveform

chapter 6: alternating current - ysl...

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