AC Circuits - Ek

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EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAOLU, Page 1

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AC Circuits

by

Prof. Dr. Osman SEVAOLU

Electrical and Electronics Engineering Department

Alternating Current AC) Circuits


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AC Circuits

What is Direct Current DC) ?

Direct Current (DC) is a current with aconstant time characteristics

Definition

Current, I

+

R1= 5 Ohms

Vs=600 V

Switch

R2= 5 Ohms

Switch is turned on at: t = 1 sec

1 2 3 4 5 6

80

0

Current(A

mp)

7

20

40

60

I = 60 A

DC (Constant) Current

Time (Sec)


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AC Circuits

What is Alternating Current AC) ?

Alternating Current (AC) is a current with

time varying characteristics

Definition

Sinusoidal AC

Non - Sinusoidal AC

0,0

1,0

2,0

3,0

4,0

5,0

0,0 1,0 2,0 3,0 4,0 5,0 6,0t (sec)

Current(Am

p)

10

- 10

0

Angle (Radians)Current(Am

p)

/2 3

/2 2


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AC Circuits

Basic Parameters of a Sinusoidal Waveform

Sinusoidal voltage is a voltagewith waveform as shown on theRHS

Definition

Sinusoidal Voltage

V(t) = V sin ( wt + )^

where

V(t) is the voltage waveform,

V is the peak value (amplitude),

w is the angular frequency, is the phase shift, i.e. angle of the

voltage at t = 0, (phase angle)

^

w = 2 f

f = 50 Hz

100

0

Angle (Radians)

Voltage (Volt)

/2 3/2 2

200

300

312

Amplitude

Phase angle


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AC Circuits

AC Alternating Current) Circuit

Positive half cycle

Negative half cycle

-25

-20

-15

-10

-5

0

5

10

15

20

25

0 0,005

Load

+

Load0,015

I (Amp)

-25

-20

-15

-10

-5

0

5

10

15

20

25

0 0,005 0,01

I (Amp)

-25

-20

-15

-10

-5

0

5

10

15

20

25

0,01 0,015 0,02

+V(t)

I(t)

V(t)

I(t)

Time (Sec) Time (Sec)


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AC Circuits

Capacitance

Capacitors store electrical charge

Definition

Storage capacity of a capacitor is

called capacitance +

_

+

_

Small Capacitance Large Capacitance

Water (hydroulic) example

Capacitance = C1 Capacitance = C2C1 < C2

_


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AC Circuits

Geometry

Control relay

Capacitor banks

Capacitor-Practical Configuration


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AC Circuits

MV Medium Voltage) Shunt Capacitor Banks


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AC Circuits

MV Medium Voltage) Shunt Capacitor Banks


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AC Circuits

Electronic Capacitors in a Motherboard


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AC Circuits

Basic Relation

Charge stored in a capacitor is

proportional to the capacitance C,Charge stored in a capacitor is

proportional to the voltage Vapplied

Basic Principle

or

Q = C Vwhere, Q is charge stored (Coulombs),

V is voltage (Volts),C is capacitance (Farads) +

_

+

1 Volt

Symbolic Representation

C = 1 Farad

+

_

+

VC

Q = 1 Coulomb1 Farad is the capacitance with a

charge of 1 Coulomb at a voltage 1 Volt

between the plates

Capacitance CVoltage Source V


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AC Circuits

Voltage - Current Relation for a Capacitor

Definition

Q = C V

The relation;

may be written in time domain as;

Q(t) = C V(t)

or differentiating both sides with respect to time

dQ(t)/dt = C dV(t)/dt

or remembering thatdQ(t)/dt = I(t)

Hence,

I(t) = C dV(t) / dt

+

_

C

I(t)

V(t)

+


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AC Circuits

Definition

The above equation may be integratedwith respect to time, yielding the

following Voltage - Current Relation for a

Capacitor

V(t) = (1/C) I(t)dt + V(0)

where V(0) is the initial voltage across

the capacitor, representing the initialvoltage due to the initial charge stored in

the capacitor

I(t)

+

_

+ C

V(t)

Switch


Vc (0)

Voltage - Current Relation for a Capacitor


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AC Circuits

Example - 1

Problem

Calculate the time waveform of the current flowingin the circuit shown on the RHS by assuming that

the capacitor is charged by the exponential

voltage V(t) shown in the figure

V(t) = 5 (1 - e-t/)

where is known as the time constant of the

circuit and given as = 10 -6sec = 1.0 sec

C = 0.1 F

+

_

+

t (sec)

V(t) = 5 (1 - e-t/

)

0,0

1,0

2,0

3,0

4,0

5,0

0,0 1,0 2,0 3,0 4,0 5,0 6,0

C

I(t)

V(t)

Vmax = Maximum voltage that can be reached = 5 Volts

Qmax = C x Vmax = Maximum charge that can be stored


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AC Circuits

Example - 1

Solution

0,0

1,0

2,0

3,0

4,0

5,0

0,0 1,0 2,0 3,0 4,0 5,0 6,0

t (sec)

I(t) = C dV(t)/dt

V(t) = 5 (1 - e-t/)

Hence,

I(t) = C d V(t) / dt

= C d 5(1 e -t/)/dt

= 0.1 x10-6( 5 /

) xe -t/

= 0.1 x10-6x5 x106x e -t/

= 0.5 xe -t/ Ampers

0,00,0 1,0 2,0 3,0 4,0 5,0 6,0

t (sec)

0,5

0,1

0,2

0,3

0,4

I(t)

+

_

+ C

I(t)

V(t)

V(t) = 5 (1 - e-t/ )


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AC Circuits

Example - 2

Problem

Current source shown in the circuit onthe RHS provides a 10 mA constant

current within the time interval;

Capacitor is initially charged to 2 Volts

voltage

t [0, 1]

Determine the voltage across thecapacitor within the time interval;

t [0, 1]

I(t) (mA)

10,0

0,0 0,2 0,4 0,6 0,8 1,0 1,2

t (sec)

8,0

6,0

4,0

2,0

12,0

14,0

C = 1000 F Vc(0) = 2 Volts

+

_

I(t)

I(t)C

Switch

Switch is turned on at: t = 0 sec,off at: t = 1.0 sec.


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AC Circuits

Example - 2

Solution

Voltage across the capacitance isexpressed as

t [0, 1]

V(t) = (1/C) I(t)dt + V(0)where,

V(0) = 2 Voltsis the initial voltage across thecapacitorHence;

C = 1000 F

Vc(0) = 2 Volts

I(t) (mA)

+

_

I(t)

I(t)C

0,0

10,0

0,0 0,2 0,4 0,6 0,8 1,0 1,2

t (sec)

8,0

6,0

4,0

2,0

12,0

14,0

V(t) (Volts)

0,0

10,0

0,0 0,2 0,4 0,6 0,8 1,0 1,2

t (sec)

8,0

6,0

4,0

2,0

12,0

14,0

1

V(t) = (1/C)

I(t)dt + 20

1

V(t) = (1/10-3) 10-2 dt + 20

= 10 t + 2 Volts


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AC Circuits

R-C Circuits

Problem

Solve the RC circuit shown on the RHS

for current I(t) flowing in the circuit whenthe switch is closed at t = 0

Differentiating both sides wrt time once;

dV(t) / dt = R dI(t) / dt + (1/C) I(t)or

d/dt I(t) + (1/RC) I(t) = (1/R) dV(t) / dt

A first order ordinary differential equation

V(t) = R I(t) + VC(t)

= R I(t) + (1/C) I(t)dt + V(0)

Writing down KVL for the circuit

Solution R

C

+

_

+

I(t)

V(0)=V0

V(t) = V sinwt^


Switch


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AC Circuits

Solution of the resulting First Order Ordinary Differential Equation

Solve the resulting first order ordinary differential equation (ODE)

Solution

dI(t) / dt + (1/RC) I(t) = (1/R) dV(t) / dt

dI(t) / dt + (1/RC) I(t) = (1/R) d/dt (V sin ( wt + ))

dI(t) / dt + (1/RC) I(t) = (V/R) w cos ( wt + ) V( t ) = V sin ( wt + )dV( t ) / dt = V w cos wt

^^

100

0

Voltage (Volt)

Angle (Radians)

/2 3/2 2

200

300

312

Amplitude

Phase angle

^

^

R

C

+

_

+ V(0)=V0

V(t) = V sinwt^


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AC Circuits


Solution

(t) dI(t)/dt+ I(t) (1/RC)(t) = (t) ( V/R ) w cos ( wt + )

(t) dI(t)/dt+ I(t) d/dt (t) = ( V/R ) (t) w cos ( wt + )

d/dt [(t) I(t)] = ( V/R ) (t) w cos ( wt + )

d/dt [(t) I(t)] dt = ( V/R ) (t) w cos ( wt + ) dt + I(0)

(t) I(t) = ( V/R ) (t) w cos ( wt + ) dt + I(0)

I(t) = I (t) -1

(t) w cos(wt + ) dt + (t)-1I(0)

^

^

^

^

^

^

Define an integration factor(t) = e t / R C

Multiply both sides of the above ODE by this factor;

V( t ) = V sin ( wt + )dV( t ) / dt = V w cos wt

^

^

100

0

Voltage (Volt)

Angle (Radians)

/2 3/2 2

200

300

312

Amplitude

Phase angle

dI(t) / dt + (1/RC) I(t) = (V/R) w cos ( wt + )^


R

C

+

_

+ V(0)=V0

V(t) = V sinwt^


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AC Circuits


Solution Continued)

Subsituting the above term into the solution;I(t)

V( t ) = V sin wt

dV( t ) / dt = V w cos wt

^

^


R

C

+

_

+ V(0)=V0

V(t) = V sinwt^

w sinwt + (1/RC) coswtI(t) = I e - t/RCwe t/RC ---------------------------------- + e - t/RCI(0)

(1/RC)2+ w 2

^

sinwt + (1/wRC) coswtI(t) = I w2----------------------------------- + e - t/RCI(0)

(1/RC)2+ w 2

^

II(t) = -------------------(sinwt + (1/wRC) coswt) + e - t/RCI(0)

(1/wRC)2+ 1

^

Steady-State Term Transient Term

Switch


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AC Circuits

Numerical Example

Now assume that the parameters of the circuit on

the RHS are as follows;

V(t) = 312 sin wt Volts

R = 10 Ohms

C = 10 Farads

I(t)

R

C

+

_

+ V(0)=V0

V(t) = V sinwt^

V( t ) = V sin wt

dV( t ) / dt = V w cos wt

^

^

Switch


I

I(t) = -------------------(sinwt + (1/wRC) coswt) + e - t/RCI(0)(1/wRC)2+ 1

^




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AC Circuits

Numerical Example

I

I(t) = -------------------(sinwt + (1/wRC) coswt) + e - t/RCI(0)(1/wRC)2+ 1

^


-20-20

-15

-10

-5

0

5

10

15

20

25

0.5 1 1.5 2 2.5 3 3.5

5

10

15

20

25

0 0.5 1 1.5 2 2.5 3 3.5 4

-30

-20

-10

0

10

20

30

40

0.5 1 1.5 2 2.5 3 3.5 4


AC Circuits


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Solution for DC Voltage Source - Two Simple Rules

Rule - 1

A capacitor acts as a DC voltage source

initially due to its initial charge

I(0) = ( V V0) / R

The current waveform will then be;

I(t) = I() + [ I(0) - I() ] e -t/

= RC = Time Constant: The time required a for

capacitor to reach 63 % of its full charge

= 2 x 1000 F = 2 x 1000 x 10-6 = 0.002 sec

Then the initial value of current will be;

I(0) = ( VDC V0) / R

C =1000 F+

_

+

I(t)VDC

R = 2

V(0)=V0V = VDC-V0

R

+

_

+

I(0) SC

0

V(t) = (1 / C) I(t)dt = V(0) = V0-

Please note that anuncharged capacitoracts effectively as a

short circuit,i.e. V0= 0

AC Circuits


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Rule - 2

A capacitor acts as an open circuit finally

to a DC voltage (or current)


I(t) = I() + [ I(0) - I() ] e -t/

I() = 0

I(t) = C d/dt V(t) = C d/dt (constant) = 0 (OC)

+

_

+

I(t)VDC

V(0)=V0

OC

+

I() = 0

R

VDC

Then the final value of current will be;

I() = 0

C =1000 F

R = 2

= RC = Time Constant: The time required for

a capacitor to reach 63 % of its full charge

= 2 x 1000 F = 2 x 1000 x 10-6 = 0.002 sec

AC Circuits


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Solution

Substituting the above expressions into

the current expression


I(t) = I() + [ I(0) - I() ] e -t/

I(t) = [( VDC V0) / R ] e-t/

The initial value of current will be;

I(0) = ( VDC V0) / R

C =1000 F+

_

+

I(t)VDC

R = 2

V(0)=V0

The final value of current will be;

I() = 0

I(t) (Amp)

0

4.0

8.0

12.0

16.0

20.0

24.0

0 0,002 0,004 0,006 0,008 0,01

% 63 of the initial value

AC Circuits


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Meaning of Time Constant

Definition

Time constant of a electric circuit is

the duration for the current to getreduced by 63 % of its initial value

Time constant

of an RC circuit issimply expressed as:

= RC

The Effect of

0

0 0,5 1 1,5 2 2,5 3 3,5 4 4,5 5 5,5 6

t (sec)

1

2

3

4

5

% 63 of the initial value

% 63 of the inital value

1

2

2 >

1

AC Circuits


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Example

Problem

Find the voltage waveform across the 1 mF

capacitor shown on the RHS, when it has aninitial charge of 6 Volts and charged by a 24Volts DC voltage source through a wire with 2Ohm resistance

C = 1 mF

R= 2 Ohm

+

_

+

I(t)

VDC= 24 Volts V(0)=V0 = 6 V

The voltage waveform will then be;

V(t) = V() + [V(0) - V()] e -t/

V(t) = 24 + ( 6 24 ) e-t/0.002

= 24 -18 e-t/0.002

Volts

Capacitor will behave as a DC source at thebeginning and as OC at the end, hence;

V(0) = V0= 6 Volts

V() = VS= 24 Voltst (sec)

V(t)(Volts)

0

4.0

8.0

12.0

16.0

20.0

24.0

0 0,002 0,004 0,006 0,008 0,01

% 63 of the final value

= RC = Time Constant: The time required

for a capacitor to reach 63 % of full charge

= 2 x 1000 F = 2 x 0.001 = 0.002 sec

AC Circuits


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Energy Stored in a Capacitor

Problem

Calculate the instantaneous energy

stored in a capacitor with a capacitanceC and an instantaneous voltage VC(t)

P(t) = VC(t) I(t)

WC(t) = P(t) dt

=

VC(t) I(t) dt

= VC(t) C dVC(t) / dt dt

= C

VC(t) dVC(t)

WC(t) = (1/2) C VC2

(t)

or

Example

A 10 F capacitor fully charged with a 12

Volts DC voltage stores an energy;(1/2) 10 x10-6x122 = 720 x10-6Joule

C+

_

+

I(t)

V(t) Vc

(t)

AC Circuits


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Example

Problem

Find the instantaneous energy in thecapacitor for the voltage shown in the

figure

Wc(t) = (1/2) 10 x 10-6 V 2sin 2wt

= 0,000005 x 3122sin2wt

= 0,4867 sin2wt= 0,4867 ( cos 2x )

V (t) = V sin wt

WC(t) = (1/2) C VC2(t)

^

^

V( t ) = V sin wt^

312

- 312

Angle (Radians)

0/2 3/2 2

Wc(t)

0,1

0,2

0,3

0,4

0,5

0,6

0/2 3/2 2

Voltage (Volts)

Angle (Radians)

C +

_

+

I(t)

C = 10 F

V(t) = 312 sin wt

Mean of Wc(t) > 0sin2wt = 1 cos2wt = 1 ( 1 + cos2wt ) / 2

= cos2wt

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AC Circuits


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Series Connected Capacitances

V1(t) = (1/C

1) I(t)dt

V2(t) = (1/C2) I(t)dt

----------- = ----------------------

V (t) = [ (1/C1) + (1/C2) ]

I(t)dt= (1/Ctot) I(t)dt

Hence,

Series connected capacitances

+

I(t)

C1

C2

V1 (t)

V2 (t)

+

+

+ +

1Ctot = ----------------------

(1/C1) + (1/C2)

Series connected capacitancesare combined in the same way asfor shunt connected resistances

V(t)

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AC Circuits


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Series and Shunt Connected Capacitances

Where,C tot= C1 + C2

is the total capacitance

Shunt connected capacitances aresimply added

Shunt connected capacitances

+

I(t)

V(t)C2C1

I2I1

Vc(t)

I1(t) = C1 d V(t) / dtI2(t) = C2d V(t) / dt

----------- = ----------------------------

I (t) = (C1 + C2) d V(t) / dt= Ctot d V(t) / dt

+ +

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AC Circuits


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Inductance

Inductance is a winding or

coil of wire around a core

Definition

Coil Core Toroidal Coil Toroidal Core

Core may be either insulator

or a ferromagnetic material

+

_

Symbolic representation

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AC Circuits


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Ferrite Core Toroidal Inductor

Ferriet core inductor has a

toroidal ferrit core inside

Definition

II

Toroidal coil

Ferrite core

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AC Circuits


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Air Core Inductor

Configuration Air core inductor has no core inside

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AC Circuits


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Basic Relation

Voltage across an inductor isproportional to the rate of change ofcurrent

Definition

V(t) = L d I(t) / dt

where, V(t) is the voltage across theinductance,I(t) is the current flowingthrough,L is the inductance (Henry)

1 Henry is a value of inductance defined as

1 Henry = 1 Voltx1 second / 1 Amp

Voltage Source V(t)

I(t)

Inductance L

+

+

_

V(t)L

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AC Circuits


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Current in an Inductance

The equation;

Definition


can be written in inverse form as

I(t) = (1/L)

V(t)dt + I(0)

where I(0) is the current initially flowingin the inductor

Voltage Source V(t)

++

Inductance L

_

I(t)

V(t) L

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AC Circuits


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I (Amp)

V(t)

I(t)

Inductance+

Phase Shift between Current and Voltage Waveforms

I(t) = (1/L) V(t) dt

= (1/L)

Vmaxsin wt dt= - (Vmax/ wL) coswt

= - Imaxcoswt

+

-25

-20

-15-10

-5

0

5

10

15

20

Time (Sec)

25

0.005 0.010 0.015 0.020

V (Volts), I (Amp)

VmaxImax

Current in an Inductance

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AC Circuits


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Series and Shunt Connected Inductors

Series connected inductors are added

V1

(t) = L1

d I(t) / dt

V2(t) = L2d I(t) / dt

----------- --------------------------

V(t) = (L1

+ L2

) d I(t) / dt

= L tot d I(t) / dt

where

Ltot

= L1

+ L2

is the total inductance

+ + +

I(t)

V(t)

L2

L1

V2 (t)

V1 (t)+

+

Series connected inductances

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AC Circuits


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Series and Shunt Connected Inductors

I1(t) = (1/L1) V(t)dt

I2(t) = (1/L2) V(t)dt

-------- --------------------I(t) = [ (1/L1) + (1/L2) ] V(t)dt

= (1/Ltot) V(t)dt

+

Shunt connected inductances

+

I(t)

V(t)L

2

L1

I2I1

Vc(t)

1Ltot = ----------------------

(1/L1) + (1/L2)

Hence,

+

Shunt connected inductances arecombined in the same way as for shuntconnected resistances

+

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AC Circuits


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Example - 3

Problem

Calculate the voltage across the 10 mHinductor with the current shown in the figure

on the RHS

0,0 2,0 4,0 6,0

t (msec)8,0 10,0 12,0 14,0

0,0

0,5

1,0

I(t) = 0 t < 1 ms

I(t) = 1/((5-1)x10-3) (t 10-3) 1 t 5 msI(t) = 1 5 t 9 msI(t) = -1/((5-1)x10-3) (t 12 x10-3) 9 t 13 msV(t) = 0 t 13 ms

I(t) (Amp)

L = 10 mH

+

I(t)

I(t)

+

L

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AC Circuits


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Example - 3

Solution


V(t) (Volts)

0,0

0,0 2,0 4,0 6,0

t (msec)

8,0 10,0 12,0 14,0

-2,5

2,5

V(t) = 0 t < 1 msV(t) = 0.01x1/((5-1)x10-3) = 2.5 V 1 t 5 msV(t) = 0 5 t 9 msV(t) = 0.01x(-1)/((5-1) x10-3) = -2.5 V 9 t 13 ms

V(t) = 0 t 13 ms

Differentiating the current waveform and

multiplying by L (L = 10-2H);

L = 10 mH

+

I(t)

I(t)

+

L

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AC Circuits


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Example - 4

Problem

Assume that the inductor shown on the RHS

is connected to a voltage source of the form

shown in the figure

Find out the inductor current waveform

assuming that the initial current in the

inductor is zero

V(t) = 0 t < 0 s

V(t) = -10 V 0 t 1 s

V(t) = 0 t 1 s

t (sec)

0,0

0,0

-10

0,5 1,0

V(t)

+

I(t)

+

L L = 10 mH

V(t) (mV)

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AC Circuits


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Example - 4

Solution

I(t) = (1/L) V(t) dt + I(0)

I(t) = 0 t < 0 s

I(t) = 1/(10x10-3)x V(t) dt= 100 x V(t) dt

= 100 x (-10 x 10-3

) t = - t 0 t 1 sI(t) = -1 A t 1 s

t (sec)

I(t) (A)

0,0

0,0

-1

0,5 1,0

V(t)

+

I(t)

+

L L = 10 mH

I(t) = (1/L) V(t) dt + I(0)

= 1/(10 x 10-3)

V(t) dt

Integrating the voltage expression;

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AC Circuits


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Energy Stored in an Inductor

Problem

Calculate the instantaneous energy

stored in an inductor with an inductanceL and an instantaneous voltage VL(t)

P(t) = VL(t) I(t)

WL(t) = P(t) dt

=

VL(t) I(t) dt

= I(t) L dI(t) / dt dt

= L

I(t) dI(t)

WL(t) = L I 2

(t)

or

+

L

_

I(t)

I(t) VL(t)

METU

AC Circuits


55/98


Example 5

Problem

Find the instantaneous energy in theinductor for the current shown in the figure

I (t) = 0 t < 1 ms

I (t) = 1/(4x10-3) (t-10-3) Amp 1 t 5 msI (t) = 1 Amp 5 t 9 ms

I (t) = 1/(4x10-3) (t - 13x 10-3) Amp 9 t 13 msI (t) = 0 t 13 ms I(t) (Amp)

0,0

0,5

1,0

0,0 2,0 4,0 6,0

t (msec)

8,0 10,0 12,0 14,0

V(t)

+

I(t)

+

L

L = 10 mH

METU

AC Circuits


56/98


Example 5

Solution

By using the above formula

WL(t) = L I2(t)

W(t) = 0 Joule t < 1 ms

W(t) = 312.5 (t-10-3)2 Joules 1 t 5 ms

W(t) = 0.01 /2 x 12 = 0.005 Joules 5 t 9 msW(t) = 312.5 (t-13x10-3)2 Joules 9 t 13 msW(t) = 0 t 13 ms

W(t) (Joule)

0,001

0,002

0,003

0,004

0,005

0,006

0,0 2,0 4,0 6,0

t (msec)

8,0 10,0 12,0 14,0

V(t)

+

I(t)

+

L

L = 10 mH

METU

AC Circuits


57/98


R-L Circuits

Problem

Solve the R-L circuit shown on the RHS

which consists of a resistance in serieswith an inductance for current waveformwhen the switch is turned on at time:t = 0 sec

A first order ordinary differential equation

V(t) = R I(t) + L dI(t) / dt

or

dI(t) / dt + (R/L) I(t) = (1/L) V(t)

Writing down KVL for the circuit

SolutionL

+

_

+

I(t)

V(t)

R

I(0)=I0

Switch


METU

AC Circuits


58/98




Solution

dI(t) / dt + (R/L) I(t) = (1/L) V(t)

dI(t) / dt + (R/L) I(t) = (1/L) V sin ( wt + )

V( t ) = V sin ( wt + )

^

100

0

Voltage (Volt)

Angle (Radians)

/2 3/2 2

200

300

312

Amplitude

Phase angle

^

L

+

_

+

I(t)

V(t)

R

I(0)=I0

METU

AC Circuits


59/98



Solution

(t) dI(t)/dt+ I(t) (1/RC)(t) = (t) ( V / L ) sin ( wt + )

(t) dI(t)/dt+ I(t) d/dt (t) = ( V / L ) (t) sin ( wt + )

d/dt [(t) I(t)] = ( V / L ) (t) sin ( wt + )

d/dt [(t) I(t)] dt = ( V / L ) (t) sin ( wt + ) dt + I(0)

(t) I(t) = ( V / L ) (t) sin ( wt + ) dt + I(0)

I(t) = I (t)-1

(t) sin (wt + ) dt + (t)-1

I(0)

^

^

^

^

^

^

Define an integration factor(t) = e t R/L

Multiply both sides of the above ODE by this factor;

^

100

0

Voltage (Volt)

Angle (Radians)

/2 3/2 2

200

300

312

Amplitude

Phase angle

dI(t) / dt + (R/L) I(t) = (V/L) sin ( wt + )^


L

+

_

+

I(t)

V(t)

R

I(0)=I0

V( t ) = V sin ( wt + )

^

METU

AC Circuits


60/98


Solution Continued)

Subsituting the integration factor(t) = e t R/L into

the above solution;

I(t) = I e t R/L

e t R/L sin (wt+ ) dt + e t R/L I(0)^


Let = 0 (for simplicity)(R/L) sinwt - w coswt

e t R/L sin wt dt = e t R/L ----------------------------------

(R/L)2+ w 2

Taken from the book : Calculus and Analytic Geometry,

Thomas,Addison Wesley, Third Ed. 1965, pp. 369


Switch

L

+

_

+

I(t)

V(t)

R

I(0)=I0

V( t ) = V sin ( wt + )^

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AC Circuits


61/98



Solution Continued)

Subsituting the above term into the solution;


(R/L) sinwt - w coswtI(t) = I e t R/L e t R/L ---------------------------------- + e t R/L I(0)

(R/L)2+ w 2

^

(R/L) sinwt - w coswtI(t) = I ----------------------------------- + e t R/L I(0)

(R/L)2+ w 2

^

I

I(t) = -------------------(sinwt - (wL/R) coswt) + e t R/L I(0)(wL/R)2+ 1

^


Switch

L

+

_

+

I(t)

V(t)

R

I(0)=I0

V( t ) = V sin ( wt + )^

METU

AC Circuits


62/98


Numerical Example

Now assume that the parameters of the circuit on

the RHS are as follows;

V(t) = 312 sin wt Volts

R = 1 Ohms

L = 10 milli Henry

Switch


I

I(t) = -------------------(sinwt - (wL/R) coswt) + e t R/L

I(0)(wL/R)2+ 1

^



L

+

_

+

I(t)

V(t)

R

I(0)=I0

V( t ) = V sin ( wt + )^

METU

AC Circuits


63/98


Numerical Example

^



I

I(t) = -------------------(sinwt - (wL/R) coswt) + e t R/L I(0)(wL/R)2+ 1

^

-250

-200

-150

-100

-50

0

50

100

150

200

250

0.5 1 1.5 2 2.5 3 3.5 4

5

10

15

20

25

0 0.5 1 1.5 2 2.5 3 3.5 4

-300

-200

-100

0

100

200

300

400

0.5 1 1.5 2 2.5 3 3.5 4

METU

AC Circuits


64/98


Solution for DC Voltage - Two Simple Rules

Rule - 1

An inductor with no initial current acts asan open circuit to a DC voltage sourceinitially

Then, the initial value of current will be;

I(0) = 0


I(t) = I() + [I(0) - I()] e-t/

I(0) = 0

= L / R = Time Constant: The time required foran inductor to reach 63 % of full current

= 1 x 2 mH = 1 x 0.002 = 0.002 Sec

L=0.002 H

OC

+ I(0) = 0

R

VDC

R = 1

+

_

+

I(t)

VDC0

I(0) = (1 / L)

V(t) dt = I0= 0 (OC)-

METU

AC Circuits


65/98


Rule - 2

An inductor acts as a short circuit to a

DC current (or voltage) finally

Then, the final value of current will be;

I() = V / R

I() = V / R

R

+

_

+

I()VDC

+

_

+ I(t)

VDC

SC

V(t) = L d/dt I(t) = L d/dt (ct) = 0 (SC)

Solution for DC Voltage - Two Simple Rules


I(t) = I() + [I(0) - I()] e-t/

L=0.002 H

R = 1

= L / R = Time Constant: The time required foran inductor to reach 63 % of full current

= 1 x 2 mH = 1 x 0.002 = 0.002 Sec

METU

AC Circuits

D fi iti f th Ti C t t


66/98


Definition of the Time Constant:


I(t) = I() + [I(0) - I()] e -t/

0

4.0

8.0

12.0

16.0

20.0

24.0

0 0,002 0,004 0,006 0,008 0,01

I(t)(Amps

)

t (sec)

% 63 of peak (24 V)

L = 2mH

+

_

+I(t)

V= 24 Volts

R = 1 Ohms

= L / R = Time Constant: The time required foran inductor to reach 63 % of full current= 1 x 2 mH = 1x 0.002 H = 0.002 Sec

METU

AC Circuits


67/98


Example - 6

Problem

Find the current waveform in the 2 mHinductor shown on the RHS when it has aninitial current of 6 Amps and connected to a24 Volt DC voltage source through a wire with1 Ohm resistance


I(t) = I() + [I(0) - I()] e t /

I(t) = 24 + (6 - 24) e-t / 0.002= 24 - 18 e-t / 0.002 Amps

Inductor has 6 Amp initial current;

I(0) = I0= 6 Amp

Inductor will be SC at the end, hence;

I() = V / R = 24 / 1 = 24 Amps

0

4.0

8.0

12.0

16.0

20.0

24.0

0 0,002 0,004 0,006 0,008 0,01

I(t)(Amps)

t (sec)

I()

I(0)

L = 2mH

+

_

+I(t)

V= 24 Volts

R = 1 Ohms

METU

AC Circuits


68/98


Example - 7

Problem

Solve the circuit shown on the RHS for

current waveform flowing in the inductor

+

_

L = 0.4 H+

I(t)V= 24 V

R1 = 10 Ohms

A

R2 = 5 Ohms

B

First take out the branch containing inductor,

and derive the Thevenin Equivalent of the LHScircuit seen from the terminals A and B

Solution

Kill the voltage source, and find Req.

R1 = 10 Ohms

+

V= 24 V R2 = 5 Ohms

AA

L = 0.4 H

B B

A

B

L = 0.4 H

R1 = 10 OhmsReq = 10 // 5

= 10 x 5 /(10 + 5)

= 10 / 3 Ohms

A

R2 = 5 Ohms

B

METU

AC Circuits


69/98


Example 7 Continued)

Solution Continued)

Open circuit terminals A B and find VAB

Form the resulting Thevenin equivalentCircuit,

Connect the inductance to the resultingThevenin equivalent circuit,

Solve the resulting circuit by using thestraightforward method described inExample 6

R1 = 10 Ohms

+

V= 24 V

L = 0.4 H+

V= 24 / 3 = 8 V

R eq= 10 // 5 = 10 x 5 /(10+5)

= 10/3 Ohms

A

B

R2 = 5 Ohms

A

B

A

B

A

B

VA-B (t)

V = 24 VxR2/ ( R1 + R2)

= 24 x 5 / 15 = 24 / 3

= 8 V

L = 0.4 H

+

V= 24 / 3 = 8 V

R eq= 10 // 5 = 10 x 5 /(10+5)

= 10/3 Ohms

A

B

METU

AC Circuits


70/98


R-L-C Circuits

Problem

Solve the following circuit which consists of a

resistance, an inductance and a capacitanceconnected in series for current waveform

V(t) = R I(t) + VL(t) + VC(t)

= R I(t) + L dI(t)/dt + (1/C)

I(t)dt + V(0)

Differentiating both sides wrt time once;

dV(t)/dt = R dI(t)/dt + Ld2I(t)/dt2+(1/C)I(t)

or

d2I(t)/dt2 + (R/L)dI(t)/dt + (1/LC) I(t) = (1/L) dV(t)/dt

A second order ordinary differential equation

Writing down KVL for the circuit;

Solution

V(t)

C+

_

+ I(t)

R

V(0)=V0

I(0)=I0

+

_

+

R L C

L

V(t)

METU

AC Circuits

Initial Conditions


71/98


Differential Equation

VC(0) = VC0

IL(0) = IL0


Initial Conditions

+

_

+

R L C

VC(0) = V(0) VL(0) VR(0)

= V(0) L d/dt IL(0) R IL(0)

The first initial condition may now be written as,

or

d/dt IL(0)= IL(0) =(1/L) [V(0) VC(0) R I L(0) ]

1

2Two initial conditions are needed for solution

V(t)C

+

_

+ V(0)=V0

I(t)

R I(0) = I0L

METU

AC Circuits

Example


72/98


R-L-C Circuit


Solve the R-L-C circuit with the given parameters

shown on the RHS for current I(t)

d2I(t)/dt2 + (2/1) dI(t)/dt + 401 I(t) = (1/1) dV(t)/dt

d2I(t)/dt2+ 2 dI(t)/dt + 401 I(t) = d/dt (10 e-4t) = - 400 e-4t

+

C = 2.494 mF

+I(t)

R = 2 L = 1 H

V(t) = 100 e- 4 t

_

1 / (LC) = 401

Writing down KVL for the circuit shown on theRHS the following ODE is obtained

METU

AC Circuits

Solution


73/98


+

_

+

R = 2

I(t)

L = 1 H

d2I(t)/dt2 + 2 dI(t)/dt + 401 I(t) = 0

First, obtain the homogeneous equation

by setting the RHS function to zero

Then, solve the characteristic equation

s2 + 2 s + 401 = 0

s1, s2= ( - b b2-4xaxc ) / (2 a)

= -1 j 20

Eigenvalues of the differential equation

V(t) = 100 e- 4 t

R-L-C Circuit

a =1 b = 2 c = 401

C = 2.494 mF

METU

AC Circuits

Solution


74/98


V(t) = 100 e- 4 t

+

_

+

R = 2

I(t)

L = 1 H

Then, the solution becomes

e j= cos+ j sin = e - t[ k1 (cos 20 tj sin 20 t )= + k2(cos 20 t + j sin 20 t ) ]

I(t) = k1 es1 t + k2e

s2 t

= k1 e(-1 j20 ) t + k2e

(-1 + j20 ) t

= k1 e - t x e j20 t + k2 e - t x e j20 t

= e - t(k1 e j20 t + k2 e

j20 t) Eulers Identity

R-L-C Circuit

= 20 t

C = 2.494 mF

METU

AC Circuits

Eulers Identity


75/98


Graphical Representaion

e j= cos+ j sin

Definition

^

1.0

cos

sin

e j = cos+ j sin = cos2+ sin2= 1

x

y

z

x2+ y2= z2

z = x2 + y2


METU

AC Circuits

Solution


76/98


Rearranging the solution terms

= e - t[ k1 (cos 20tj sin 20t )

= + k2(cos 20t + j sin 20t ) ]

= e

- t

[ (k1 + k2) cos 20t + (k2 k1) sin 20t ]

A B

R-L-C Circuit

I(t) = e - t( A cos 20t + B sin 20t)

Hence, the sinusoidal solution (decayingsinusoidal term) becomes;

V(t) = 100 e- 4 t

+

_

+

R = 2

I(t)

L = 1 H

Unknown coefficients to determined

C = 2.494 mF

METU

AC Circuits

Solution


77/98


Nonhomogeneous Term

Transient Term)

Now the nonhomogeneous (transient) term in thesolution is to be determined V(t) = 100 e- 4 t

+

_

R = 2

+

I(t)

L = 1 H

Definition of the Nonhomogeneous Term

General form of the nonhomogeneous (transient)term in the solution may be expressed as

In(t) = c e 4 t

where, In(t) is the nonhomogeneous term in the solution,c is an unknown coefficient to be determined

C = 2.494 mF

METU

AC Circuits

Solution


78/98


Substitute the nonhomogeneous solution;

In(t) = c e 4 t

to the given differential equation;

and solve it for the unknown coefficient c

d2I(t)/dt2 + 2 dI(t)/dt + 401 I(t) = - 400 e-4t

Nonhomogeneous Term

Transient Term)

d2(ce 4 t)/dt2 + 2 d(ce 4 t)/dt + 401 ce 4 t = -400 e-4t

16c e 4 t+ 2c(-4 e 4 t ) + 401 c e 4 t = - 40 e-4 t

V(t) = 100 e- 4 t

+

_

+

R = 2

I(t)

L = 1 H

C = 2.494 mF

These terms cancel

METU

AC Circuits

Solution


79/98


Transient Term

16 c - 8 c + 401 c = - 400

409 c = - 400 or c = - 400 / 409 = - 0.97799

Transient term then becomes;

In

(t) = - 0.97799 e 4 t

d2(ce 4 t)/dt2 + 2 d(ce 4 t)/dt + 401 ce 4 t = -400 e-4t

16c e 4 t+ 2c(-4 e 4 t ) + 401 c e 4 t = - 40 e-4 t

These terms cancel

V(t) = 100 e- 4 t

+

_

+I(t)

R = 2 L = 1 H

C = 2.494 mF


80/98

METU

AC Circuits

Solution


81/98


V(t) = 100 e- 4 t

+

_

+

R = 2

I(t)

L = 1 H

I(t) = - 0.97799 e 4 t

+ e- t

( A cos 2t + B sin 2t)

Transient Term

Determination of the Unknown

Coefficients

The above solution must satisfy the giveninitial conditions;

Decaying sinosoidal Term

C = 2.494 mF

IL (0) = IL0= 32.5 A

VC(0) = VC0 = 22 Volts

METU

AC Circuits

Solution


82/98



Coefficients

IL (0) = IL0= 32.5 A


Substitute the given initial conditions intothe complete solution equation and solve forthe unknown coefficients A and B;

V(t) = 100 e- 4 t

+

_

+

R = 2

I(t)

L = 1 HI(t) = - 0.97799 e 4 t

+ e- t


Transient Term Decaying sinosoidal TermC = 2.494 mF

METU

AC Circuits

Solution


83/98



Coefficients

Substitute the given initial conditions into thecomplete solution equation and solve for theunknown coefficients A and B;

A = 0.97799

IL(0) = - 0.97799 e0 + e 0( A cos 0 + B sin 0)

= - 0.97799 + A = 0

IL(t) = - 0.97799 e 4 t + e - t( A cos 20t + B sin 20t)

V(t) = 100 e- 4 t

+

_

+

I L(t)

R = 2 L = 1H

C = 2.494 mF

IL (0) = IL0= 32.5 A


METU

AC Circuits

Solution


84/98


V(0) = 100 e- 4 t

= 100 e- 4 x 0

= 1000 0


Coefficients

IL (0) = IL0= 32.5 A


IL(t) = - 0.97799 e 4 t

+ e- t


d/dt IL(0) = IL(0) = (1/L) [ V(0) VC(0) R IL(0) ]

= (1/1) (100 22 2 * 32.5)= 13 Amp/sec

V(t) = 100 e- 4 t

+

_

+

I L(t)

R = 2 L = 1 H

C = 2.494 mF

METU

AC Circuits

Solution


85/98



Coefficients

d/dt IL(t) = 0.97799 x 4 e 4 t

- e- t

(A cos 20t +B sin 20t)+ e - t( -20 A sin 20 t + 20 B cos 20t )

d/dt IL(0) = 0.97799 x 4 e0- e 0(A cos 0 + B sin 0)

+ e 0(-20 A sin 0 + 20 B cos 0 ) = 13

= 0.97799 x 4 A + 20B = 13

= 0.97799 x 4 0.97799 + 20B = 13

B = (13 - 3 x 0.9799) / 20 = 0.5033

V(t) = 100 e- 4 t

+

_

+

I L(t)

R = 2 L = 1 H

C = 2.494 mF

METU

AC Circuits

Solution Terms


86/98


IL(t) = - 0.9799 e 4 t + e - t( 0.97799 cos 20t + 0.5033 sin 20t)

General form of the Solution

Decaying sinosoidal TermTransient Term

+

_

V(t) = 100 e- 4 t

+

I(t)

R = 2 L = 1 H

C = 2.494 mF

METU

AC Circuits

Solution Terms


87/98


Transient Term

Transient Term

+

_

V(t) = 100 e- 4 t

+

I(t)

R = 2 L = 1 H

C = 2.494 mF

IL(t) = - 0.9799 e 4 t + e - t( 0.97799 cos 20t + 0.5033 sin 20t)

-1,20

-1,00

-0,80

-0,60

-0,40

-0,20

0,000,0 0,2 0,4 0,6 0,8 1,0 1,2 1,4 1,6

METU

AC Circuits

Solution Terms


88/98


Sinosoidal Terms

Sinusoidal Terms

0.9779 cos 20t + 0.5033sin 20t

0.9779 cos 20t

0.5033 sin 20t

+

_

V(t) = 100 e- 4 t

+

I(t)

R = 2 L = 1 H

C = 2.494 mF

IL(t) = - 0.9799 e 4 t + e - t( 0.97799 cos 20t + 0.5033 sin 20t )

-1,20

-0,80

-0,40

0,00

0,40

0,80

1,20

0 0,1 0,2 0,3 0,4 0,5 0,6 0,7 0,8 0,9 1

METU

AC Circuits

Solution Terms


89/98


Exponentially Decaying Sinusoidal Term

V(t) = 100 e- 4 t

+

_

+

I(t)

R = 2 L = 1 H

C = 2.494 mF


-1,20

-0,80

-0,40

0

0,40

0,80

1,20

0,5 1 1,5 2 2,5 3 3,5 4

e - t

METU

AC Circuits

Solution Terms


90/98


Overall Solution

V(t) = 100 e- 4 t

+

_

+

I(t)

R = 2 L = 1 H

C = 2.494 mF


-1,60

-1,20

-0,80

-0,40

0,0

0,40

0,80

0,5 1,0 1,5 2,0 2,5 3,0 3,5 4,0

METU

AC Circuits

The Concept of RMS Value


91/98


Definition: RMS Root Mean Square)

Justification is based on the principle ofequating the currents calculated in both cases.

First, calculate the power dissipated (lost asheat) in the resistance R in the DC circuitshown on the RHS

VR = R xIDCP = VRxIDC

= R xIDC2

VDC

_

+

IDC

VR

RMS value of an AC current is the value of DCcurrent that would dissipate the same amountof power as the AC current on a resistance R

Justification

+

R

METU

AC Circuits


I 2(t) (I i t )2I(t) I i t


92/98


Now, calculate the power dissipated (lostas heat) in the resistance R in the AC

circuit shown on the RHS

V (t) = I(t) xR

P (t) = V(t) xI(t)

= R I(t)2

= R (Vmax/ R sin wt )2

= R (I maxsin wt )2

R

V(t) = Vmaxsin wt

_

+

I(t)

V(t)

+

V(t)

I(t) = (Vmax/ R) sin wt

Imax

-2,0

-1,5

-1,0

-0,5

0

0,5

1,0

1,5

2,0

2,5

Angle (Radians)

/2 3/2 20

I 2(t) = (Imaxsin wt )2I(t) = Imaxsin wtMS Root Mean Square)

METU

AC Circuits



93/98


Problem

Let us now calculate the average of P(t)

Pavg = (1/T) P(t) dt

= (1/T)

R I(t)2 dt

= R (1/T)

I(t)2dt

= R Irms2

where I(t)rms = ((1/T) I(t)2dt)1/2

I(t)rms

=

( 1/T ) I(t)2dt

0

0,5

1,0

1,5

2,0

2,5

Angle (Radians)

/2 3/2 20

Irms = ( 1/T ) I(t)2dtI2(t) = ( Imaxsin wt )2

METU

AC Circuits

Definition of RMS Value

I2(t) ( I i t )2


94/98


Problem

RMS value of an AC current is the valueof DC current that would dissipate the

same amount of power as AC current ona resistance R

_

+

IDC

R

+

VR

_

+

I(t)

V(t) +

VR(t)

Pavg= R IDC2 Pavg= R I(t)rms

2=

VDC

R

I2(t) = ( Imaxsin wt )2

0

0,5

1,0

1,5

2,0

2,5

Angle (Radians)

/2 3/2 20

Irms = ( 1/T ) I(t)2dt

METU

AC Circuits

RMS Value of a Sinusoidal Waveform


95/98


RMS value of a sinusoidal waveform maybe expressed as 0.7071 multiple of thepeak value of the waveform as follows;

I (t) = I maxsin wt

Irms = ((1/T) I(t)2dt)1/2

= ((1/T) (I maxsin wt)2dt)1/2

= ((1/T) (Imax

2 sin2w td t )1/2

= ((1/T) Imax2 (1 cos2wt) dt )1/2

= ((1/T) Imax2 ( d t cos2wt dt ))1/2

= ((1/T) Imax2( T / 2 x0 ))1/2

= ((1/T) T Imax2/ 2 )1/2

= ( Imax2/ 2 )1/2

= Imax/ 2 = I maxx0.7071-2,0

-1,5

-1,0

-0,5

0

0,5

1,0

1,5

2,0

Current (Amp)

Angle (Radians)

/2 3

/22

sin2wt = 1 cos2wt

= 1 (1 + cos 2wt) / 2

= 1 cos 2wt

= cos2wt

METU

AC Circuits

RMS Value of a Sinusoidal Waveform


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Calculate the RMS value of thesinusoidal voltage waveform shown onthe RHS

Vrms = Vmax/ 21/2 = V maxx0.7071

= 312 x 0,7071 = 220 Volts

Problem

Domestic rms voltage level in Turkey

312

- 312

0

V

oltage(Volts)

Angle (Radians)

/2 3/2 2

Vrms = 220 Volts

220

METU

AC Circuits

Example - 8

V = ( 1/T ) V(t)2 dt


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Problem

Calculate the RMS value of a the voltagewaveform shown on the RHS

Vrms = ( 1/T ) V(t)2dt

-4

-2

0

2

4

0,1 0,2 0,3 0,4 0,5 0,6

Time (msec)Voltage(Volts)

Voltage2(Volts)

0

8

16

24

32

0,1 0,2 0,3 0,4 0,5 0,6

Time (msec)

V2(t) = 16

_

+

I(t)

V(t) +

VR(t)

Vrms = ( 1/T ) V(t)2dt

Vrms = ( 1/0.3) ( 42dt + (-4)2dt)0

0.1

0.1

0.3

= 4 Volts

METU

AC Circuits

Any Questions Please ?


98/98


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