ALTERNATING CURRENT CIRCUITS

No. of

lectures allocated

Actual No. of lectures

dates :

3 30/5/09-3 /6/09

33.1 Alternating Current Sources Generator or dynamo is a device that operate according to the electromagnetic induction and Faraday's law. It is a device that produces a current by converting the mechanical energy into electrical energy. A simplified schematic diagram of a generator is shown in the Figure. It consists of a loop of wire that can rotate, by an external means, in a magnetic field. The ends of the loop are connected to slip rings that rotate with the loop. The slip rings are connected to the external circuit by two fixed brushes. As the loop rotate, the magnetic flux through it changes with time inducing an emf and so a current in the external circuit. Suppose that the loop, with area A, rotates with constant angular speed ω, and suppose that θ is the angle between

Figure 33.1 Schematic diagram of a generator.

Figure 33.2 The emf produced by a generator as a function of time.

ε

εmax

-εmax

t

the magnetic field and the area of the loop. The magnetic flux through the loop at any time is then ( )tBABAm ωθ coscos ==Φ Therefore, the induced emf in the loop is given by ( ) ( )ttAB

dtd m ωωω εε sinsin max==Φ

−= 33.1

With ωε AB=max . This means that the emf varies sinusoidally with time forming a sinusoidal wave, as shown in Figure 33.2. Because of that a source of AC is represented in the circuits by the symbol The frequency of commercial generators in our country and most of the world is 50 Hz, while it is 60 Hz in some other countries like the USA and Canada. Note that the frequency f is related to the angular speed ω through the relation fπω 2= .

Figure 14.7 (a) DC current versus time. (b) AC current versus time.

I

Io

-I

t

(b)

I

t

(a)

Ohm's law, RVI = , is still hold for AC circuits. Now from Equation 14.13 we have

( )tRR

I ωεε

sinmax==

Or ( )tII m ωsin= 33.2 The voltage across an element in an AC circuit is given by ( )tVV m ωsin= 33.3 Im and Vm are called the peak current and the peak voltage, respectively. As it is clear from Figure 14.7(b), the average value of an AC is zero. This doesn't mean, however, that no power is needed or that no heat is produced in a resistor. Electrons for AC move forth and back and so produce heat. It is known that the power dissipated in a resistor is given by ( )tRIRIP m ω222 sin==

Since the current is now squared, we see that the power is always positive. The quantity ( )tω2sin varies between 0 and 1; and it is easy to show that its average is simply 0.5. That is, the average value of 2

212 is mII , and

therefore, the average power dissipated in a resistor in AC circuits is then

R

VRIP mm

2

212

21 == 33.3

That is, what is important for calculating the average power is the mean value of the square of the current or voltage. The square root of each of these values is known as the rms (root-mean-square) and is given by

2m

rmsII = 33.4

2m

rmsVV = 33.5

The rms values of V and I are sometimes known as the effective values. They are useful because they can be substituted directly into the famous power formula IVP = to get the average power, that is,

R

VRIVIP rmsrmsrmsrms

22 === 33.6

From which we can recover Equation 33.3. It is the rms values that are specified or measured. Therefore, ammeters and voltmeters are designed to read the corresponding rms values. When we say that the standard voltage in our country is 220 V, we mean that the rms value of V is 220V. The peak value of such a voltage is, from Equation 14.20 V3112 == rmsm VV Phasors: It is a counterclockwise-rotating vector representing an AC quantity such that its length represents the maximum value of the quantity and its projection onto the vertical axis represents the instantaneous value of the quantity. Example 33.1 A 900.-W microwave oven is designed to operate at 220. V. Calculate its resistance and the peak current when it is operating. Solution The average power of the oven is 900 W and

V220=rmsV . From Equation 14.21, we have

( )Ω=== 8.53

900220 22

PVR rms

Again from Equation 14.21 we obtain A09.4

220900

===rms

rms VPI

Therefore, from Equation14.19 we get A78.52 == rmsm II

33.2 Resistors in AC Circuits

Consider the circuit that consists of a resistor an connected in series with an AC source. Let vR, be the voltage at some instant across the resistor. This voltage must equal to the voltage of the AC source, that is tVv mR ωsin= 33.7 The current passing through the resistor is

tVv m ωsin=

R

tI

RtV

Rvi m

mR ωω sinsin

=== 33.8

As it is clear from Eqs. 33.7 & 33.8 that The current i and voltage vR across a resistor in a pure resistive AC circuits are in phase. 33.3 Inductors in AC Circuits

Consider the circuit that consists of an inductor connected in series with an AC source. Let vL, be the voltage at some instant across the inductor. Again this voltage must equal to the voltage

iR, vR

Vm

t

Im

vR

iR

ω t

iR, vR

Im

Vm

tVv m ωsin=

L

of the AC source, that is tVv mL ωsin= 33.9 To find the current passing through the inductor we have

dtdiLvL = ⇒

)cos()sin(1 tL

VdttL

VdtvL

i mmL ω

ωω −=== ∫∫

Using the identity ( ) )cos(2sin tt ωπω −=− ⇒ ( ) ( )2sin2sin πωπω

ω−=−= tIt

LVi m

m 33.10

With

L

mm X

VI =

Where

LXL ω= 33.11 Is called the inductive reactance. Now comparing Eqs. 33.9 & 33.10 we conclude that

The current i lags behind the voltage vL across an inductor in a pure inductive AC circuits by 90o . 33.4 Capacitor AC Circuits

Consider the circuit that consists of an capacitor connected in series with an AC source. Let vC, be the voltage at some instant across the capacitor. Again this voltage must equal to the voltage of the AC source, that is

iR, vR

Vm

t

Im vR

iR

ω t

iR, vR

Im

Vm

tVv m ωsin=

C

tVv mC ωsin= 33.12 To find the current passing through the inductor we have CCvq = ⇒ )cos( tCV

dtdvC

dtdqi m

C ωω===

Using the identity ( ) )cos(2sin tt ωπω =+ ⇒

( ) ( )2sin2sin πωπωω +=+= tItCVi mm 33.13 With

C

mm X

VI =

Where

CXC ω

1= 33.14

Is called the capacitive reactance. Now comparing Eqs. 33.12 & 33.13 we conclude that

The current i leads the voltage vC across a capacitor in a pure capacitive AC circuits by 90o . 33.5 SERIRS RLC AC CIRCUITS

Consider the circuit which consists of a resistor, an inductor, and a capacitor, connected in series with an AC source. Let vR, vL, and vC be the voltage at some instant across each element, respectively. Since they are connected in series, the voltage across the combination is CLR vvvv ++= 33.15

tVv m ωsin=

L C

R

Figure 14.9 A series RLC circuit.

iR, vR

Vm

t

Im

vR

iR

ω t

iR, vR

Im

Vm

This voltage must equal to the voltage of the AC source, that is CLRm vvvtV ++=ωsin 33.16 Even though the instantaneous across the combination is given by Equation 33.16, a voltmeter does not read ( ) ( ) ( )rmsCrmsLrmsR VVV ++ when connected across the combination. What will read then? Let us see. Supposing the current passing through each element (which is the same due their series connection) to be given as ( )φω −= tII m sin 33.17 where the constant φ is called the phase angle between the current and the applied voltage. Now we have ( )φω −== tRIIRv mR sin 33.18

+−=

2sin π

φωtXIv LmL 33.19

−−=

2sin π

φωtXIv CmC 33.20

To solve Eq. 3.16 we use the phase diagram by letting the phasor representing I to be along the horizontal axis, as shown. Now from the diagram we conclude that ( )22

CLRm VVVV −+= 33.21 And

R

CLV

VV −=φtan 33.22

Using the relations RIV mR = , LmL XIV = , and CmR XIV = , Equation 33.21 becomes

φ

Im

Vm

VR

VL

VC

VR

VL-VC Vm

φ

( ) ZIXXRIV mCLmm =−+= 22 33.23 With ( )22

CL XXRZ −+= 33.24 is called the impedance of the circuit and has the unit of Ohm. Equation 14.22 now reads

R

XX CL −=φtan 33.25

Note that if the circuit consists only of a resistor with the source, ( )0== CL XX , we have RZ = and the phase angle is zero, that is If the circuit contains only an inductor, ( )0== CXR , we have LXZ = and o90=φ , that is If the circuit contains only a capacitor, ( )0== LXR , we have CXZ = and o90−=φ , that is In general if CL XX > the phase angle is positive and the current lags behind the applied voltage. On the other hand, if CL XX < the phase angle is negative and the current leads the applied voltage. When CL XX = , then phase angle is zero. In this case, the impedance equals

the resistance and the current has its peak value given by RVm . The frequency ωo at which this occurs is called the

resonance frequency of the circuit. Using the condition CL XX = we get

CL

oo ω

ω1

=

And

LCo1

=ω 33.26

Example 33.4 In a series RLC circuit we have

V120=mV Ω= 200R , μF0.4=C , Hzf 60= . Find L such that the voltage across the capacitor lags the applied voltage by 30o. Solution Since the angle φ is between the applied voltage and the voltage across R, and since the angle between VR and VC is 90o ⇒ ooo 609030 −=−=φ Now Ω=

×== − 663

)104)(60(211

6πωCXC

200

66373.1tan −=−⇒

−= LCL X

RXX

φ ⇒

Hf

LLXL 83.0120317

2317317 ===⇒=Ω=

ππω

Example 33.5 In a series RLC circuit we have

V150=mV Ω= 425R , μF5.3=C , HL 25.1= 1377 −= sω . a) Determine Z. b) Find the maximum current in the circuit. c) Find the phase angle. d) Find the maximum voltage across each element. Solution The reactances are given by

( )( ) Ω=== 47125.1377LXL ω and ( )( ) Ω=×== − 758105.337711 6CXC ω . Therefore, the

impedance is ( ) ( ) ( ) Ω=−+=−+= 513758471425 2222

CL XXRz b) The maximum current is given by A292.0

513150

===Z

VI mm

c) The phase angle is given by oCL

RXX 34

.425758417tan −=⇒

−=

−= φφ

Note that since the circuit is more capacitive ( LC XX > ), the current leads the applied voltage by the angle 34o d) For the maximum voltage across each element we have V124== RIV mR , V138== LmL XIV , and V221== CmR XIV 33.6 Power in AC Circuits The instantaneous power delivered by Ac source is )sin()sin( maxmax tVtIiVP ωφω −== ⇒

φωωφω sin)cos()sin(cos)(sin maxmax2

maxmax ttVItVIP −= ⇒

( ) [ ] φωωφω sin)cos()sin(cos)(sin2maxmax avravravr tttVIP −=

Knowing that( ) 2

12 )(sin =avrtω and [ ] 0)cos()sin( =avrtt ωω ⇒

φcos21

maxmaxVIPavr = 33.27

Or φcosrmsrmsavr VIP = 33.28 The quantity φcos is called the power factor given by

max

max

maxcos

VRI

VVR ==φ ⇒

RIP rmsavr

2= 33.29

Example 33.6 In a series RLC circuit we have V150=mV Ω= 425R , μF5.3=C , HL 25.1= 1377 −= sω .

What is the power delivered to the circuit. Solution From the previous example we have Z=518 Ω. Now A206.0

2==

ZVI m

rms ⇒

WRIP rmsavr 1.18)425()206.0( 22 ===

33.7 Resonance in a Series in RLC Circuits From Eq. 3.29 we have

( )22

22

22

CL

rmsrms

rmsrmsavr XXR

RVRI

zRV

RIP−+

====

But

( ) ( )222222

2 1oCL L

CLXX ωωω

ωω −=

−=− ⇒

( )222222

22

o

rmsavr

LR

RVPωωω

ω

−+= 33.30

With

LCo1

=ω is the resonance frequency

From Eq, 33.30 is clear that the average power is maximum when oωω = . At resonance the average power is given by

R

VP rmsavr

2=

The average power versus the frequency is plotted for two different values of the frequency.

The sharpness of the curve is measured by a factor known as the quality factor defined as

ωω∆

= oQ 33.31

With ω∆ is the width of the resonant power curve at half maximum and given by

LR

=∆ω ⇒

R

LQ oω= 33.32

Example 33.7 In a series RLC circuit we have

V20=rmsV Ω= 150R , mHL 2= 15000 −= sω . What is the value of C for which the current is maximum. Solution The current is maximum when oωω =

Now FC

CLCo µω 2102

1500013

=⇒×

=⇒=−