3. sinusoidal steady state circuit analysis (3.1 study the ac basic circuits)

1

Define, analyze and calculate the response of RLC (resistance R and reactance X).

Define and analyze the impedance and impedance diagram for RLC elements.

ELECTRICAL CIRCUITS ET 201

2

14.2 The Derivative

• To understand the response of the basic R, L and C elements to a sinusoidal signal, you need to examine the concept of the derivative.

• The derivative dx/dt is defined as the rate of change of x with respect to time.

• If x fails to change at a particular instant, dx = 0, and the derivative is zero.

• The derivative dx/dt is actually the slope of the graph at any instant of time.

3

14.2 The Derivative• The derivative dx/dt is zero only at the positive and

negative peaks (ωt = π/2 and 3π/2) since x fails to change at these instants of time.

• For the sinusoidal waveform, the greatest change in x will occur at the instants ωt = 0, π, and 2π. The derivative is therefore a maximum at these points.

4

14.3 Response of Basic R, L and C Elements to a Sinusoidal Voltage or Current

Resistor• For power-line frequencies and frequencies up to a few

hundred kilohertz, resistance is, for all practical purposes, unaffected by the frequency of the applied sinusoidal voltage or current.

• For this frequency region, the resistor R can be treated as a constant.

Riv

ResistorGiven

Where;

For a given

Where; 5


tItRV

Rvi m

m sinsin

tVv m sin

RVI m

m

tVtRIRtIiRv mmm sinsinsin

;sin tIi m

RIV mm

6


Resistor

• For a purely resistive element, the voltage across and the current through the element are in phase, with their peak values related by Ohm’s law.

tVv mR sin

tIi mR sin

7


Inductor• The inductive voltage is directly related to the frequency f

(or more specifically, the angular velocity of the sinusoidal ac current through the coil) and the inductance of the coil L.

• The voltage across an inductor is directly related to the rate of change of current through the coil.

dtdiLv L

L

8


tLItIdtdL

dtdiLv mmL

L cossin

tIi mL sin

Or; 90sincos tVtLIv mmL

Where; mm LIV

Hence; Lm

m XLIV

InductorGiven

XL : Inductive reactance

Inductor

• For an inductor, vL leads iL by 90°, or iL lags vL by 90°.

9


tIi mL sin

90sin tVv mL

10


Capacitor• For a particular capacitance, the greater the rate of

change of voltage across the capacitor, the greater the capacitive current.

• The capacitive current is directly related to the rate of change of the voltage across the capacitor.

dtdvCi C

C

11


tCVtVdtdC

dtdvCi mm

CC cossin

tVv mC sin

90sincos tItCVi mmC

mm CVI

1C

m

m XCI

V

CapacitorGiven

Or;

Where;

Hence; XC : Capacitive reactance

12


Capacitor

• For a capacitor, iC leads vC by 90°, or vC lags iC by 90°.

tVv mC sin 90sin tIi mC

13


• If the source current leads the applied voltage, the network is predominantly capacitive.

• If the applied voltage leads the source current, the network is predominantly inductive.

14


Resistor: Inductor: Capacitor:vR and iR in phase vL leads iL by 90° iC leads vC by 90°

Relationship between differential, integral operation in phasor listed as follow:


dtvdtdv

j

j1

16

Summary of voltage-current relationshipElement Time domain Frequency domain

R

L

C

Riv RIV

;dtdiLv LIjV

;dtdvCi Cj

IV


dtiC

v 1

dtvL

i 1

17


Example 14.1(a)

The voltage across a 10 resistor is given by the expression;

Find the expression for the current i through the resistor and sketch the curves for v and i.

tv 377sin100

18


Example 14.1(a) – solution

A 1010100

RVI m

m

Hence;

A 377sin 10sin ttIi mm

ttVv m 377sin100sin

V 100mV

and f 2rad/s 377

19


Example 14.1(a) – solution (cont’d)

The alternative waveform;

t (ms)16.78.35

0

100 V

10 A

v , i

v

i

In phase

20


Example 14.1(b)The voltage across a 10 resistor is given by the expression;

Find the expression for the current i through the resistor and sketch the curves for v and i.

60377sin25 tv

21


Example 14.1(b) – solution

A 5.21025

RVI m

m

Hence;

A 60377sin5.2sin ttIi mm

60377sin25sin ttVv m

V; 25mV

and

rad/s 37760

22


Example 14.1(b) – solution (cont’d)

The alternative waveform;

t (ms)16.67

2.780

25 V

2.5 A

v , i

vi In phase

12.58.334.17-4.17

23


Example 14.3(a)

The current a 0.1 H coil is given by the expression;

Find the expression for the voltage v across the coil and sketch the curves for v and i.

ti 377sin10

24


Example 14.3(a) – solution

vL leads iL by 90°,

Hence, if the current is;

the voltage will be;

where; and;

tIi mL sin

90sin tVv mL

Lmm XIV fLLX L 2

25


Example 14.3(a) – solution (cont’d)From the expression; ti 377sin10

A; 10mI rad/s; 377

ms; 67.616011

f

T

Hence;

7.371.0377LX L

Hz; 602377

f

Hence the expression for the voltage is;

V 3777.3710 Lmm XIV


26


Example 14.3(a) – solution (cont’d)

The waveforms;

t (ms)16.67

8.33 12.54.17-4.17

ti 377sin10 90377sin377 tv

27


Example 14.3(b)

The current a 0.1 H coil is given by the expression;

Find the expression for the voltage v across the coil and sketch the curves for v and i.

70377sin7 ti

vL leads iL by 90°,



where; and;

28


Example 14.3(b) – solution

tIi mL sin

90sin tVv mL

Lmm XIV fLLX L 2

29


Example 14.3(a) – solution (cont’d)From the expression; 70377sin7 ti

A; 7mI rad/s; 377

ms; 67.616011

f

T

Hence;

7.371.0377LX L

Hz; 602377

f


V 9.2637.377 Lmm XIV

20377sin9.26390sin ttVv m

70

30


Example 14.3(b) – solution (cont’d)

The waveforms;

t (ms)

16.67

8.33 12.54.170.933.24

v leads i by 90 4.17 ms

70377sin7 ti 20377sin9.263 tv

31


Example 14.5The voltage across a 1 F capacitor is given by the expression;

Find the expression for the current i through the capacitor and sketch the curves for v and i.

V 400sin30 tv

iC leads vC by 90°,

Hence, if the voltage is;

the current will be;

where; and;

32


Example 14.5 – solution

tVv mC sin

90sin tIi mC

C

mm X

VI fCC

XC 211

33


Example 14.5 – solution (cont’d)From the expression; tv 400sin30

V; 30mV rad/s; 400 Hz; 63.72400

f

ms; 7.517.63

11

fT

2500

10140011

6CXC

Hence the expression for the current is;

mA 12250030

C

mm X

VI

mA 90400sin 1290sin ttIi m

Hence;

34


Example 14.5 – solution (cont’d)

The waveforms;

t (ms)

15.7

7.85

11.783.933.93

i leads v by 3.93 ms 90

tv 400sin30 mA 90400sin 12 ti

35


Example 14.6The current through a 100 F capacitor is given by the expression;

Find the expression for the voltage v across the capacitor.

V 60500sin40 ti




where; and;

36



tIi mC sin

90sin tVv mC

Cmm XIV fCC

XC 211

37


Example 14.6 – solution (cont’d)From the expression; 60500sin40 ti

A; 40mI rad/s; 500 Hz; .6972500

f

ms; 57.216.79

11

fT

2010100500

116C

XC

60


Hence; V 8002040 Cmm XIV


38


Example 14.7(a)Determine the type of element in the box (C, L or R) and calculate its value if;

and V 40sin100 tv A 40sin20 ti

Solution

The voltage and current are in phase .The element is a resistor (R).

520

100

m

m

IVR

39


Example 14.7(b)Determine the type of element in the box (C, L or R) and calculate its value if;


Solution

The voltage leads the current by 90The element is an inductor (L).

; 2005

1000

m

mL IVX H 53.0

377200

LXL

40


Example 14.7(c)Determine the type of element in the box (C, L or R) and calculate its value if;


Solution

The voltage lags the current by 90 The element is a capacitor (C).

; 5001

500

m

mC IVX F 74.12

50015711

CXC

41


Example 14.7(d)Determine the type of element in the box (C, L or R) and calculate its value if;

and V 20cos50 tv A 110sin5 ti

Solution V 110sin50 9020sin50 20cos50 tttv

The voltage and current are in phase .The element is a resistor (R).

105

50

m

m

IVR

42

Impedance Z• It is a ratio of the phasor voltage V to the phasor

current I.• Unit in ohms (Ω).

IVZ [Ω]

15.2 Impedance and the Phasor diagram

Impedance ZImpedance Z has two components:• Real component (ZRe) : Resistance, R

• Imaginary component (ZIm) : Reactance, X

• Reactance can be inductor, L and capacitance, C.• Positive X is for L and negative X is for C.

4343

jXRZ [Ω]


444444

Admittance Y• It is the reciprocal of impedance Z.• Unit in siemens (S).

jXR

VI

ZY

1

1

[S]



Impedance for Resistor, Rv and i are in phase;

Or;

In phasor form;

Where;

RVI m

m

RIV mm

tVv m sin 0VV

2mVV

Impedance for Resistor, R Applying Ohm’s law and phasor algebra;

Since v and i are in phase;

Hence;

Therefore;

RR RV

R

0R

V

ZVI

RRV 0

00 R

0R


0 R RZ0

RVI


• The boldface Roman quantity ZR, having both magnitude and an associate angle, is referred to as the impedance of a resistive element.

• ZR is not a phasor since it does not vary with time.

• Even though the format R0° is very similar to the phasor notation for sinusoidal current and voltage, R and its associated angle of 0° are fixed, non-varying quantities.

0 R RZ

Impedance for Resistor, R

Example 15.1

Find i in the figure.

Solution

tv sin100 V 07.70 V

A 014.4105

07.70

RZVI

Applying Ohm’s law;


A 014.41 I

Inverse-transform;


A sin20 ti

V = 70.7 V

I = 14.14 A


vR and iR in phase

Example 15.2

Find v in the figure

Solution

A 30sin4 ti A 3083.2 I



0RIR IZV 023083.2

V 3066.5


Inverse-transform;

V 3066.5 V V 30sin8 tv

V = 5.66 V

I = 2.83 A

30


vR and iR in phase

Impedance for Inductor, LvL leads iL by 90°,

By phasor transformation;

Where;

2 fLLX L

0VV

2mVV


tVv m sin

Impedance for Inductor, LApplying Ohm’s law and phasor algebra;

Since v leads i by 90;

Hence;

Therefore;

LL

LLLL

XV

XV

X

0

0L

VZVI


900 L90L

9090

LX LL

Z 90

LXVI

Example 15.3


Solution

V sin24 tv V 097.16 V


903097.16

90

LL XV

ZVI


A 9066.5


V = 16.97 V

I = 5.66 A


Inverse-transform;

A 9066.5 I A 90sin8 ti

vL leads iL by 90°

Example 15.4

Find v in the figure.

Solution

A 30sin5 ti A 3054.3 I


9043054.390 LL XI IZV


A 12014.14


V = 14.14 V

I = 3.54 A

30120


Inverse-transform;

V 12014.14 V A 120sin20 tv

vL leads iL by 90°

Impedance for Capacitor, C

By phasor transformation;

Where;

2

11fCC

XC



0VV

2mVV

tVv m sin

Impedance for Capacitor, CApplying Ohm’s law and phasor algebra;

Since i leads v by 90;

Hence;

Therefore;

CC

CCCC

XV

XV

X

0

0C

VZVI


900 C90C

90190

C

XCC

Z

90CXVI

Example 15.5


Solution

V sin15 tv V 061.10 V


By Ohm’s law;

90

CC XV

ZVI

A 9031.5902

061.10

Example 15.5 – Solution (cont’d)

A 9031.5 I A 90sin5.7 ti

Inverse-transform;

V = 10.6 V

I = 5.3 A


iC leads vC by 90°

Example 15.6

Find v in the figure.


Solution

A 60sin6 ti A 60244 .I

By Ohm’s law;

90CC XI IZV 905.06024.4

V 15012.2


V 150sin3 tv V 150122 .V

Inverse- transform;


iC leads vC by 90°

64

Impedances and admittances of passive elements

Element Impedance, Z Admittance, Y

R

L

C

R

Lj

15.2 Impedance and the Phasor Diagram

0 R

90 CX

90- LX

Lj1R1

0 1R

90- 1

LX

90 1

CXCj

Cj1

CXLX CL

1 ;

15.2 Impedance and the Phasor diagramImpedance Diagram• An angle is associated with

resistance, inductive reactance and capacitive reactance, each can be placed on a complex plane diagram.

• Combining different types of elements give total impedance from -90o to 90o.

Inductive reactance

Capacitive reactance

resistance

T angle = 0o Resistive in nature

T closer to 90o Inductive in nature

T closer to -90o Capacitive in nature


Impedance Diagram• For any configuration

(series, parallel, series-parallel, etc.), the angle associated with the total impedance is the angle by which the applied voltage leads the source current.

• For inductive networks, T will be positive.

• For capacitive networks, T will be negative.

3. sinusoidal steady state circuit analysis (3.1 study the ac basic circuits)

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