Phy 1 (2)

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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 = =

Q.2.

(d.c current)2  (rms value of a.c.current)2 (10)2  (5 / 2 )2 =

15 2

.

XL = 2fL = 2  50  0.7 = 220 . R 2  XL2 = 220 2 

z=

Power factor, cos  =

220 1 R = = . z 220 2 2



Q.3.

(a) v avg = veff =

1 v v v m sin d  m [  cos ]0  m 2 0 2 



  v2 1   ( v m sin  )2 d  0  m 2  0 4 



vm 2 (b) When switch is closed 2 & 3 – series  2 + 3 = 5 and 6 & 9 - series  6 + 9 = 15 5  12 60 5 & 12 – parallel   5  12 17 60 60 315 15 & series  15   17 17 17 10 5 & 10 parallel  3 10 315 (10 / 3)  (315 / 17) &  are parallel  = 2.825H 10 315 3 17  3 17 1 I (t) = E1  E 2  R 1 = (25 3 + 25 6 sin t) 50 3 I (t) = (1 + 2 sin t) 2 Heat produced in one cycle of AC.

 veff =

Q.4.

=

2 /  2

 I ( t )Rdt 

0

3  50 4

2 / 

 1  2 sin

0

2



t  2 2 sin t dt

http://www.rpmauryascienceblog.com/ 75  2 2  3   J 2     2 No. of cycle in 14 minute is N = 14  60  50 3 Total heat produced =  14  60  50 2 Total heat produced = 3/2  14  60  50 = 63000 J

=

Q.5.

Q.6.

(a) 0 =

1 LC

1



iC

50  10  2  200  10  6 2  0 = 100  (= frequency of impressed voltage) hence net current through resistance is zero. (b) v = 5 sin 100 t.

Comparing, E = 100

2 sin (100 t) with E = Emax sin t,

we get, Emax = 100 2 V and  = 100 Emax (rad/sec.) 1 1 XC =   10 4  C 100  10  6 as ac instrument reads rms value, the reading of ammeter will be, E E Irms = rms  max  10 mA 2 2 Xc

Q.7.

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

. . . . . (i) . . . . . (ii) . . . . . (iii)

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

1 2

vi

iL

http://www.rpmauryascienceblog.com/ (b)

Q.8.

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]

In one complete cycle Iav = 0 /2

 0

Irms =

 2I0   t      

2

/2

 dt 0

I0

Irms = Q.9.

3

.

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.  40 -4 -t/2 i= e  t / 2 = - 2  10 e amp. 0.2  106 where –ve sign indicates current is flowing in opposite direction. to our convention. 

 Wdissipated =





i2Rdt  ( 4  10  8 )(2  10 5 )e  t dt



0

=8

 

 10-3 e t 0

0

= 8  10-3 J.

(80)2  (60)2  100 volt,

Q.10. v = p.f. =

80 = 0.8 lagging 100

Q.11. The rms value of the current is 1.5 mA 1 1    6.67k C 300  0.5  106 20 The rms voltage across the capacitor is 1.5  103   103  10 V 3 The impedance of the circuit is

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

|Z| =

Q.12. 2f =



10  103



2

2

400  20  = 1.2  104   103  = 103  100  3 9  

1 1   LC (0.5  10 3  5  106

http://www.rpmauryascienceblog.com/ 10 4 Hz  1 2 1 2 cV  Li 2 2 5  10-6  4  104 = 0.5  10-7 i2 i = 20 Amp.

 f=

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 di L  iR  0 dt At t = 0, i = 0, di 60 - 0.008 0 dt di 60 = 7500 A/s  dt 0.008 di Where = 500A/s, an equation yields, dt 60 - (0.008)(500) - 30i = 0 30 i = 60 - 4  i = 1.867 A di For the final current = 0, and dt L(0) - 30 IF = 0 IF = 2A Q.16. Applying KVL to the circuit at time 't' 0.2 di 2i + = 20 10 dt 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. Q.18. Constant value in the cycle therefore

http://www.rpmauryascienceblog.com/ vavg = v0 vrms = v0

(80)2  (60)2  100 volt,

Q.19. v =

80 = 0.8 lagging 100

p.f. =

Q.20. When connected to d.c. source V 12 R=  = 3  …(1) I 4 When connected to a.c. source V 12 Z = rms  = 5  …(2) I rms 2.4 2

R 2  L  …(3)

Z=

From (1), (2) and (3) L = 0.08 H with condenser

1   R   L   C  

Z =

=

2

2

1   3   50  0.08  6  50  2500  10  

2

2

=5

R  0.6 Z'  P = Vrms.Irms cos  = 12  2.4  0.6 = 17.8 volt cos  =

L 100  10 3 = 2  10-3 sec.  R 50 v   = I0 (1 – et/) = (1  e t /  ) R

Q.21. (a)  L =

0.001

 dI v t /  100 210  3  e = . e 3 dt .R 2  10  50 = 103 (0.606) = 606 amp/sec. 

(b) Heat produced, H = I20 ( t / )2Rdt 

 0

 Irms =

(1/ 3)I20R R



I0 3

.

Q.22. Let effective resistance is r.

12 I0R 3

http://www.rpmauryascienceblog.com/ 3 =

vc = 1

Vc rI 3 rI

… (i)

vc (rI  10I) 3 3r = r + 10 r=5  capacitive nature of box.

=

Q.23. The angular resonance frequency of the circuit is given by: 1 1 =   104 rad / s 3 6 LC 10  10  10 At a frequency that is 10% lower than the resonance frequency, the reactance XC, of the capacitor is: 1 1 XC =    111 C 10 4  10 6  0.9 And that of the inductance is XL = L = 104  0.9  102 = 90 The impedance of the circuit is |Z| =

2

R 2   XL  XC  

2

32   21   21.2

15 A  0.7A 21.2 The average power dissipated is: 12 1 2 i R   0.7   3  0.735 W 2 2 The average energy dissipated in 1 cycle is 2 0.735  J 0.9  104 or 5.13  104 J

The current amplitude, i =

Q.24.

1 2 1 2 Li0  Cv 0 2 2 C i0 = v 0 L q 5.0  10 6 v0 = 0   1.25  10  2 volt C 4.0  10  4 4.0  10 4 -2  8.33  10  4 A i0 = 1.25  10 0.09

umax =

1 2 1 2 Li0  Cv 0  3.125  10  8 J 2 2 2

3.125  10

-8

 8.33  10 4  1 1 q2 = (0.09)    2 2 2 (4.0  104 )  

q = 4.33  10-6 C.

http://www.rpmauryascienceblog.com/ 1

Q.25. At resonance frequency f0 =

2 LC i0 = v0/R From given condition 1 v0 v0  2 R 1 2 R 2  ( L  ) C 1 L   R. C 1 2f1L R … (i) 2f1C 1 2f2L  R … (ii) 2f2C solving equations (i) and (ii) we get, R f1 – f2 = . 2L T

 Q.26. vrms =

0

T

 dt 0

T

vmean = vmean =

2

T

V 2dt

V 2   t dt T 0

 

T

 dt

=

V 3

0

V

 T tdt 0

T

V 2

Q.27. Impedance Z=

I cos 

(R 2  2L2 ) 

= (50) 2  (2  50  0.7) 2 I sin  = 225.4 ohms V 200  Current I   = 0.887 amp Z 225.4 I L 2  50  0.7 and tan  = = 4.4  R 50 or  = 77 12’.  Power component of current IR = I cos  = 0.887cos 77 12’ = 0.197 amp. and wattless components of the current IL = I sin= 0.887sin7712’ = 0.865 amp.

V

http://www.rpmauryascienceblog.com/ 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  dQ  2 Rate of production of heat    i1R1  24.5 (J)  dt 



E 1  e Rt / L = i0 1  eRt / L R Rt i = i0/2  = ln 2 L t = 0.0014 sec. 1 2 Energy stored = Li 2 = 0.00045 (J).

(ii) i =







R3 2 (1) I1R1 I1 I2

R1

E1

2

12V E2

S A

3V

(I1-I2) (2) R2 B 2

I2