UNIT-1 SOLVED PROBLEMS
1.3.
An electron with a velocity of 3 x 105 ms-1 enters an electric field of 910 v/m making an
angle of 600 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= 3 105 m
;E=
950V / m.;
sec.
Component of velocity in y direction = V Sin 

= 60
5
𝑉𝑌 = 3 × 105 𝑆𝑖𝑛 60𝑜 = 3  10  0.866 ; Electron feels a retarding force due to electronic field and
at maximum height final velocity will be zero.
eE 1.6 1019  910
u
V  u  at (or )0  at (or )t  ; a  
 1.6 1014 m sec2
31
a
m
9.110
5
3 10  0.866
V sin 
For a projectile time of ascent =
=
 1.62 109.sec.
14
a
1.6 10
1.4) The magnetic flux density B=0.02 wb/m 2 and electric field strength E=105 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


  velocity of translation of center of the
m
B
rolling circle.
u 
B

Q  

B
e
w B e m
mB 2
105
=
 1420 106
11
2
1.76 10  (0.02)
Z  2 Q  2    1420  106  8922  106  0.8922cm.
1.5) Two parallel plates of a capacitor are separated by
4cms. 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 . If
the 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:
t

vx = 1.76 1017 t 2 ; Distance x  v xdt
o

t
x  1.76  1017 t 2 
o
When t  40  10


 37546 1010  3.754 106  3.754m
3
1.76
 1017  40  103
3
9, Velocity
vx  1.76  1017  40  109


2
 2816  101  281.6m / 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 is 1000 V . [Apr/May’ 03, May’ 06,
Aug’ 06]
SOLUTION:
lLVd
lL
;
S
2dVa
2dVa
d
= 0.5 cm;
Vd
L
= 20 cm
Va
= 1000 V.
l = 2 cm
D=
S
D?
= 25 V
 2  102  20  102 
lL
5
=
 = 40 10  0.04
2dVa  2  0.5  102  1000 
cm/V
tan  
D S .Vd 0.04  25  102


 0.05
L
L
20  102
tan   0.05
OR   2.860
2eVa
= 21.7610111000 1.876 10 7 m/sec.
m
Resultant velocity
v
1.876 107 1.876 107
= 1.8778 x 107 m/sec.
vr  ax 

o
Cos Cos 2.80
.999
Velocity vax 
1.7)
When an electron is placed in a magnetic field with a period of rotation T=
35.3
1012 sec so
B
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 is 900 v and B=0.01 wb/m 2 .
b.
What is the time period of rotation?
SOLUTION
35.3
1012 sec.
B
Va = 900 V; B= 0.01 Wb/m;
T = ?, r = ?
12
35.3 10
35.3
T
1012 sec. =
 3530 1012 Sec.
B
0.01
Given
T
[May’ 06]
2eVa
mV
;
V 
V  2 1.76 1011  900  17.79 106 m / sec.
eB
m
17.79 106
r
 1010.8 105  10.11mm
11
1.76 10  0.01
r
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.
S= 1 cm ; u  0 ;
Given:
Final Velocity V 
2eVa
m

Va  1000V ;
2 1.76 1011 1000 = 1.876 107 m / sec
Electron follows Newton Law’s of motion.
=
at
or a  v
V  u  at
s  ut  12 at 2  12 at 2 
1
2
t
 v / t  t 2  vt / 2 ;
t
2S
2  102

 1.066  109  1.066n sec.
7
v 1.876  10
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
plates. Calculate.
a.
Velocity of electron on reaching the field
b.
Acceleration due to deflection field
c.
Deflection produced on the screen In cm.
d.
Deflection sensitivity in cm/V.
[Jun’ 01]
SOLUTION
Va  600V ; l  1.5cm; d  0.8cm; L  20cm;Vd  20V
lLVd
D
2dVa
Part (a) velocity of electron reaching the field
2eVa
Vox = 14532721 m/sec.
vox 
 2  1.76  1011  600
m
Part (b) Acceleration due to electric field.
eV
1.76  1011  20
ay  d 
 4.4  1014 m / sec2
2
md
0.8  10
Part (c) deflection produced in cm.
lLVd 1.5  102  20  102  20
D

 6.25  103 m  .625cm
2dVd
2  0.8  102  600
Part (d) deflection sensitivity in cm.
D 0.625

 S  0.03125cm / V .
Vd
20
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
Va  400V ; L  19.4cm; l  1.27cm ;
Electro static deflection: D=
s
S
lLVd
;
2dVa
D
; B  ? for Vd  30V
Vd
D
1.27 X 102 X 19.4 X 10 2 X 30
 0.019m  1.9cm
2 X 0.475 X 10 2 X 400
d 0.019

 6.3  104 m / V  0.06cm
Vd
30
Electro magnetic deflection:
D
B
lLB
e
;

m
Va
B
D Va
lL
e 1
m 2
.
0.019 X 400

1.27 X 102 X 19.4 X 102
1.76 X 1011
2
0.38
 5.199 X 104 wb/ m 2
730.88
1.11) In a parallel place diode, the cathode and anode are spaced 5mm apart and the anode is kept at
200V 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 2X106m/s in the direction towards the anode.
SOLUTION
D=5mm
v  ?; t  0.5ns
Vd=200V
when (a) u=0, (b) u=2x106m/s
Electrons follow Newton’s laws of motion v=u+at
v
200
Where a  e . d ; a  1.759 1011 
 7.036 1015 m / sec 2
m d

3
5 10
case (a) when initial velocity u=0
V=at = 7.036x1015x0.5x10-9= 3.518x106 m/s =3518 km/sec.
Case(b) V=u + at = 2x106 + 3.518x106 =5.518x106 m/sec or 5518 km/sec.
Distance  S  ut  1 at 2
2

Case(a): u=0;  S  1 at 2  1  7.036 1015  0.5 109
2

2
 0.8795 103 metres
2
Case(b): S  2x106  0.5x109  0.8795 103  1x103  1.8795 103 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. The
distance between the anode screen distances is 20cm.
Anode
Given: B = 0.6 Gauss V = 400 Volts;screen distance
= 20 cm.
To calculate:
Deflection
D=?
To find out D, we must know radius R.
v ( m)
R is given by, R 
we know m, e & B, v to be found.
e( B )
2eVa
(or )5.93  105 Va
m
Velocity of electron along x-ox is, vox  5.93  105 400 = 5.93 x 105 x 20 = 118.6 x 105 m/sec.
= 1.19 x 107 m ./sec.
B
= 0.6G.;
1 Wb / m2
=
104 Gauss
-4
2
B
= 0.6 x 10
Wb / m
= 6.0 x 10-5 wb / m2
v.m
v
1.19 107
R
=
( e / m = 1.759 x 1011)


11
5
eB (e / m) B 1.759 10  6 10
= 0.112 x 10 = 1.12 m = 112 cm.
In  PYZ, YZ2 = PZ2 + PY2
YZ
= OP = 112 cm; PY = 20 cm.; PZ
= OZ – OP
(Deflection D = OP)
(YZ)2 = (PZ)2 + (PY)2  112 2 = (112 – D)2 + 202  1122 = 1122 – 224D + D2 + 400
or D2 – 224D + 400
=0
v is given by v 
This is in the form of ax2 + bx + c = 0, quadratic equation. Where x 
b  b2  4ac
2a
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 are
5cm., 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:
 t 
10 X  7 
 10  =2 x 109 t V/m
 = (Voltage X time) / Distance 
5 X 102
We know acceleration a = dv /d t. = f/m. = e/m.
Or
a = (1.76 x 1011) ( 2 x 109t)
= 3.52 x 1020 t m /sec2
t
We know, v =

o
l
a dt
=

o
3.52 X 10 20 t dt = 1.76X1020 t2 m/sec
At t
v
b)
= 50 n.sec.
= 50 x 10-9
= 1.76 x 1020 (50 x 10-9)2
Distance travelled
= 4.4 x 105 m /sec.
t
t
o
0
x   v.dt   1.76  1020 t 2  5.87  1019 t 3
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 1019 t3
x
0.05
or
(distance x = 5 cm) t
= 9.46 x 10-8 sec.
t3 

19
5.87 10
5.87 1019
1.14 Derive the expression for acceleration, velocity and displacement of a changed particle in an
electgric field
SOLUTION:
(i) F  ma  eE  a  eE --------(1);
m
t
(ii) Velocity v  
adt & a  dv ; v  dx ;(or ) ;
dt
o
dt
2
d x
a  dv  d  dx  
   (2)
dt dt  dt  dt 2
Substituting value of a from Eqn.(2) in to Eqn (1)=
d 2 x eE

   (3)
2
m
dt
t
t
eE
V   adt   eE dt 
t  C1 (C1 is constant)
om
m
o
Average velocity  eE
m
dx eE
dx eE

t  C1 At + =0; x  0; C1  0
V

 t    (4)
OR
dt m
dt m
t
dx t eE
t dt


m
dt
o
o
eEt 2
Distance x 
   (5)
2m
x
eE t 2
 C2
m 2
If C2 is taklen as 0
eE 2
t    5
Distance x 
2m
UNIT – III SOLVED PROBLEMS
1 Q) A 15 – 0 – 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)
Solution: I DC
where f is power line frequency.
4 fc
Vrms = 15V, Assume voltage drop (rms) across diode as
1V.
Vrms across load = 14V.
4170 Idc
4170 Idc
Vdc = Vm = 19.8 C
10,000
100 Idc ; 19.8 (or) Idc ; 0.198 Amp
VDC  Vm 
Vdc = 19.8 Volts.
Vm = Vrms 2 = 142 = 19.8 volts.
2 Q)
A FWR is used to supply power to a 2000 load, choke of 20H and capacitor of 16f are
available. Compute ripple factor using filter 1 (i) one inductor (ii) one capacitor (iii)
single L – type.
Solution:
i)
(ii)
(iii)
(iv)
3 Q)
RL = 2000; C = 16f; L = 20H.
One Inductor Filter
RL
2000
1
r


 6.25 103  0.006
16000 L 16000  20 160
One capacitor filter:
2410
2410
r

 75.31103  0.075
CRL 16  2000
Single LC Type Filter
0.83
0.83
r

 0.0025
LC 20 16
 Selection
3330
3330
r

 325 106  .000325
CC1 L1 RL 16 16  20  2000
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.
(JNTU – 2000)
Solution: Given VDC = 9V,
To find: L, C.
f = 60Hz.; IL 100mA, r = 2% = 0.02
r  0.83 when f = 60 Hz. → (1)
→ (1)
2 1
1
.
.
3 2 wc 2 wL.
Where LC = RL / 1130.
0.83
LC 
 41.5 → (2)
Or
0.02
Critical Inductance (value of Inductance for which diode
conducts continuously)
Else r 
V
RL
9
.; RL  DC 
 90
1130
I DC 100mA
Lc = 41.5
(2)
41.5
C 
 521 f
79.6 103
Transformer rating – Vrms = ?; Diode ratings PIV= Vm ;
L = 796mH;
C = 521f
LC 
4 Q)
current rating = load current
A FWR operating at 50 Hz i.e., to provide DC current of 50mA at 30V with a 80f, C – type
filter. 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,
IDC = 50mA
VDC = 30V.
C = 80f
To find Vm = ? ratio of surge to mean current
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--------------------V -V
Current through the diode at any time S C
R d +R S
When Vc = 0, Id will be max =
VS max
Vm

Rd  RS Rd  RS
This is called surge current.
V  Vdc
Mean diode current I d  S
Rd  RS
Ratio of surge current to mean current =
=
 R  Rs 
Vm
 d
 Rd  Rs  Vm  Vdc
Vm
47.17
47.17


 2.75
Vm  VDC 47.17  30 17.17
Note: Designer must cater for 3 times the required average current.
2410
Part III :
Ripple factor r 
f = 60 Hz.
CRL
Or
Ripple factor r 
1
4 3 fcRL
Vdc 300V

 30V 2 = 600
Idc 50mA
1
106
r

 0.006
4 3.50.80 106  600 16627687.75
RL 
5 Q)
For a FWR circuit AC voltage input to transformer primary is 115V. Transformer secondary
voltage is 50V, RL =25. Determine
i)
Peak DC component, RMS and AC component of load voltage
ii)
Peak DC component, RMS and AC component of load current.
(June 2002)
Solution:
Given Vrms = 50V, to find Vm, Vr, Im, Ir.
V 

RL = 25  Vm  Vdc  r 
2

Part (i)
Vm
VRms 
orVm  2;VRms  2,50  70.7Volts
V2
V
r  r , r of FWR (without filter) = 0.48
Vdc
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
 peak
66.6
Vrms =
=
= 47 Volts
2
2
Part – II
Vm 70.7
Im 

 2.828 Amps.
RL
25
Vr 21.6

 0.864 Amps.
RL
25
Peak DC current component = 66.6/25 = 2.664 Amps.
Runs DC current component = 47/25 = 1.8 Amps.
Ir 
Calculate ripple factor of capacitor filter with peak rectified voltage of 20V and C = 50 f and
IDC = 50mA.
(June 2004)
Solution: Vm = 20V, Idc = 50mA, C = 50f.
VDC
Vr
Vm = 20V
6 Q)
VDC  Vm 
Idc
4 fc
Suppose f = 50Hz, Then VDC  20 
50  103
4  50  50  106
1000
 20  5  15volts
200
VDC
15
150  1000
RL 


 300
Idc
50mA
50
1
1  106
100
1
r



= 0.192
4 3 fcRL 4 3  50  50  300 4 3  25  3 3 3
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: -
= 20 
For drawing DC load line we must know
a) VCE max when Ic = 0;
b) Ic max when VCE = 0
VCE max = VCC = 20V
IC  I E  IC max 
VCC
20

 9.52mA
RC  RE 2000  100
VCE max 20

 10V
2
2
Ic max 9.52

 4.76mA
=
2
2
Quiescent VCEQ =
Quiescent ICQ
IC
4.76
 0.0952mA ; Quiescent IEQ = ICQ + IBQ = 4.855mA.

50
Find out stability factor of the circuit given below:
(May 2005)
Quiescent IBQ =
Q)

Stability factor of self-biased Circuit given by:
RB
RE
S  1   
;
 RB 
1   

 RE 
1
R1 R2 5  50
50

k  4.5k  4500
11
R1  R2 5  50
RB
4500

 45
RE
100
 1  45 
S   50  1 
  24.54
 1  50  45 
RB 
For the circuit shown, determine the value of Ic and VCE. Assume VBE = 0.7V and  = 100
(Sep.06)
Vin 
Vcc .R2
10  5k
50


 3.33volts
R1  R2 10  5  k 15
R th 
10  5
50
k
k  3.33k .
10  5
15
Vth
= IBRB + VBE + IERE
= IBRB + VBE + (+1)IBRE
Vth – VBE = IB(RB+( + 1)RE)
3.33  0.7
Vth  VBE
=
3.3K  101  500
RB     1 RE
2.63
2.63
IB 

 48.88 A.
3300  50500 53800
IC   .I B  4888 A.
I E  IC  I B  4888  48.88  49.6 A
IB 
Part (b) VCE = ?;
VCC = ICRC + VCE + IERE  VCE = VCC – ICRC – IERE
VCE = 10 – 4888 x 10-6x 103 – 4937 x 10-6 x 500
= 10 – 04.888 – 2.468
= 2.64 volts;
IC = 4.89 mA;
Q3)
VCE = 2.64 Volts.
For the JFET shown in the circuit with the voltage divides bias as shown below.
Calculate VG, VS, VD and VDS if VGS = -2V.
(Sep. 2006)
Solution:
VDD .R2
15  4k
15


 3.75V
R1  R2 12  4  k 4
Since gate circuit is negligible, Voltage drop across RG = 0
VG 
VGS = VG – IdRs.- 2 V = 3.75 - IdRS
IdRS = 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 = VDD – IDRL = 15 – 2.875 = 12.125V.
4Q) For the circuit shown, calculate VE, IE, Ic and Vc.
Assume VBE = 0.7V.
(Sep. 2006)
Solution:
VB = VBE + VE or VE = VB – VBE = 4 – 0.7 = 3.3V
V
3.3
IE  E 
 1mA
RE 3.3k
Since  is not given, assume Ic  IE = 1mA.
VC = VCC – ICRL = 10 – 1 x 10-3 x 4.7 x 103 = 5.3 volts
5 Q)
In the circuit shown, if IC = 2mA and VCE = 3V, calculate R1 & R3
Solution:
Ic 2mA
IB  
 0.02mA
 100
I E  IC  I B  2  0.02  2.02mA
VE  I E RE  2.02mA  500  1.01volts
VR 2  VE  VBE  1.01  0.6  1.61volts
VR 2 1.61

 0.161mA
R2 10k
VR1 = VCC – VR2 = 15 – 1.61 = 13.39 volts
V
13.39
R1  R1 
 73.97k 
I  I B  0.161  0.02  mA
VR3 = VCC – VE – VCE;
VCE = 3V
VR3 = 15 – 1.01 – 3 = 10.99 volts
V
1099
R3  R3 
 5.49k 
IC
2mA
I
(Sep. 2006)