Chapter 6

CP3 1 FYSL

Chapter 6: Alternating Current

6.1 Alternating Current

L.O 6.1.1 Define alternating current (AC)

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 .

Extra Knowledge:

Alternating current can be used to create a changing magnetic field, and changing magnetic

fields can be used to create alternating current. This relationship between alternating current

and magnetic fields makes three important devices possible: alternator, motor and transformer.

L.O 6.1.2 Sketch and interpret sinusoidal AC waveform

L.O 6.1.3 Use sinusoidal voltage and current equations

tII o sin

tVV o sin

Phase (Phase denotes the particular point in the cycle of a

waveform, measured as an angle in degrees/ radian)

Chapter 6

CP3 2 FYSL

Terminology in AC

Peak (maximum) current ( I0 )

Definition: Magnitude of the maximum current.

Peak (maximum) voltage ( V0 )

Definition: Magnitude of the maximum voltage.

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)

Angular frequency ( )

Unit: radian per second (rad s-1)

Equation: f 2 or T

2

Example

Question Solution

Figure shows the variation of voltage with

time for a sinusoidal AC current. Determine

a. the frequency

b. the phase angle

c. the peak-to-peak voltage and

d. write the expression for the graph

The current in an AC circuit is given by the

expression:

tI 50sinmA 25

Sketch a I-t graph for the AC circuit.

Determine the current when t = 50 s.

fT

1

Chapter 6

CP3 3 FYSL

6.2 Root Mean Square (rms)

L.O 6.2.1 Define root mean square (rms) current and voltage for AC source

L.O 6.2.2 Use rms voltage and current equations

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.

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.

**Equations Irms and Vrms are valid only for a sinusoidal alternating current and

voltage**

The average power: Relationship between Pave and P0

2

0

rms

II

2

0

rms

VV

Average value for one complete cycle is zero

R

VRIVIP

2

rms2

rmsrmsrmsave 22

1

22

0

00

00

ave

PVI

VIP

000 VIP

acdc power average power rms

2

ave

2

ave

2

III

RIRI

rms

2

ave

2

ave

2

VVV

R

V

R

V

Chapter 6

CP3 4 FYSL

Example

Question Solution

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

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.

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.

Exercise

Question

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

a. the reading of the AC ammeter in series with the resistor

b. the peak current and mean power

Answer:

An AC current is given as I = 5 sin (200t) where the clockwise direction of the current is

positive. Find

a. the peak current

b. the current when t = 1/100 s

c. the frequency and period of the oscillation.

Answer: 5 A; 4.55 A; 31.88 Hz; 0.0314 s

Chapter 6

CP3 5 FYSL

6.3 Resistance, Reactance and Impedance

L.O 6.3.1 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

L.O 6.3.3 Define and use capacitive reactance, inductive reactance, impedance and

phase angle

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).

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)

θ

OP

ONsin

Phasor diagram

(not including the circle) Sinusoidal Waveform

Instantaneous value

Chapter 6

CP3 6 FYSL

Pure resistor in the AC circuit

The current flows in the resistor is

The voltage across the resistor VR at any instant is

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

The instantaneous power

The average power

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

ωtII sin0

IRV R

ωtRIV sin0R

VωtVV sin0R

00 VRI

tageSupply vol:V

and

IV ωtωt

0

0

0

rms

rms

I

V

I

VR

tVIP

tVtIP

R

VRIIVP

2

00

00

22

sin

sinsin

000

2

0

2

rmsave2

1

2

1

2

1PIVRIRIP

Chapter 6

CP3 7 FYSL

Pure capacitor in the AC circuit

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

The voltage across the capacitor VC

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 current flows in the ac circuit is

The phase difference between V and I is

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.

CCVQ

tCVQ sin0

dt

dQI

tCVdt

dI sin0

tdt

dCVI sin0

)cos(0 ωtCVI 00 ICV

2sin0

ωtII

and

2

2

Δ

ωttΔ

ωtVV sin0C

Chapter 6

CP3 8 FYSL

The capacitive reactance in a pure capacitor is

The instantaneous power

The average power

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.

The relationship between capacitive reactance XC and frequency f

fCCX

CV

V

I

V

I

VX

2

11C

0

0

0

0

rms

rms

C

Definition

tPP

tVIP

tVtIP

R

VRIIVP

2sin2

1

2sin2

1

sincos

0

00

00

22

0ave P

circuitcapacitor purein CXZ

Chapter 6

CP3 9 FYSL

Pure inductor in the AC circuit

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

The current flows in the ac circuit is

When the current flows in the inductor, the back emf caused by the self-induction is produced

and given by

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:

The phase difference between V and I is

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.

ωtII sin0

dt

dILε B

ωtIdt

dL sin0B

ωtωLI cos0B

ωtωLIV cos0B

2sin

2sin

0

0

ωtVV

ωtωLIV 00 LIV and

2

2

Δ

ωtωtΔ

Chapter 6

CP3 10 FYSL

The inductive reactance in a pure inductor is

The instantaneous power

The average power

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.

The relationship between inductive reactance XL and the frequency f

Definition

tPP

tVIP

tVtIP

R

VRIIVP

2sin2

1

2sin2

1

cossin

0

00

00

22

0ave P

circuitinductor purein LXZ

fLLX

I

LI

I

V

I

VX

2L

0

0

0

0

rms

rmsL

Chapter 6

CP3 11 FYSL

L.O 6.3.2 Use phasor diagram to analyze voltage, current, and impedance of series

circuit of RL, RC and RLC

RC Circuit

The total p.d (supply voltage), V across R and C is equal to the vector sum of VR and VC as

shown in the phasor diagram.

From the phasor diagrams, the current I leads the supply voltage V by ϕ radians where

Graph of Z against f

2

C

2

2

C

2

2

C

2

R

XRIV

IXIRV

VVV

2

C

2 XRI

VZ

I leads V by ϕ

R

CtanV

V

R

X Ctan or

The resistance, R

is independent of

frequency

Chapter 6

CP3 12 FYSL

RL Circuit

The total p.d (supply voltage), V across R and L is equal to the vector sum of VR and VL as

shown in the phasor diagram.

From the phasor diagram, the supply voltage V leads the current I by ϕ radians where

Graph of Z against f

V leads I by ϕ

2

L

2

2

L

2

2

L

2

R

XRIV

IXIRV

VVV

2

L

2 XRI

VZ

R

LtanV

V

R

X Ltan or

The resistance, R

is independent of

frequency

Chapter 6

CP3 13 FYSL

RCL Circuit (VL > VC)

The total p.d (supply voltage), V across L, R and C is equal to the vector sum of VL, VR and

VC as shown in the phasor diagram.

From the phasor diagram, the supply voltage V leads the current I by ϕ radians where

Graph of Z against f

V leads I by ϕ

2

CL

2

2

CL

2

2

CL

2

R

)(

)(

XXRIV

IXIXIRV

VVVV

2

CL

2 XXRZ

VL > VC

R

CLtanV

VV

IR

XXI CLtan

or

VL < VC ?

VL = VC ?

Chapter 6

CP3 14 FYSL

L.O 6.3.4 Explain graphically the dependence of R, XC, XL and Z on f and relate it to

resonance

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:

at low frequency, impedance Z is large because 1/ωC is large.

at high frequency, impedance Z is high because ωL is large.

at resonance frequency, impedance Z is minimum (Z = R; XC = XL) and I is maximum

CL

XX

1

CL

LCf

CfLf

2

1

2

12

r

r

r

Resonance

frequency

2

CL

2 XXRZ

02

min RZ

RZ min

max min, , 1

rmsrms

rms

rms

IZZ

I

Z

VI

Chapter 6

CP3 15 FYSL

Example

Question Solution

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.

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.

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.

Chapter 6

CP3 16 FYSL

Question Solution

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.

Exercise

Question

A rms voltage of 12.2 V with a frequency of 1.00 kHz is applied to a 0.29 mH inductor.

a. What is the rms current in the circuit?

b. Determine the peak current for a frequency of 2.50 kHz.

Answer: 6.70A; 3.78 A

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.

Answer: 6.38×10-3 A; 10.2 V; 57.9° or 1.01 rad

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.

Answer: 64.9 Ω , 1.85 A, 67.3o

Chapter 6

CP3 17 FYSL

6.4 Power and Power Factor

L.O 6.4.1 Apply average power, instantaneous power and power factor equations in

AC circuit consisting of RL, RC and RLC in series

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)

For RCL circuit:

From the diagrams above,

Substitute (2) into (1):

The term cos ϕ is called the power factor.

The power factor (cos ϕ ) can vary from a maximum of +1 (or 100%) to a minimum of 0.

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

resonance (RCL).

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

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

rmsrmsave RVIP Vrms across resistor

(1)

Z

R

V

V cos and cos R

(2)

cos

or

cos

rmsrmsave

2

ave rms

VIP

ZIP

aapparent PP a

avecosP

PRearranging

r

2

ave rms PRIP

Power used by the load

Power delivered to the load

Chapter 6

CP3 18 FYSL

Example

Question Solution

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.

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?

Exercise

Question

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.

Answer: 4.4 A, 61.3o , 0.48, 0.23 kW

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.

Answer: 2.94 x 10-3 H , 4.84 Ω , 0.163

tV 300sin300