AC Circuits.pdf

Vidyamandir Classes

AC Circuits

AC Circuits ALTERNATING CURRENT

Section - 1

Most of the electric power generated and used in the world in the form of a.c. i.e., alternation current. The magnitude of alternating current changes continuously with time and its direction is reversed periodically. It is represented by I = I0 sin t

or

I = I0 cos t

Here, I is instantaneous value of current i.e., magnitude of current at any instant of time t and I0 is the peak value of maximum value of a.c. It is also called amplitude of a.c.,  is called angular freqneucy of a.c. Also,  

2  2 v T

where, T is the time period or period of a.c. It is equal to the time taken by the a.c. to go through one complete cycle of variation.

The term used for a.c. holds equally for alternation voltage which may be represented by V = V0 sin t

or

V = V0 cos  t

In general, I  I 0 sin(t   ) V  V0 sin(t   )

Self Study Course for IITJEE with Online Support

Section 9

1

Vidyamandir Classes

AC Circuits

Average and RMS Value of Alternating Current : Average Current (Mean Current) As we know an alternating current is given by i = i0 sin (t + )

. . . .(i)

The mean or average value of ac over any time t is given by t

 i dt 0

iavg = t

 dt 0

Using equation (i) t

 i 0 sin t    iavg =

0 t

 dt 0

In one complete cycle average current T

i 0  cos  t     iavg = –   T    0

=–

i 0  cos T     cos     T   

i 0  cos  2     cos    0 (As T = 2) T    Since ac is positive during the first half cycle and negative during the other half cycle so iavg will be zero for long time also. Hence the dc instrument will indicate zero deflection when connected to a branch carrying ac current. So it is defined for either positive half cycle or negative half cycle. Now to find mean value of current i = i0 sin t for positive half cycle. i.e. from t = 0 to t = T/2 

T 2



i 0 sin t

iavg  0 T 2



 dt

2i0



 0.637 i 0

. . . .(5)

0

2

Section 9


Vidyamandir Classes

AC Circuits

Similarly vavg =

2v0

 0.637 v0



. . . .(6)

R.M.S. Value of Alternating Current The notation rms refers to root mean square, which is given by square root of mean of square current. i.e., irms =

2 iavg T 2

 i dt

i2

i20 T 1T 2 2 =  i0 sin (t  ) dt =  [1  cos 2(t  )]dt avg = 2T 0 T0  dt 0 T

0

T

i20  sin 2(t  )  = t   2T  2 0

i20  sin (4  2)  sin 2   i20 T   =   2T  2 2  irms =

i0 2

 0.707 i0

. . . .(7)

Similarly the rms voltage is given by Vrms =

V0 2

 0.707 v0

. . . .(8)

The significance of rms current and rms voltage may be shown by considering a resistance R carrying a current i = i0 sin (t + ) The voltage across the resistor will be V = Ri = (i0 R) sin (t + ) The thermal energy developed in the resistor during the time t to t + dt is i2 R dt = i02 R sin2 (t + ) dt The thermal energy developed in one time period is T

T

U =  i Rdt = R  i02 sin 2 (t  )dt 2

0

0

1 T  2 2   = 2 RT i sin (  t   ) dt 0 = RT  T   i rms  0


. . . .(9)

Section 9

3

Vidyamandir Classes

AC Circuits

It means the root mean square value of ac is that value of steady current, which would generate the same amount of heat in a given resistance in a given time. So in ac circuits, current and ac voltage are measured in terms of their rms values. Like when we say that the house hold supply is 220V ac it means the rms value is 220V and peak value is 220 2 = 311 1V.

Illustration - 1

If the voltage (in volts) in an ac circuit is represented by the equation, V = 220 2 sin (314 t – ), (where t is in seconds). Calculate (a) peak and rms value of the voltage (b) frequency of ac. SOLUTION : (a)

The rms value of voltage

For ac voltage, V = V0 sin (t – )

Vrms =

The peak value V0 = 220 2 = 311 1V,

(b)

2

; Vrms = 220 V

As  = 2f, 2f = 314 314

i.e.,

Illustration - 2

V0

f=

2 

= 50 Hz

The electric current in a circuit is given by i = i0 (t/T) for some time. Calculate the rms

current for the period t = 0 to t = T for i0  20 3 A. SOLUTION : The mean square current is

 avg i2



1T 2 i0 (t / T )2 dt   T 0

Thus, the rms current is

i02 T 2 i02 t dt   3 T3 0

irms =

2 iavg .

i0 3

= 20 A

RC Circuit When an AC voltage V(t) is applied across an RC circuit, the current I (t) is also an alternating current.

4

Let

I = Im cos t

then

VR = Im R cos t VC 

q 1  1    I dt   sin t  I m C C  C 

VC 

1   I m cos   t   C 2 

Section 9

 1   C  X C called reactance of capacitor   


Vidyamandir Classes

AC Circuits

Total voltage across combination V  VR  VC

(Instantaneous)

 I m R cos  t 

1   I m cos  t   C 2 

V (t )  I m R 2 



1 2 2

cos(t   )

 C

1  RC

where

  tan 1

The ratio

V  Z is called impedance I

V V 1 Z  m  rms  R 2  2 I m I rms  C2



Voltage across RC combination lags behind current by the phase angle 1   tan 1  RC Note : All values drawn in the phasor diagram are rms values. Vrms Vm  Z. I rms Im

Vm   Vrms  2    Im    I rms  2   

(a) Voltage across R is in phase with I.  (b) Voltage across C lags behind I by 2 (c) V across RC combination lags behind I by tan

1

1  RC

 In general, rms value of a periodic function f (t ) 

1 T 2 f (t ) dt T 0

RL Circuit I (t) = Im cos t ;

VR = Im R cos t

  dI   ( L ) I m sin t   L I m cos  t   2 dt  V(t) = VR(t) + VL(t) VL  L

[  L  X L called reactance of inductor] Self Study Course for IITJEE with Online Support

Section 9

5

Vidyamandir Classes

AC Circuits V (t )  VR (t )  VL (t )

(Instantaneous)

   I m R cos t   L I m cos  t   2  V (t )  I m R 2   2 L2 cos (t   )

  tan1

L R

Voltage across RL combination leads I by tan1 Impedance

V V Z  m  rms  Im I rms

L R

R 2   2 L2

Phasor Diagram (Using rms values of VR, VL, V and I) (a)

VR across R is in phase with I.

(b)

VL across L leads I by /2

(c)

V across RL combination leads I by tan

1

L R

LCR Circuit I (t) = Im cos t V = VR + VC + VL  I m R cos t  I m

(a)

L 

1     cos  t    I m  L cos  t   C 2 2  

1 C

1 2  V (t )  I m R 2    L   cos ( t   ) ; C  

1   L   C    tan1   R    

6

Section 9

V leads I by .


Vidyamandir Classes

(b)

L 

AC Circuits

1 C

 1  V (t )  I m R 2     L  cos (t   ) ;  C 

(c)

 1   L     tan1  C  R     1 L  C V(t) = Im R cos t

1   R2    L  C  

Z 

V lags behind I by .

V and I are in same phase. or

Z 

 1  R2    L   C 

2

Z=R

Average Power across R Pav 

1 T 1 T 2 V (t ) I (t ) dt   I m R cos 2 t dt  0 0 T T

2 Im 2 Pav  R  I rms R T

Average Power across L Pav 

1 T V (t ) I (t ) dt T 0

I2 T    m   L cos   L   cos t T 0 2 



Pav = 0 acros L.

During one half cycle, L stores energy and during next half cycle it supplies this energy back to the circuit.

Average Power across C Pav 

1 T V (t ) I (t ) dt T 0

I2 T 1    m cos  t   cos t  0 2 0 C 2 

C also stores and gives back this energy alternately during each half cycle.


Section 9

7

Vidyamandir Classes

AC Circuits In general, average power in a circuit Pav = Vrms Irms cos 

If Vrms is the total supply voltage and  is the phase difference between current and voltage. (cos  is also known as power factor).

Resonance in LCR circuit When  L 

1 and Z = R, it is called resonant condition. C

Resonance frequency : 0 

1 LC



v0 

1 2 LC

Voltage supplied across combination is in same phase with current  (a)

 = 0° and cos  = 1 Resonance occurs when XL – XC = 0

(b) (c)

(d)

or

i.e. power factor is unity.

1 LC



Current reaches a maximum value of

V at resonance. R

V2 Power dissipated is maximum and is rms (current and power are maximum because Z is minimum). R V V2 Irms   rms , P  rms R Z Z Current is in phase with voltage ( = 0).

Parallel AC Circuits RC Circuit Using phasor diagram I 2  I C2  I R2

8

(rms values)



2 2  Vrms   Vrms   Vrms   Z   R   X       C 



1  Z

1 R2

Section 9



1 X c2

2

1    Xc   c  


Vidyamandir Classes Let

AC Circuits

V  Vm sin  t V I  m sin(t   ) z

I V / xc R tan   c    R c IR V / R Xc



  tan 1 R c

Current through source leads v by 

L – R Circuit Using phasor diagram I 2  I c2  I R2 (rms values) 2



2  Vrms   Vrms  2  Vrms       Z      XL   R 



1  z

Let

1 R

2



1

( xL  L )

xL2

V  Vm sin  t V I  m sin( t   ) z I R R tan   L   IR XL L



  tan 1

R L

Current thorugh source lags voltage by  ,

LCR Circuit : (A) 

IC  I L

(rms)

X L  XC I 2  I R2   IC  I L 

2

(rms values)

2 2  Vrms   Vrms   Vrms Vrms    Z   R   X  X       C L 



1  Z

 1 1     2 X R  C XL  1

2

2


Section 9

9

Vidyamandir Classes

AC Circuits Let V  V0 sin t V I  0 sin(t   ) Z 1 1  I I X XL tan   C L  C 1 IR R

Current though source leads voltage by  . (B)

IC  I L 

(rms) X L  XC I 2  I R2   I L  IC 

2

(rms value)



2 2  Vrms   Vrms   Vrms Vrms   Z   R   X  X       L C 



1  Z

 1 1     R2  X L X C  1

2

2

Let V  V0 sin t V I  0 sin(t   ) Z

1 1  I I X XC tan   L C  L 1 IR R Current through source lags voltage by  .

(C)



XC  X L I  I R (rms) Vrms Vrms  Z R

10

Section 9


Vidyamandir Classes  Let

AC Circuits

ZR V  V0 sin t

V I  0 sin  t ,  0 Z Current though source and voltage are in same phase.

Illustration - 3 For the circuit shown in figure. Current in inductance is 0.8 A while in capacitance is 0.6 A. The current drawn from the source is x  10–1. Find the value of x ?

SOLUTION : In this ac circuit  = 0 sin t is applied across an inductance and capacitance in parallel, current in inductance will lag the applied voltage

 while across the capacitor will lead both by I = IL – IC

2

(rms)

= 0.8 – 0.6  0.2 

2 10

 x2


Section 9

11

Vidyamandir Classes

AC Circuits

SUBJECTIVE SOLVED EXAMPLES Example - 1

In the given circuit, calculate capacitive reactance and impedance maximum and rms current phase angle and voltages R across and C.

(a) (b) (c)

SOLUTION : 1 1   10  (a) X C  C 2 vC R 2  X C2  12.2 

Z 

(b) peak value of current V 200 V Im  m   16.4 A Z 12.2 

rms value of current

V I I rms  rms or m Z 2 (c) phase angle

  tan 1

Example - 2

VR(rms) = Irms R = 11.6 × 7 = 81.2 V VC ( rms )  I rms

1  11.6  10  11.6  10  116 V C

V 200 Vrms  m   141.5V 2 2 (It can be seen that V2rsm = V2Rrms + V2Crms). Note that :

Total

V I (t )  m sin (2 v t   ) Z

I(t) = 16.4 sin (3.14t 

11 ) 36

1 10  tan1  55  RC 7 (V lags or I bads by 55°)

Find the rms and instantaneous current and voltage across R and L.

SOLUTION : Z 

2

2 2

R  L

where XL = 100 × 20 × 10–3 = 6.28  is the inductive reactance.

12

Subjective Solved Examples

Z  92  6.282 10.97  V 200 / 2  12.89 A (a) I rms  rms  Z 10.97


Vidyamandir Classes

AC Circuits

V 200 I peak  I m m  Z 10.97

(b) VR(rms) = Irms R = 12.89 × 9 = 116 V (c) VL(rms) = Irms L = 12.89 × 6.28 = 80.94 V

 18.23 A(or I m  I rms 2)

  tan 1

V(total ) 

L 6.28  tan 1  35 R 9

200  141.4 V 2

2 2  L2L ) (check that Vrms  VR rms rms

V leads or I lags by 55°. I(t) = Im sin (100t – 35°)



= 18.23 sin (100t – 35°)

Example - 3 (a) (b) (c) (d) (e) (f)

Find : Instantaneous current I(t) rms values of I and V across elements power factor power input resonant frequency impedance, current and power at resonance conditions.

Given that XC = 4, XL = 8. SOLUTION : (a) Z 

R 2  ( X L  X C )2

XL = L = 8

V(rms) = Irms R = 120 V

Z  32  (8  4) 2  5 

I V( rms )C  rms  160V C V(rms) = Irms L = 320 V

XC 

1  4 wC

  cos 1 As

1 (b) I rms  m  40 A 2

X  XC R  tan1 L  53 Z R

XL > XC , V leads or I lays by 53° I (t )  I m sin(2 vt   ) (  53) V  m sin(314t   ) (  53) Z  84.6sin(314t   ) (  53)


2 Check that V(2rms ) R  V(2rms ) L  Vrms C

= 200 V = Vrms (c) Power factor = cos  = cos 53°  0.6 (d) Power input = power dissipation = Vrms Irms cos  = 200 × 40 × 0.6 = 4800 W (or I2rms R)


13

Vidyamandir Classes

AC Circuits (e) V0 

1 2 LC

 35.4 Hz

Power = Vrms Irms cos  = Vrms Irms

(f) Z0 = R = 3 at resonance

 200 

V V 200 I rms  rms  rms  A at resonance Z R 3

200 40000  W 3 3

Example - 4 A box P and a coil Q are connected in series with an AC source. The emf of the source is 10 V. Box P contains a capacitance of 1F in series with a resistance of 32. Coil Q has a self-inductance 4.9 mH and a resistance of 68 in series. The frequency is adjusted so that the maximum current flows in P and Q. Find the impedance of P and Q at this frequency. Also find the voltage across P and Q respectively. SOLUTION : As this circuit is a series LCR circuit, current will be  5924  77 maximum at resonance, 12  R 2   L 2  1 Z  and Q  2 i.e,     LC 

with

1



 

4.9  103 106

I

V

R So the impedance,



10

 32  68





12

105

 5 2   10 2     68    4.9  103   7     

7 rad/s

1

A

10

 9524  97.6 

and hence, VP  IZ P 

12

2   1   2 Z P   R1      C     

and

VQ  IZQ 

1 10 1 10

  77   7.7 N   97.6   9.76V

12

2    2  7   32     106   5   10   

14



Vidyamandir Classes

AC Circuits

Example - 5 A current of 4A flows in a coil when connected to a 12 V DC source. If the same coil is connected to a 12 V, 50 rad/s AC source a current of 2.4 A flows in the circuit. Determine the inductance of the coil. Also find the power developed in the circuit if a 2500 F capacitor is connected in series with the coil. SOLUTION : XL 4 In case of a coil as Z  R 2   2 L2 X L   L, L    0.08  But as  50 V V Now when the capacitor is connected to the above i.e., I  2 2 2 Z circuit in series, R  L 1 1 So when DC is applied  As X C  V C 50  2500  106 I R 103   8 12 i.e., R   3 125 4 and when AC is applied, So, Z  R 2   X L  X C  2 V 2 I  32   4  8   5 Z V 12  V   12  I    2.4 A  5  i.e., Z      and hence,  Z 5  I   2.4  So, Pav.  Vrms I rms cos  or R 2  X L2  52 (As Z  R 2  X L2 ) R  I  Z  I    rms rms   So, X L2  52  R 2  52  32  4 2 Z i.e., XL = 4

2 2 i.e., Pav.  I rms R   2.4   3  17.28W

Example - 6 For a resistance R and capacitance C in series, the impedance is twice that of a parallel combination of the same elements. What is the frequency of applied emf ? SOLUTION : As shown in figure, in case of series combination,

12

2   1   2 2 2 Zs  R  X C  R       C     In case of parallel combination,

 1  1   Z P  R 2  X C  1

i.e.,


  

12 2


  

15

Vidyamandir Classes

AC Circuits i.e.,

R

ZP 

2 2 2



1  C R

For

 2 R 2C 2  1  2C 2 1  RC  f  2 RC

ZS = 2ZP 

Z S2  4 Z P2

 4

R2 1   2 R 2C 2

Example - 7

In the given circuit, AC source has   100 rad / s. Considering the inductor and capacitor to be ideal, find : (a) The current through the circuit, I (b) The voltage across 100 resistor (c)

The voltage across 5 resistor

SOLUTION : (C)

I2 

V ( L)2  R22

 0.2 2

I 2 lags behind V by 45 as  L  R2 L  R2 So I1 leads I 2 by 45  45  90  I1 

V 2

 0.1 2

 1  2   R1  c 

1 R I1 Lead V by 45 as c

I  I12  I 22  (0.1 2)2  (0.2 2) 2

 0.1 10  0.316 P.D. across R1  100  0.1 2  10 2 V P.D. across R2  50  0.2 2  10 2 V

NOW ATTEMPT OBJECTIVE WORKSHEET TO COMPLETE THIS EBOOK

16



Vidyamandir Classes

AC Circuits

THINGS TO REMEMBER 1.

Most of the electric power generated and used in the world is in the form of a.c. i.e., alternating current. The magnitude of alternating current changes continuously with time and its direction is reversed periodically. It is represented by I  I 0 sin  t In general,

or I  I0 cos t

I  I 0 sin(t   ) V  V0 sin(t   )

2.

Average Current (Mean Current) As we know an alternating current is given by i  i0 sin(t   )

…(i)

The mean or average value of ac over any time t is given by t

 i dt iavg  0t

0

 i dt 0

3.

R.M.S. Value of Alternating Current T

i

2

dt

i 2avg  0T

 dt 0

i irms  0 2

4.

RC Circuit VR  I m R cos t


Things to Remember

17

Vidyamandir Classes

AC Circuits

VC  

1   I m cos   t   C 2 

V (t )  I m R 2 

1 2 2

cos(t   )

 C

V V 1 Z  m  rms  R 2  I m I rms  2C 2

  tan 1

5

1  RC

RL Circuit VR  I m R cos t

  VL   L I m cos     2  V (t )  I m R 2   2 L2 cos(t   )

  tan 1

L R

V V Z  m  rms  R 2   2 L2 I m I rms

6.

LCR Circuit (a)

L 

1 C 2

1   V (t )  I m R 2    L   cos(t   ); C   1  L   C   tan 1  R  

18

Things to Remember

    



V leads I by  .


Vidyamandir Classes

(b)

L 

AC Circuits

1 C 2

 1  V (t )  I m R 2     L  cos(t   );  C   1   L     tan 1  C  R    

(c)

L 

V lags behind I by  .

1 C

V (t )  I m R cos t

V and I are in same phase.

1   Z  R2    L  C  

7.

or

 1  Z  R2   L  C 

2



ZR

Average Power across R I2 2 Pav  m R  I rms R 2 Average Power across L Pav  0 Average Power across C Pav  0

8.

Resonance in LCR Circuit 1 and Z = R, it is called resonance condition. L Resonance Frequency :

When  L 

0 

1 LC



v0 

1 2 LC


Things to Remember

19

Vidyamandir Classes

AC Circuits (a)

Resonance occurs when X L  XC  0

or



1 LC V at resonance. R

(b)

Current reaches a maximum value of

(c)

V2 Power dissipated is maximum and is rms (current and power are maximum because Z is minimum). R 2 Vrms Vrms I rms   , P R Z Z2

(d)

9.

Current is in phase with voltage (  0).

Parallel AC Circuits RC Circuit I 2  I C2  I R2

(rms values)

V I  m sin(t   ) z

1  Z

1 R

2



1 X c2

  tan 1 R c

10.

L-R Circuit I 2  I c2  I R2 (rms values)

V I  m sin( t   ) z

20

  tan 1

R L

1  Z



1 R

2

1 X L2

Things to Remember


Vidyamandir Classes

11.

AC Circuits

LCR Circuit (A)

IC  I L (rms) V I  0 sin(t   ) Z

1 1  I I X XL tan   C L  C 1 IR R 1  Z

(B)

 1 1     R2  X C X L  1

2

IC  I L (rms) V I  0 sin(t   ) Z

1 1  I I X XC tan   L C  L 1 IR R 1  Z

(C)

 1 1     R2  X L X C  1

2

IC  I L (rms) V I  0 sin t ,   0 Z Z=R


Things to Remember

21

Vidyamandir Classes

My Chapter Notes


Vidyamandir Classes

Illustration - 1


Vidyamandir Classes


Vidyamandir Classes


AC Circuits.pdf

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