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Solution

Q.1. Given current in a mixture of a d.c. component of 10 A and an alternating current of maximum value 5A.

R.M.S. value = (d.ccurrent)2(rmsvalueofa.c.current)2

= (10)2(5/ 2)2 = 2 15 . Q.2. XL = 2fL = 2  50  0.7 = 220 . z = L2 2 X R  = 220 2  Power factor, cos  =

z R = 2 220 220 = 2 1 . Q.3. (a) vavg =           

m 0 0 m m v ] cos [ 2 v d sin v 2 1 veff = 4 v 0 d ) sin v ( 2 1 m2 0 2 m             

  veff = 2 vm

(b) When switch is closed

2 & 3 – series  2 + 3 = 5 and 6 & 9 - series  6 + 9 = 15 5 & 12 – parallel  17 60 12 5 12 5    15 & 17 60 series  15  17 315 17 60  5 & 10 parallel  3 10 3 10 & 17 315  are parallel  17 315 3 10 ) 17 / 315 ( ) 3 / 10 (   = 2.825H Q.4. I (t) =

E1 E2

R 1  = 50 1 (25 3 + 25 6 sin t) I (t) = 2 3 (1 + 2 sin t)

Heat produced in one cycle of AC.

=

          / 2 0 2 / 2 0 2 50 1 2sin t 2 2sin tdt 4 3 Rdt ) t ( I

(2)

= J 2 3 2 2 2 75            

No. of cycle in 14 minute is N = 14  60  50 Total heat produced =

2 3

 14  60  50

Total heat produced = 3/2  14  60  50 = 63000 J

Q.5. (a) 0 = 6 2 2 10 200 10 50 1 LC 1       

0 = 100  (= frequency of impressed voltage)

hence net current through resistance is zero. (b) v = 5 sin 100 t.

iC

vi

iL

Q.6. Comparing, E = 100 2 sin (100 t) with E = Emax sin t,

we get, Emax = 100 2 V and  = 100 Emax (rad/sec.)

XC = 6 104 10 100 1 C 1      

as ac instrument reads rms value, the reading of ammeter will be, Irms = 10mA X 2 E 2 E c max rms  

Q.7. Charged stored in the capacitor oscillates simple harmonically as Q = Q0 sin (t  ) Q0 = 2  10-4 C  = 6 3 5 10 10 2 1 LC 1       = 104 s-1 Let at t = 0, Q = Q0 then, Q(t) = Q0 cos t . . . (i) i(t) =  dt dQ Q0sin t . . . (ii) and dt ) t ( di = - Q0 2 cos t . . . (iii) (a) Q = 100 C or 2 Q0 from (i) equation cos t =

2 1

from equation (iii)

      dt di = Q0 2 cos t = 2  10-4 (104)2  2 1 dt di = 104 A/s

(3)

(b) Q = 200 C or Q0 then cos t = 1 i.e., t = 0, 2 ………

from equation (ii) I(t) = - Q0  sin t

I(t) = 0

[sin 0 = sin 2 = 0]

Q.8. In one complete cycle Iav = 0 Irms =

               2 / 0 2 / 0 2 0 dt t I 2 Irms = 3 I0 . Q.9. When ‘S’ is closed Applying Kirchoff’s law 0 = iR + (1/C)  idt i = - (q0 / RC) e-t/RC = - (40/R)e-t/RC but RC = 0.2  106  10  10-6 = 2 sec.  i = 6 10 2 . 0 40   t/2 e = - 2  10-4 e-t/2 amp.

where –ve sign indicates current is flowing in opposite direction. to our convention.  Wdissipated =

       0 t 5 8 0 2 e ) 10 2 )( 10 4 ( Rdt i dt = 8  10-3

 

et 0 = 8  10-3 J. Q.10. v = (80)2(60)2 100volt, p.f. = 100 80 = 0.8 lagging

Q.11. The rms value of the current is 1.5 mA

The impedance of the capacitor is given by |ZC| = 6

1 1

6.67k C  300 0.5 10   

  

The rms voltage across the capacitor is 1.5  103  20 103 10 V

3   The impedance of the circuit is

|Z| =

2 2 3 20 3 10 10 10 3       = 103 100 400 9   = 1.2  104 Q.12. 2f = 3 6 1 1 LC  (0.5 105 10   

(4)

 f = 4 10 Hz  2 2 1 1 cV Li 2 2 5  10-6  4  104 = 0.5  10-7 i2 i = 20 Amp. Q.13. XL = L =   Impedance =Z = 3.3  Power factor = R/ Z = 1/3.3 Q.14. C= 0.014200  F

For minimum impedance, L = 1/ C L= 0.36 mH.

Q.15. The loop equation

L iR 0 dt di   At t = 0, i = 0, 60 - 0.008 0 dt di  008 . 0 60 dt di  = 7500 A/s Where dt di

= 500A/s, an equation yields, 60 - (0.008)(500) - 30i = 0 30 i = 60 - 4  i = 1.867 A For the final current

dt di

= 0, and L(0) - 30 IF = 0

IF = 2A

Q.16. Applying KVL to the circuit at time 't'

2i + dt di 10 2 . 0 = 20

solving this differential equation : i = 10 (1+9e-100t )

Q.17. Current in inductance will lag the applied voltage while across the capacitor will lead.

IL = Imax sin (t - /2) = - 0.4 cos t for inductor

IC = Imax sin (t + /2) = +0.3 cos t for capacitor

so current drawn from the source. I = IL + IC = - 0.1 cos t

Imax = |I0| = 0.1 A.

(5)

vavg = v0 vrms = v0 Q.19. v = (80)2(60)2 100volt, p.f. = 100 80 = 0.8 lagging

Q.20. When connected to d.c. source

R = V I 

12

4 = 3  …(1) When connected to a.c. source Z = V I rms rms  12 2 4. = 5  …(2) Z = R2  

 

L 2 …(3) From (1), (2) and (3) L = 0.08 H with condenser Z = 2 2 C 1 L R           = 2 6 2 10 2500 50 1 08 . 0 50 3            = 5  cos  = R Z' 0 6 .  P = Vrms.Irms cos  = 12  2.4  0.6 = 17.8 volt Q.21. (a) L = 50 10 100 R L  3  = 2  10-3 sec.   = I0 (1 – et/) = (1 e ) R v t /      e t/ R . v dt dI = 210 3 001 . 0 3 .e 50 10 2 100      = 103 (0.606) = 606 amp/sec. (b) Heat produced, H =

    R I 3 1 Rdt ) / t ( I 20 0 2 2 0  Irms = 3 I R R I ) 3 / 1 ( 20 0    .

(6)

3 = rI Vc … (i) vc = 3 rI 3 1 = ) I 10 rI ( vc  3r = r + 10 r = 5 

 capacitive nature of box.

Q.23. The angular resonance frequency of the circuit is given by:

 = 4 3 6 1 1 10 rad / s LC  10 10 10   

At a frequency that is 10% lower than the resonance frequency, the reactance XC, of the capacitor

is: XC = 4 6 1 1 111 C 10 10 0.9     

And that of the inductance is XL = L = 104  0.9  102 = 90

The impedance of the circuit is |Z| = 2

2 2

 

2

L C

R  X X  3  21  21.2

The current amplitude, i = 15 A

21.2  0.7A The average power dissipated is:

2

2

1 1

i R 0.7 3 0.735 W

2  2  

The average energy dissipated in 1 cycle is

4 2 0.735 J 0.9 10    or 5.13  104 J Q.24. 20 Cv20 2 1 Li 2 1  i0 = v0 L C v0 = 4 2 6 0 1.25 10 10 0 . 4 10 0 . 5 C q         volt i0 = 1.25  10 -2 4 4 10 33 . 8 09 . 0 10 0 . 4      A umax = Cv 3.125 10 J 2 1 Li 2 1 2 8 0 2 0     3.125  10-8 = 2 4 2 4 1 8.33 10 1 q (0.09) 2 2 2 (4.0 10 )            q = 4.33  10-6 C.

(7)

Q.25. At resonance frequency f0 = LC 2 1  i0 = v0/R

From given condition 2 1 2 2 0 0 ) C 1 L ( R v R v      L - 1 R C   . 2f1L - R C f 2 1 1   … (i) 2f2L - R C f 2 1 2    … (ii)

solving equations (i) and (ii) we get, f1 – f2 = L 2 R  . Q.26. vrms =

       T 0 T 0 2 2 T 0 T 0 2 dt dt t T V dt dt V = 3 V vmean = T tdt T V T 0

vmean = 2 V Q.27. Impedance Z= (R2 2L2) = (50)2 (2500.7)2 = 225.4 ohms  Current 4 . 225 200 Z V I  = 0.887 amp and tan  = 50 7 . 0 50 2 R L     = 4.4 or  = 77 12’. I cos  I sin  I V 

 Power component of current IR = I cos  = 0.887cos 77 12’ = 0.197 amp. and wattless components of the current IL

(8)

Q.28. (i) Applying Kirchhoff's law to meshes (1) and (2)

2I2 + 2I1 = 12 - 3 = 9

and 2I1 + 2(I1 -I2) = 12

solving I1 = 3.5 A, I2 = 1A

P.D.between AB = 2I2 + 3 = 5 volt

Rate of production of heat iR 24.5 dt dQ 1 2 1         (J) R2 B 2 R3 R1 2 12V E1 E2 A S 3V I1R1 (I1-I2) (1) (2) I1 I2 I2 2 (ii) i =

1 e Rt/L

R E  = i0

1eRt/L

i = i0/2  L Rt = ln 2 t = 0.0014 sec. Energy stored = 2 1 Li2 = 0.00045 (J).

References

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