AC ANALYSISTime domain analysis

We are surrounded by sinusoidal signals:1. Household power supply (50

Hz)2. FM Radio (MHz)3. Cellular Phones (UHF:400 to

1000 MHz)4. Television

Advantages of AC:• Less loss in transmission• Carrier waves are modulated by

voice or video signals of lower frequency

Phasors:We will use the complex

exponential to represent currents and voltages.

This is known as the phasor representation.

Reason: The voltage across an inductor (and the current through a capacitor), involves derivatives.

If we use sinusoidal functions to describe – we have mixture of sine and cosine terms: not easy to handle

For a resistor, the current and voltage (across it) are in phase.What about an inductor?Let the current through the inductor be given by: tjIeti )(The voltage across the inductor is given by:

)()( tLijIeLjdtdiLtv tj

)2/()( tjLIetvTherefore the voltage across the inductor leads the current by /2.

For a capacitor, if the voltage across the capacitor is given by:

tjVetv )(

)()( tCvjVeCjdtdvCti tj

)2/()( tjCVetiTherefore the current in a capacitor leads the voltage by /2These relations can be represented by phasor diagrams.

Imaginary axis

Real axis

I

jLI

t

Phasor diagram for inductor

Real axis

Imaginary axis

I

V=I/jC

t

Phasor diagram for capacitor

For the resistor, inductor and capacitor, we get a phasor equation of the form:

V = Z IZ is the impedance of the element.

For a resistor: ZR = RFor a capacitor: ZC = 1/jCFor an inductor: ZL = jL

Note that the impedance of a capacitor decreases with increase in frequency and that of an inductor increases with increase in frequency.

The reciprocal of impedance is called admittance.

The advantage of the phasor notation is that we can avoid writing down differential equations and solving them.

The source supplies AC given by Vs = 6sin2t = 6cos(2t-900)

1/jC = -j1 W

~Vs =6/-900 = 2 rad/s

½ F

KVL for this circuit can be written as:

Vs = I – jI6/-900 = (1-j)I

= √2(-450)I

00

0

4523452

906

I

Voltage across capacitor = -jI 013523

Real axis

Imaginary axis

Vo = 3√2/-1350 I=3√2/-4501350

Vs =6/-900

Once we have obtained the solution by phasor method, we can write down the time dependence of the voltages and currents

V0 = 3√2cos(2t-1350)i(t) = 3√2cos(2t-450)

Application of Thevenin’s theorem to AC circuits:

5cos(4t-300)V

5 mF

50 HZL~

200 W

Voc

V

Applying KCL at the node labeled Voc :

20050305

jVV

jV ococ

3Voc +V = 20 /-300

Applying KCL at node V:

200200V

jVVoc

ocVj

V

1

1

Substituting this in the other equation, we have:

030201

3

j

VV ococ

030201

34

jjVoc

0302034

1

jjVoc

Multiplying the numerator and denominator by the complex conjugate of the denominator:

0302025

7

jVoc

Convert the complex number in brackets to polar form.

000 87.2124302013.852

ocV

Therefore:Voc = 4√2cos(4t-21.870)

To find the Thevenin equivalent impedance, set the independent source to zero.

Net impedance of the resistor and the inductor is 200+200j. This is in parallel with the capacitor.

jjjjZo 50200200)50)(200200(

jjjZo 34))(200200(

jj

jZo 56834

)200200(

Zo = 40√2/-81.870 W

Ex. 4.6Application of Mesh analysis and Cramer’s rule to AC circuits.

9cos(5t)V1/15

F~

6 W

V

V1

- +

3 W I1 I2

For the mesh on the left:0)(369 211 IIjI

03)(9 221 IIIj

For the mesh on the right:-3v1 +3I2 =0

03)31( 12 IIj

Solve using Cramer’s rule

01 49.23.1

53)23(3

jI

02 6.1524.1

53)27(9

jI

Power:If a source is supplying a

sinusoidal voltage, then the average voltage supplied by the source over one cycle is zero.

However, the average power supplied by the source is non-zero.For an arbitrary element (could be resistor/capacitor/inductor), the power absorbed by the element is: )cos()cos()()()( tItVtitvtp

cos2cos2

tVI

T

ave dttpT

P0

)(1

T

ave dttVIT

P0

cos2cos2

1

The first integral can be shown to be equal to zero and the second integral is non-zero, givingPave = (VI/2 )cosf

Therefore if the element is:A resistor: Pave = VI/2 an inductor: Pave = 0A capacitor: Pave = 0Recall that, the impedance of an inductor or a capacitor is complex.The real part of the impedance is called resistance and the imaginary part is called reactance.

Maximum Power transfer:Given a circuit with a

Thevenin equivalent output impedance Z0 = Ro +jXo what value of load impedance ZL = RL +jXL will lead to a maximum transfer of power?

The condition is:ZL = RL +jXL = Ro – jX0 = Zo

* If ZL is restricted to be real (only resistive loads), then the condition is:

22ooL XRR

Effective value of a sinusoid:We define Ie ,the effective

value of an ac current, as the dc current that would yield the same power absorption (on the average) as the ac current.

Therefore: 22

21 RIRIe

IIe 21

VVVe 707.02

1

For household power supplies, the usual value (220 Volts) which is quoted is the effective value and not the peak value of the sinusoidal voltage. Therefore the peak value is 220/0.707 = 311.17 V.

Why is the effective voltage in India 220 V and 115 V in the US?

The effective value is same as the RMS value.

The RMS or root mean squared value of the power dissipated in a resistor is:

T

e dttRiT

RI0

22 )(1

T

e dttRIT

RI0

222 cos1 2IIe

The quantity IeVe is known as the apparent power, but we know that the average power is:Pave = (VI/2 )cosf = Ve Ie cosf

We use the unit Volt-Ampere (VA) for the apparent power to distinguish it from the actual (average power).

cospowerapparent power average factor Power

The angle is known as the power factor angle.

If power factor is negative, it indicates that the load is capacitive and if pf is positive it indicates an inductive load.

For an inductive load, the pf is called lagging (current lags the voltage)

for a capacitive load, the pf is leading (current leads).

Power factor correction:It is desirable to have a power

factor of around 1 (no lead or lag).

If we have capacitive or inductive loads (due to induction motors etc), then we can apply power factor correction to compensate for the lead or lag in the load.

This is achieved by adding capacitors or inductors as the case may be.

1. Regulator 2. Network connection points3. Fuses 4. Contactors 5. Capacitors Transformer 400/230 Volts

What is power factor correction?Ex 4.11A large consumer of electricity

(like our campus!) requires 10 kW of power by using 230 V rms at a pf angle of 600 lagging (pf = 0.5).

The current drawn by the load will be :

AVPI

ee 87

5.023010000

cos

But the transmission line from the power station to the campus will have some resistance (say 0.1 W).Therefore the power station has to generate

Ve = 0.1 IL + VL = 8.7/-600 + 230 /00

= 234 /-1.840

The loss in transmission is Ie

2 R = 757 W. If the power factor can be corrected, the current and hence the transmission loss will reduce.

What is a Single Phase three-wire circuit?