S28._AlternatingCurrentCircuits

Short Version : 28. Alternating Current Circuits

28.1. Alternating Current

Reminder: All waves can be analyzed in terms of sinusoidal waves (Fourier analysis Chap 14).





Sinusoidal wave (Chap 13) :

2 2

1 1 1

 sin 2 x dx   cos x dx 

2



2 0

2 0

2



Vp sin 

Vp Ip

Vrms  I rms 

2 2



Angular frequency :   2 f [] = rad/s





=/6 V  Vp sin  t  V  I  I p sin  t  I 



 = phase

Example 28.1. Characterizing Household Voltage



Standard household wiring supplies 110 V rms at 60 Hz.

Express this mathematically, assuming the voltage is rising through 0 at t = 0.





Vp  2 Vrms  156V



  2 f  2  60 Hz   377 s 1



V  Vp sin  t   



V  t  0  0  0  V p sin 



  0 V  V p sin  t  156 sin 377 t  V

28.2. Current Elements in AC Circuits





 Resistors

 Capacitors

 Inductors

 Phasor Diagrams

 Capacitors & Inductors: A Comparison

Displacing Functions



g g is moved to the right (forward) by  to give f.

f



f  x   g  x  



x x





sin cos Displacement: sin is cos moved forward by /2.

Phase: sin lags cos by /2.



 

sin  x   cos  x  

 2





  Derivative: moves sinusoidal functions backward by /2.

sin x  cos x  sin  x  

 2 phase is increased by /2.



  Integral: moves sinusoidal functions forward by /2.

 sin x dx   cos x  sin  x  

 2 phase is decreased by /2.

Resistors







V V

I  p sin  t

R R

+ Vp

V I  I p sin  t Ip  I & V in phase

 R

Vrms

I rms 

R

Capacitors



q  C V  C V p sin  t



+

d q C dV

V I  C V p  cos  t

 dt dt



 

I  C V p  sin   t   I leads V by 90

 2



I peaks ¼ cycle before V Vp

I p  C Vp  

XC



1

XC  Capacitive reactance

C



 XC   

X C   as   0 Static: open ckt.



X C  0 as    HF: short ckt.

Inductors

I dI

EL   L   V   V p sin  t

dt

+  1 Vp Vp

Vdt cos  t

V



L

+

I

L



L  sin  t d t   L



Vp  

I sin   t   I trails V by 90

L  2

Vp Vp

Ip  

I peaks ¼ cycle after V L XL



XL   L Inductive reactance





XL  



X L  0 as   0 Static: short ckt.



X L   as    HF: open ckt.

Table 28.1. Amplitude & Phase in Circuit Elements





Circuit Element Peak Current vs Voltage Phase Relation

Vp

Resistor Ip  V & I in phase

R

Vp Vp

Capacitor Ip   V lags I 90

XC 1/ C

Vp Vp

Ip   V leads I 90

Inductor XL L

Example 28.2. Equal Currents?



A capacitor is connected across a 60-Hz, 120-V rms power line,

and an rms current of 200 mA flows.

(a) Find the capacitance.

(b) What inductance, connected across the same powerline,

would result in the same current?

(c) How would the phases of the inductor & capacitor currents compare?





200  103 A

(a) Vrms

I

 rms C

I rms

  4.42  106 F  4.42  F

C Vrms  120 V   2 60 Hz 



Vrms   L I rms Vrms 120 V

(b) L   1.59 H

I rms   200  10 A  2 60 Hz 

3









(c) Capacitor: IC leads V by 90.  IC leads IL by 180.

Inductor: V leads IL by 90.

Phasor Diagrams

Phasor = Arrow (vector) in complex plane. Length = mag. Angle = phase.



V  I X V  V e i  V e it









V leads I by 0. V leads I by 90. V leads I by 90.

( same phase ) ( V lags I by 90 )

Capacitors Revisited

I

q CV  C Vp ei t



+

Vp e i t d q C dV

V I  C Vp i  ei t

 dt dt





I i  CV   CV e

i

2

I leads V by 90





Taking the real part as physical



V  V p cos  t



I  CV p  sin  t  CV p  cos   t  

 

 2



Taking the imaginary part as physical



V  V p sin  t V I Z



I  CV p  cos  t  CV p  sin   t  

  Z 

1

i

 2 Impedance

C

Inductors Revisited

I dI

EL   L  V   Vp e i  t

dt

+  1 Vp Vp

Vdt e e it

it

Vp e i t V L I  dt 

 + L L i L





I

V V i 

 e 2 I lags V by 90

i L L



Taking the real part as physical



V  V p cos  t

Vp Vp  

I sin  t  cos   t  

L L  2



Taking the imaginary part as physical



V  V p sin  t

Z  Li

Vp Vp  

I  cos  t  sin   t  

L L  2

Capacitors & Inductors: A Comparison









C  L translator:

EB

q  B

VI

ZY

Table 28.2. Capacitors & Inductors

Capacitor Inductor



C

q B

Defining relation L

V I

dV d I

Defining relation; differential form I C V L

dt dt



Opposes change in V I

1 1

Energy storage UE  C V 2 UB  L I2

2 2



Behavior in low freq limit Open circuit Short circuit



Behavior in high freq limit Short circuit Open circuit



Reactance 1 / XC   C XL   L

Admittance / Impedance YC  i  C  1 / Z C ZL  i  L



Phase I leads by 90 V leads by 90

Application: Loudspeaker Systems









Loudspeaker







Q  1 

C passes High freq V   IC R   R  IC

C iC 

d IL

V L  ILR  i  L  R  I L

dt



Loudspeaker system with

high & low frequency filters.

L passes low freq

28.3. LC Circuits

I



+

V



Analyzing the LC Circuit

I



+ 1 1

V U  UB  UE  L I2  CV2



2 2 q dI

L 0

C dt

dU dI dV

0 LI CV

dt dt dt



dq dI d2 q

V

q dV 1 d q

 I  

C dt C dt dt dt d t2



d q d2 q q 1dq

0L C

d t d t2 CC dt

d2 q q

L  0

d t2 C

1

q  q p cos  t 

LC

Resistance in LC Circuits – Damping

+ VR 

I

1 1

U L I L  C VC2

2



 + 2 2

VC VL

+  dU dI d VC

 I R

2

LI  C VC

dt dt dt



dq

VC 

q dVC I

 I

C dt C dt



dI I dI q

 LI  q  I 2R  0 L   IR  0

dt C dt C



d2q dq q

L 2 R  0

dt dt C



q t   qp e R t / 2 L cos  t (see next page)

Resistance in LC Circuits – Damping

+ VR 

I



q dI dq

+ + I RL 0 I 

VC VL C dt dt

 



 d2q dq q

L 2 R  0

dt dt C









q t   qp e R t / 2 L cos  t (see next page)

Solutions to Damped Oscillator



d2q dq q

L 2 R  0

dt dt C



 1

Ansatz: q  q0 e a t  L a 2  R a   q0 e a t  0

 C



1

L a2  R a  0

C

1  L R

a  R  R2  4   i 4L  R2C

2L  C 2L





q  e R t / 2 L  c ei  t  c ei  t 

2

1 L 1  R 

 4  R2  

2L C 

LC  2 L 

 e R t /2 L  A cos t  B sin t 

2

 R  0 

1

 q0 e  R t / 2 L cos  t

   

2

0

 2L  LC

28.4. Driven RLC Circuits & Resonance



+ VR 

I dI q

V I RL  0

dt C

+ +

VL I

V  I R  i d L I  0

  i d C



 VC +   1 

V  I  R  i  d L  

  d C  





Driven damped oscillator :

Long time: oscillates with frequency d.  0 2

Z  R  i d L  1  2 

Resonance if d = 0.  d 

Resonance in the RLC Circuit



  0  

2

02 VL  VC  V  I R

V  I  R  i d L  1  2   VL  i d L I VC  VL 2

  d   d





VC & VL are 180 out of phase.









VC p VL p 1

IC p   IL p   VL p  VL p if  d L i.e., d  0

2 2



1 / d C d L d C

Frequency Response of the RLC Circuit

Series circuit  same I phasor for all.



VR in phase with I.

VC lags I by 90.

VL leads I by 90.



Q dI  1 

I R L  0  I R  i  L

High Q

C dt  iC 

 1 

Z  R  i  L 

Low Q  C

2

 1 

Vp  I p Z Z  R2    L 

 C

See Prob 71 for definition of Q.



1

VL p  VC p  L

tan   X  XC C L 02 

 L   1   2 

VR p R R 

R 

At resonance,  = 0.

Example 28.4. Designing a Loud Speaker System

Current flows to the midrange speaker in a loudspeaker system through a

2.2-mH inductor in series with a capacitor.



(a)What should the capacitance be so that a given voltage produces the

greatest current at 1 kHz ?

(b) If the same voltage produces half this current at 618 Hz,

what is the speaker’s resistance ?

(c) If the peak output voltage of the amplifier is 20 V,

what will be the peak capacitor voltage be at 1 kHz ?





1

(a) Greatest I is at resonance: 02 

LC



1 1

C   11.5  106 F  11.5  F

L  2.2  10 H  2  10 Hz 

2 3 3 2

0

(b) If the same voltage produces half this current at 618 Hz, what is the speaker’s resistance ?



2

 1  Z R

Vp  I p Z Z  R2    L  At resonance:

 C

2

Ip  1 

Z  Ip R R2    L    2R

2  C



1  1  1  

  2  618 Hz   2.2  10 H  

3 1

R  L   C   

3  

3  2  618 Hz  11.5  10 F  

6



 8.0 



(c) If the peak output voltage of the amplifier is 20 V,

what will be the peak capacitor voltage be at 1 kHz ?





Peak voltage is at resonance (1 kHz). Vp  I p R



Vp 1 20 V 1

VC p  I p X C    35V

R C 8.0   2  103 Hz  11.5  106 F 

Capacitor: 28.5. Power in AC Circuits

I leads V by 90 ,  P  = 0







PIV   I p sin  t     V p sin  t 







Resistor:

P  I p sin  t     V p sin  t 

I & V in phase ,  P  > 0

 I p Vp sin  t    sin  t



 I p Vp sin  t cos  cos  t sin   sin  t



 I p Vp  sin  t

2

cos   cos  t sin  t sin  

1

I & V out of phase ,  P   P  I p V p cos   I rms Vrms cos 

2

Power factor



Dissipative power = I2 R

 large power factor reduces I & hence heat loss.

Conceptual Example 28.1. Managing Power Factor



You’re chief engineer of a power company.

Should you strive for a high or a low power factor on your lines?





P  I rms Vrms cos



Power factor





Generator : fixed Vrms .

To maintain fixed , Irms cos  = const.

Smaller power factor  higher Irms .

 higher power loss.

Ans.: keep power factor close to 1.

Making the Connection



Transmission losses on a well-managed electric grid average about

8% of the total power delivered.



How does this figure change if the power factor drops from 1 to 0.71?



P  I rms Vrms cos



I

To deliver the same power I new   1.4 I

0.71



PL  I rms R PL, new  1.4  PL

2

Transmission losses:

2

  2 PL



( doubles to 16% )

28.6. Transformers & Power Supplies

Transformer: pair of coils wound

on the same (iron) core.

Vsec ondary Nsec ondary



Vprimary N primary



Works only for AC.

Direct-Current Power Supplies

Diode passes + half of each cycle









Diode





Diode cuts off  half of each cycle









RC (low freq) filter