EDC - Important Problems

8/10/2019 EDC - Important Problems

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UNIT-1 SOLVED PROBLEMS1.3. An electron with a velocity of 3 x 105ms-1enters an electric field of 910 v/m making an

angle of 060 with the positive X direction. The direction of the electric field is in the positive

Y direction. Calculate the time required to reach the maximum height.

[Nov 04, Jun 05, May 06, Aug 07]

SOLUTION:Given V= 53 10

sec.m ;E= 950V / m.; = 60

Component of velocity in y direction = V Sin

=

53 10 0.866 ; Electron feels a retarding force due to electronic field andat maximum height final velocity will be zero.

( )0 ( ) ua

V u at or at or t ;19

14 2

31

1.6 10 9101.6 10 sec

9.1 10

eEa m

m

For a projectile time of ascent =sinV

a

=

59

14

3 10 0.8661.62 10 .sec.

1.6 10

1.4) The magnetic flux density 2B=0.02 wb/m and electric field strength 5E=10 v/m are uniform

fields, perpendicular to each other. A pure source of an electron is placed in a field. Determine the

minimum distance from the source at which an electron with 0V will again have 0V in its trajectory

under the influence of combined electric and magnetic fields.[May 03, Nov 05, May 06, Aug

06, May/Aug 07]

SOLUTION:

Angular velocity of rolling circle w,

eB

m B

velocity of translation of center of the

rolling circle.

2

B

u BQ

ew B e mmB

=5

6

11 2

101420 10

1.76 10 (0.02)

6 62 2 1420 10 8922 10 0.8922 .Z Q cm

1.5) Two parallel plates of a capacitor are separated by4cms. An electron is at rest initially at the bottom

plate. Voltage is applied between the plates, which increases linearly from 0v to 8v in 0.1 m.sec . Ifthe top plate is +ve, determine

i. The speed of electron in 40 n.sec.

ii. The distance traversed by the electron in 40 n.sec.[Nov 03, May 06, Aug 06, May 07]

SOLUTION:


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xv =17 21.76 10 t ; Distance

t

xdt

o

x v

3

17 2 17 31.761.76 10 10 40 103

t

o

x t 10 637546 10 3.754 10 3.754 m

When9,40 10t

Velocity

217 9 1

1.76 10 40 10 2816 10 281.6 /xv m Sec

1.6) In a electro static deflecting CRT the length of the deflection plates is 2 cm , and spacing

between deflecting is 0.5 cm, the distance from the centre of the deflecting plate to the screen is

20 cm . The deflecting voltage is 25 V . Find the deflection sensitivity, the angle of deflection and

velocity of the beam. Assume final anode potential is1000 V . [Apr/May 03, May 06,

Aug 06]

SOLUTION:

D=2

d

a

lLV

dV;

2a

lLS

dV

l= 2 cm ?D

d = 0.5 cm;

L = 20 cm dV = 25 V

aV = 1000 V.

2 a

lLS

dV =

2 2

2

2 10 20 10

2 0.5 10 1000

= 540 10 0.04

cm/V2

2

. 0.04 25 10tan 0.05

20 10

dS VD

L L

tan 0.05 OR 2.86

0

Velocity2 a

axeV

vm

= 11 72 1.76 10 1000 1.876 10 /sec.m

Resultant velocity7 71.876 10 1.876 10

.9992.80

axr o

vv

Cos Cos

= 1.8778 x 107m/sec.

1.7) When an electron is placed in a magnetic field with a period of rotation 1235.3

T= 10 secB

so

that the trajectory of an electron is a circle?

a. What is the radius described by an electron placed in a magnetic field, perpendicular

to its motion, when the accelerating potential is900 v and 2B=0.01 wb/m .

b. What is the time period of rotation? [May 06]

SOLUTION

Given 1235.3

10 sec.TB

Va= 900 V; B= 0.01 Wb/m; T = ?, r = ?

1235.3 10 sec.TB

=12

1235.3 103530 10 .

0.01Sec


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2 aeVmVr V

eB m ; 11 62 1.76 10 900 17.79 10 / sec.V m

65

11

17.79 101010.8 10 10.11

1.76 10 0.01r mm

1.7) The distance between the plates of a plane parallel capacitor is 1 cm. An electron

starts at rest at the negative plate. If a direct voltage of 1000V is applied how

long will it take the electron to reach the positive plate?

[May 02, Nov 04]

- +

- +

- - 1000 V

1 Cm.

Given: S= 1 cm ; 0u ; 1000aV V ;

Final Velocity2eVa

mV 112 1.76 10 1000 = 71.876 10 / secm

Electron follows Newton Laws of motion.

V u at = at or va t

2 21 1 12 2 2

/s ut at at v t 2 / 2t vt ;2

9

7

2 2 101.066 10 1.066 sec.

1.876 10

St n

v

1.9) . An electrostatic cathode ray tube has a final anode voltage of 600V. The

deflection plates are 1.5cm long and 0.8cm apart. The screen is at a distance of

20cm from the centre of plates. A voltage of 20V is applied to the deflection

plat es . Cal cula te .

a. Velocity of electron on reaching the field

b. Accelerati on due to def lect ion fi eld

c. Deflection produced on the screen In cm.

d. Deflection sensitivity in cm/V. [Jun 01]

SOLUTION

600 ; 1.5 ; 0.8 ; 20 ; 20V l cm d cm L cm V V a d V

2

lLVdD

dVa

Part (a) velocity of electron reaching the field

112 2 1.76 10 600aoxeV

v

m

Vox= 14532721 m/sec.

Part (b) Acceleration due to electric field.11

14 2

2

1.76 10 204.4 10 / sec

0.8 10

d

y

eVa m

md

Part (c) deflection produced in cm.2 2

3

2

1.5 10 20 10 206.25 10 .625

2 2 0.8 10 600

d

d

lLVD m cm

dV

Part (d) deflection sensitivity in cm.

0.6250.03125 / .

20d

DS cm V

V


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1.10). The electrons emitted from the thermonic cathode of a cathode ray Gun are accelerated by a

potential of 400V. The essential dimensions are L=19.4cm, l=1.27cm and d=0.475 cm. Determine

deflection sensitivity. What must be the magnitude of transverse magnetic field acting over the whole

length of the tube in order to produce the same deflection as that produced by a deflection potential of

30V? [Nov 04]

SOLUTION

400 ;a

V V 19.4 ;L cm 1.27l cm

;d

DS

V ;

?B for 30

dV V

Electro static deflection: D=2

d

a

lLV

dV;

2 2

2

1.27 10 19.4 10 300.019 1.9

2 0.475 10 400

X X X XD m cm

X X X

0.019 46.3 10 / 0.0630

ds m V cm

Vd

Electro magnetic deflection:

lLB eD

mVa ;

1112 2 2

0.019 400

. 1.76 101.27 10 19.4 10

2

a

em

D V XB

lL XX X X

40.38 5.199 10730.88

B X wb/ 2m

1.11) In a parallel place diode, the cathode and anode are spaced 5mm apart and the anode is kept at200V d.c. with respect to cathode. Calculate the velocity and the distance traveled by an electron after

a time of 0.5ns, when:

(a)The initial velocity of an electron is zero and(b)The initial velocity is 2X10

6m/s in the direction towards the anode.

SOLUTIOND=5mm ?; 0.5v t ns Vd=200V when (a) u=0, (b) u=2x10

6m/s

Electrons follow Newtons laws of motion v=u+at

Where .e dm

va

d ; 11 15 2

3

2001.759 10 7.036 10 / sec

5 10a m

case(a) when initial velocity u=0

V=at = 7.036x1015

x0.5x10-9

= 3.518x106m/s =3518 km/sec.

Case(b) V=u + at = 2x106+ 3.518x10

6=5.518x10

6m/sec or 5518 km/sec.

Distance 21

2

S ut at

Case(a): u=0; 2

2 15 9 31 12 2

7.036 10 0.5 10 0.8795 10S at metres

Case(b): 6 9 3 3 32 10 0.5 10 0.8795 10 1 10 1.8795 10S x x x m 1.12) Calculate the deflection of a cathode ray beam caused by earth magnetic field. Assume that

the tube axis is normal to the field, of strength 0.6 Gauss. The anode potential is 400 V. Thedistance between the anode screen distances is 20cm.


5/12

AnodeGiven: B = 0.6 Gauss V = 400 Volts; screen distance = 20 cm.

To calculate:

Deflection D = ?

To find out D, we must know radius R.

R is given by,( )

( )

v mR

e B we know m, e & B, v to be found.

v is given by52 ( )5.93 10 a

eVav or V

m

Velocity of electron along x-ox is, 55.93 10 400vox = 5.93 x 105x 20 = 118.6 x 105m/sec.

= 1.19 x 107m ./sec.

B = 0.6G.; 1 Wb / m2 = 10

4Gauss

B = 0.6 x 10-4

Wb / m2 = 6.0 x 10

-5wb / m

2

R =7

11 5

. 1.19 10

( / ) 1.759 10 6 10

v m v

eB e m B

( e / m = 1.759 x 1011

)

= 0.112 x 10 = 1.12 m = 112 cm.

In PYZ, YZ2= PZ2+ PY2YZ = OP = 112 cm; PY = 20 cm.; PZ = OZOP (Deflection D = OP)

(YZ)2

= (PZ)2

+ (PY)2

1122

= (112D)2

+ 202

1122

= 1122

224D + D2

+ 400or D

2224D + 400 = 0

This is in the form of ax2+ bx + c = 0, quadratic equation. Where

2 4

2

b b acx

a

D can be found by above equation , As D = 1.8 cm.

MOTION OF ELECTRON IN VARYING ELECTRIC FIELD.

Example: - An electron starts at rest on one plate of a plane parallel capacitor whole plates are5cm., apart. The applied voltage is zero at the instant the electron is released, and it increases linearly

from zero to 10V in 0.1 sec.a) If the opposite plate is positive, what speed will the electron attain 50 n sec..

b) Where will it be at the end of this time?

c)

With what speed will the electron strike the positive plate?

Solution: -

a) Magnitude of field intensity will be gradually increasing with time and given by:

= (Voltage X time) / Distance7

2

1010

5 10

tX

X

=2 x 109t V/m

We know acceleration a = dv /d t. = f/m. = e/m.Or a = (1.76 x 10

11) ( 2 x 10

9t) = 3.52 x 10

20t m /sec

2

We know, v =t

o

a dt =203.52 10

l

o

X t dt = 1.76X1020

t2m/sec


6/12

At t = 50 n.sec. = 50 x 10-9

v = 1.76 x 1020

(50 x 10-9

)2 = 4.4 x 10

5m /sec.

b) Distance travelled 20 2 19 3

0

. 1.76 10 5.87 10

t t

o

x v dt t t

At t = 50 n sec., x = 5.87 x 1019

(50 x 10-9

)3 = 7.32 X 10

-3m = 0.732 cm.

c) To find the speed at which the electron strikes the positive plate, we must find the time

taken to reach the plate (i.e., to travel the distance x)

x = 5.87 x 10

19

t

3

or 3

19 19

0.05

5.87 10 5.87 10

xt

(distance x = 5 cm)t = 9.46 x 10-8sec.

1.14 Derive the expression for acceleration, velocity and displacement of a changed particle in anelectgric field

SOLUTION:

(i) F ma eE eEm

a --------(1);

(ii) Velocityt

o

v & ; ;( )dv dxdt dt

adt a v or ;

22

(2)dv d dxdt dt dt

d xadt

Substituting value of a from Eqn.(2) in to Eqn (1)=2

(3)2

d x eE

mdt

1

tt eEeEV adt dt t C o m m

o

(C1is constant)

Average velocity eEm

OR 1dx eE t Cdt m

At + =0; 0; 01x C (4)dx eE V tdt m

t

o

t

o

dx

dt

eEt

mdt

Distance2

(5)2

eEtx

m

2

22

eE tx C

m

If 2C is taklen as 0

Distance 2 52

eEx t

m

UNITIII SOLVED PROBLEMS

1 Q) A 150 15 volts (rms) ideal transformer is used with a FWR with diodes having fwd

drop of 1 volt. The load resistance in 100 and capacitor of 10,000 f is used on filter.

Calculate the Dc load current and voltage. (JNTU 2005)


7/12

Solution: -

4

DC

DC m

IV V

fc where f is power line frequency.

Vrms = 15V, Assume voltage drop (rms) across diode as

1V.

Vrms across load = 14V.

dc dcdc m

dc dc

dc

4170 I 4170 IV = V - = 19.8 -C 10,000

100 I ; 19.8 (or) I ; 0.198 Amp

V = 19.8 Volts.

Vm = Vrms 2 = 142 = 19.8 volts.

2 Q) A FWR is used to supply power to a 2000 load, choke of 20H and capacitor of 16f are

available. Compute ripple factor using filter 1 (i) one inductor (ii) one capacitor (iii)

single Ltype.

Solution: RL= 2000; C = 16f; L = 20H.

i) One Inductor Filter

32000 1 6.25 10 0.00616000 16000 20 160

LRrL

(ii) One capacitor filter:

32410 2410 75.31 10 0.07516 2000

L

rCR

(iii) Single LC Type Filter

0.83 0.830.0025

20 16

r

LC

(iv) Selection6

1 1

3330 3330325 10 .000325

16 16 20 2000L

rCC L R

3 Q) Design a full wave rectifier with an LC filter to provide 9V DC at 100mA with a max

ripple of 2%. Given line frequency f = 60Hz. (JNTU2000)

Solution: -Given VDC= 9V, f = 60Hz.; IL100mA, r = 2% = 0.02

To find: L, C.

0.83r when f = 60 Hz. (1) (1)

Else2 1 1

. .3 2 2 .

rwc wL

Where LC= RL/ 1130.

Or0.83

41.50.02

LC (2)

Critical Inductance (value of Inductance for which diode

conducts continuously)


8/12

S C

d S

V -VCurrent through the diode at any time

R +R

9

.; 901130 100

DCL

C L

DC

VRL R

I mA

Lc = 41.5 - (2)

3

41.5521

79.6 10C f

Transformer ratingVrms = ?; Diode ratings PIV= Vm ; current rating = load current

L = 796mH; C = 521f

4 Q) A FWR operating at 50 Hz i.e., to provide DC current of 50mA at 30V with a 80f, Ctypefilter. Calculate (i) Vm the peak secondary voltage of the transformer (ii) Ratio of surge to

mean currents of diode (iii) The ripple factor of the output.

(JNTU 2002)

Solution: -

Given f = 50 Hz, C = 80fIDC= 50mA To find Vm= ? ratio of surge to mean current

VDC= 30V.

Part (i) Vm= ?VDC= 0.636 Vm. Or Vm= VDC/0.636 = 30/0.636 = 47.17 volts.

Part ii Ratio of surge to mean current

----------------------------------------X-------------------------X--------------------------------------Surge Current

It is the current flowing through the diode, when the power supply is just switched on i.e., at

time t = 0+. At time t = 0+. Voltage across capacitor will be zero.

--------------------------------X-------------------------------------X --------- X---------------------

When Vc = 0, Id will be max = maxS m

d S d S

V V

R R R R

This is called surge current.

Mean diode currentS dc

d

d S

V V

I R R

Ratio of surge current to mean current =

d smd s m dc

R RV

R R V V

=47.17 47.17

2.7547.17 30 17.17

m

m DC

V

V V

Note: Designer must cater for 3 times the required average current.

Part III : Ripple factor2410

L

rCR

f = 60 Hz.


9/12

Or Ripple factor1

4 3 Lr

fcR

2300 3050

L

Vdc V R V

Idc mA = 600

6

6

1 100.006

16627687.754 3.50.80 10 600r

5 Q) For a FWR circuit AC voltage input to transformer primary is 115V. Transformer secondary

voltage is 50V, RL=25. Determinei) Peak DC component, RMS and AC component of load voltageii) Peak DC component, RMS and AC component of load current.

(June 2002)

Solution:

Given Vrms = 50V, to find Vm, Vr, Im, Ir.

RL= 252

rV

Vm Vdc

Part (i)

2

2; 2,50 70.7Rms RmsVm

V orVm V VoltsV

,rV

rVdc

r of FWR (without filter) = 0.48

Vdc = 0.637

Vm= 0.637x7.07=45 volts

Vr = , Vdc = 0.48 x 45 = 21.6 volts

Peak DC component = VDC + Ripple voltage = 45+21.6 = 66.6 volts

66.6V = = = 47 Volts2 2

rms peak

PartII

m

70.7I 2.828 .

25

21.60.864 .

25

L

L

VmAmps

R

VrIr Amps

R

Peak DC current component = 66.6/25 = 2.664 Amps.

Runs DC current component = 47/25 = 1.8 Amps.

6 Q) Calculate ripple factor of capacitor filter with peak rectified voltage of 20V and C = 50 f andIDC= 50mA. (June 2004)

Solution: -

Vm = 20V, Idc = 50mA, C = 50f. VDCVr Vm = 20V

4DC

IdcV Vm

fc

Suppose f = 50Hz, Then3

6

50 1020

4 50 50 10DC

V


10/12

=1000

20 20 5 15200

volts

15 150 1000300

50 50L

VDCR

Idc mA

61 1 10 100 1= 0.192

4 3 4 3 50 50 300 4 3 25 3 3 3Lr

fcR

1Q) Determine the quiescent current and collector to emitter voltage for germanium

transistor with = 50 in self-biasing arrangement draw the circuit with given component

value VCC= 20V, RC= 2K, RE= 100 , R2= 5K (Also find out stability factor).

(May 2005)Solution: -

For drawing DC load line we must knowa) VCEmax when Ic = 0; b) Ic max when VCE= 0

VCEmax = VCC= 20V

20max 9.52

2000 100

CC

C E C

C E

VI I I mA

R R

Quiescent VCEQ =max 20

102 2

CEV V

Quiescent ICQ =max 9.52

4.762 2

IcmA

Quiescent IBQ =4.76

0.095250

CI mA

; Quiescent IEQ= ICQ+ IBQ= 4.855mA.

Q) Find out stability factor of the circuit given below: (May 2005)

Stability factor of self-biased Circuit given by:

1

1

1

B

E

B

E

R

RS

R

R

;

1 2

1 2

BR RR

R R

5 50 50 4.5 45005 50 11

k k

4500 45

100

B

E

R

R

1 45

50 1 24.541 50 45

S


11/12

For the circuit shown, determine the value of Ic and VCE. Assume VBE = 0.7V and = 100

(Sep.06)

2

1 2

. 10 5 503.33

10 5 15

10 5 503.33 .

10 5 15

cc

in

th

V R kV volts

R R k

R k k k

Vth = IBRB + VBE+ IERE

= IBRB + VBE + (+1)IBRE

VthVBE= IB(RB+(+ 1)RE)

1th BE

B

B E

V VI

R R

=

3.33 0.7

3.3 101 500K

2.63 2.6348.88 .

3300 50500 53800. 4888 .

B

C B

I A

I I A

4888 48.88 49.6E C BI I I A

Part (b) VCE= ?; VCC= ICRC+ VCE+ IERE VCE= VCCICRCIEREVCE= 104888 x 10

-6x 10

34937 x 10

-6x 500 = 1004.8882.468

= 2.64 volts;

IC= 4.89 mA; VCE= 2.64 Volts.

Q3) For the JFET shown in the circuit with the voltage divides bias as shown below.Calculate VG, VS, VDand VDSif VGS= -2V. (Sep. 2006)

Solution:

2

1 2

. 15 4 153.75

12 4 4

DDV R k

VG VR R k

Since gate circuit is negligible, Voltage drop across RG= 0

VGS= VGIdRs.- 2 V = 3.75 - IdRSIdRS= 3.75 + 2 = 5.75V = Vs.

Id= 5.75/1k = 5.75mA.Voltage drop across RL=IDRL= 5.75 x 10-3 x 500 = 2.875V

VDS= VDD IDR2 IDRS. = 15 2.875 8.75 = 6.375 volts

VD= VDDIDRL= 152.875 = 12.125V.

4Q) For the circuit shown, calculate VE, IE, Ic and Vc.

Assume VBE= 0.7V.

(Sep. 2006)

Solution:


12/12

VB= VBE+ VE or VE= VBVBE= 40.7 = 3.3V

3.31

3.3

E

E

E

VI mA

R k

Since is not given, assume Ic IE= 1mA.VC= VCCICRL= 101 x 10

-3x 4.7 x 10

3= 5.3 volts

5 Q) In the circuit shown, if IC= 2mA and VCE= 3V, calculate R1& R3 (Sep. 2006)

Solution:

20.02

100B

Ic mAI mA

2 0.02 2.02E C BI I I mA

2.02 500 1.01E E EV I R mA volts

2 1.01 0.6 1.61R E BEV V V volts

2

2

1.610.161

10

RVI mAR k

VR1= VCCVR2= 151.61 = 13.39 volts

1

1

13.3973.97

0.161 0.02

R

B

VR k

I I mA

VR3= VCCVEVCE; VCE= 3V

VR3 = 151.013 = 10.99 volts

3

3

10995.49

2

R

C

VR k

I mA

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EDC - Important Problems