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 value5A.

R.M.S. value = 22 )current.c.aof valuerms()currentc.d(

= 22 )2/5()10( =2

15.

Q.2. XL = 2fL = 2 50 0.7 = 220 .

z = 2L2 XR = 220 2

Power factor, cos =z

R =

2220

220 =

2

1.

Q.3. (a) vavg =

m00

mm

v]cos[

2

vdsinv

2

1

veff =4

v0d)sinv(

2

1 2m

0

2m

veff =2

vm

(b) When switch is closed2 & 3 – series 2 + 3 = 5 and 6 & 9 - series 6 + 9 = 15

5 & 12 – parallel 17

60

125

125

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) = 21

EER

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 dttsin22tsin21504

3Rdt)t(I

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= J2

322

2

75

No. of cycle in 14 minute is N = 14 60 50

Total heat produced = 2

3

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

Q.5. (a) 0 =62

21020010

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 =4

610

10100

1

C

1

as ac instrument reads rms value, the reading of ammeter will be,

Irms = mA10X2

E

2

E

c

maxrms

Q.7. Charged stored in the capacitor oscillates simple harmonically asQ = Q0 sin (t )

Q0 = 2 10-4 C

=63 105102

1

LC

1

= 104 s-1

Let at t = 0, Q = Q0 then,Q(t) = Q0 cos t . . . . . (i)

i(t) = dt

dQQ0sin t . . . . . (ii)

anddt

)t(di = - Q0

2 cos t . . . . . (iii)

(a) Q = 100 C or2

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


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(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 cycleIav = 0

Irms =

2/

0

2/

0

2

0

dt

tI2

Irms =3

I0 .

Q.9. When ‘S’ is closed Applying Kirchoff’s law0 = iR + (1/C) idti = - (q0 / RC) e

-t/RC = - (40/R)e-t/RC but RC = 0.2 106 10 10-6 = 2 sec.

i =6102.0

40

2/te = - 2 10-4 e-t/2 amp.

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

Wdissipated =

0

t58

0

2 e)102)(104(Rdti dt

= 8 10-3 0te = 8 10-3 J.

Q.10. v = 100)60()80( 22 volt,

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| = 61 1

6.67kC 300 0.5 10

The rms voltage across the capacitor is 1.5 103 320

10 10 V3

The impedance of the circuit is

|Z| = 2

23 32010 10 10

3

= 3

40010 100

9 = 1.2 104

Q.12. 2f =3 6

1 1

LC (0.5 10 5 10


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f =410

Hz

2 21 1cV Li2 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= 0.014200 F

For minimum impedance, L = 1/ C

L= 0.36 mH.

Q.15. The loop equation

L 0iRdt

di

At t = 0, i = 0,

60 - 0.008 0dt

di

008.0

60

dt

di = 7500 A/s

Wheredtdi = 500A/s, an equation yields,

60 - (0.008)(500) - 30i = 030 i = 60 - 4 i = 1.867 A

For the final currentdt

di = 0, and

L(0) - 30 IF = 0IF = 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 inductorIC = Imax sin (t + /2) = +0.3 cos t for capacitorso current drawn from the source.I = IL + IC = - 0.1 cos tImax = |I0| = 0.1 A.

Q.18. Constant value in the cycle therefore


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

Q.19. v = 100)60()80( 22 volt,

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

Irms

rms

12

2 4. = 5 …(2)

Z = R L22

…(3)

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

Z =2

2

C

1LR

=

2

6

2

10250050

108.0503

= 5

cos =R

Z'. 0 6

P = Vrms.Irms cos = 12 2.4 0.6 = 17.8 volt

Q.21. (a) L =50

10100

R

L 3 = 2 10-3 sec.

= I0 (1 – et/) = )e1(

R

v /t

/t

eR.

v

dt

dI

=

3102

001.0

3 e.50102

100

= 103 (0.606) = 606 amp/sec.

(b) Heat produced, H =

RI3

1Rdt)/t(I 20

0

220

Irms =3

I

R

RI)3/1( 020

.

Q.22. Let effective resistance is r.


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3 =rI

Vc … (i)

vc = 3 rI

3

1

= )I10rI(

vc

3r = r + 10r = 5 capacitive nature of box.

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

= 43 6

1 110 rad /s

LC 10 10 10

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

XC = 4 61 1

111C 10 10 0.9 And that of the inductance isXL = L = 10

4 0.9 102 = 90 The impedance of the circuit is

|Z| = 2 22 2

L CR X X 3 21 21.2

The current amplitude, i =15

A21.2

0.7A

The average power dissipated is:

221 1i R 0.7 3 0.735 W

2 2

The average energy dissipated in 1 cycle is

4

20.735 J

0.9 10

or 5.13 104 J

Q.24. 2020 Cv

2

1Li

2

1

i0 = v0 L

C

v0 = 24

6

0 1025.1100.4

100.5

C

q

volt

i0 = 1.25 10-2

44

1033.809.0

100.4

A

umax = J10125.3Cv2

1Li

2

1 820

20

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.


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Q.25. At resonance frequency f 0 =LC2

1

i0 = v0/RFrom given condition

21

22

00

)C

1L(R

vRv

L -1

RC

.

2f 1L - RCf 2

1

1

… (i)

2f 2L - RCf 2

1

2

… (ii)

solving equations (i) and (ii) we get,

f 1 – f 2 =L2

R

.

Q.26. vrms =

T

0

T

0

2

2

T

0

T

0

2

dt

dttT

V

dt

dtV

=3

V

vmean = T

tdtT

VT

0

vmean =2

V

Q.27. Impedance Z= )LR( 222

= 22 )7.0502()50(

= 225.4 ohms

Current4.225

200

Z

VI = 0.887 amp

and tan = 507.0502

R

L

= 4.4or = 77 12’.

I cos

I sin

I

V

Power component of current IR = I cos = 0.887cos 77 12’ = 0.197 amp.and wattless components of the current IL= I sin= 0.887sin7712’ = 0.865 amp.


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http://www.rpmauryascienceblog.com/ Q.28. (i) Applying Kirchhoff's law to meshes (1) and (2)

2I2 + 2I1 = 12 - 3 = 9and 2I1 + 2(I1 -I2) = 12solving I1 = 3.5 A, I2 = 1AP.D.between AB = 2I

2 + 3 = 5 volt

Rate of production of heat 5.24Ridt

dQ1

21

(J) R2 B

2

R3

R1

2 12V

E1

E2

AS

3V

I1R1

(I1-I2)(1)

(2)

I1

I2 I2

2

(ii) i = L/Rte1R

E = i0 L/Rte1

i = i0/2 L

Rt = ln 2

t = 0.0014 sec.

Energy stored =2

1 Li2

= 0.00045 (J).

Phy 1 (2)

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