(delhi region) - nims dubai4. electrons are emitted from a photosensitive surface when it is...

PYSP 2005 / Class XII 1

Assessments by Vriti

PHYSICS 2005 (Delhi Region)

Time Allowed: 3 Hours M. Marks: 70

General Instructions:

(a) All questions are compulsory. (b) There are 30 questions in total. Question 1 to 8 carry one mark each, questions 9 to 18 carry two marks

each, questions 19 to 27 carry three marks each and questions 28 to 30 carry five marks each. (c) There is no overall choice. However, an internal choice has been provided in one question of two marks,

one question of three marks and all three questions of give marks. You have to attempt only one of the given choices in such questions.

(d) Use of calculators are not permitted. (e) You may us the following physical constraints whenever necessary.

c = 3 108 ms

-1

h = 6.6 10-34

Js

e = 1.6 10-19

C

o = 4 10-7

TmA-1

Mass of neutron mn= 1.675 10-27

kg

Boltzmann's constant K = 1.38 10-23

J K-1

Avogadro's number N = 6.023 1023

/mole

1. An electric dipole of dipole moment 20 10-6

cm is enclosed by a closed surface. What is the net flux coming out of the surface ?

Sol. Zero, as from Gauss law = enclosed

0

q

0

q q0

E E

2. An electron beam projected along +X axis, experiences a force due to a magnetic field along the +Y axis. What is the direction of the magnetic field?

Sol. +Z-axis (From Fleming left hand Rule)

3. The power factor of an a.c. circuit is 0.5. What will be the phase difference between voltage and current in this circuit?

Sol. Power factor, cos =.5 = 1

2

phase diff. = 60o

4. Electrons are emitted from a photosensitive surface when it is illuminated by green light but electron emission does not take place by yellow light. Will the electrons be emitted when the surface is illuminated by (i) red light, (ii) blue light?

Sol. Electron will be emitted the surface is eliminated by blue light but not with red light.

5. What should be the length of the dipole antenna for a carrier wave of frequency 3 108 Hz?

Sol. Length of antenna 8

8

v 3 10L 1m

f 3 10

6. Define ‘electric line of force’ and give its two important properties.

Sol. Electric line of force (def)

An electric line of force is an imaginary straight or curved path along which a unit positive charge is supposed to move when free to do so in an electric field.

Two important properties are

(a) Two electric lines of force never intersect each other.



(b) Electric lines of force do not pass through a closed conductor, i.e., no lines of force exist within the conductor.

7.(a) Why does the electric field inside a dielectric decrease when it is placed in an external electric field?

(b) A parallel plate capacitor with air between the plates has a capacitance of 8 pF. What will be the capacitance if the distance between the plates be reduced by half and the space between them is filled with a substance of dielectric constant K = 6?

OR

Three point charges of +2 C, -3 C and -3 C are kept at the vertices. A, B and C respectively of an equilateral triangle of side 20 cm as shown in the figure. What should be the sign and magnitude of the charge to be placed at the mid-point (M) of side BC so that the charge at A remains in equilibrium?

M

+2 C A

3 C 3 C

20cm

Sol.(a) When dielectric is exposed to external electric field, polarization of electric dipoles starts. Thereby

net 0 PE E E

E0

+

+

+

+

+

+

+

+

+

+

+

EP

-

-

-

-

-

-

-

-

-

-

-

-

-

-

-

-

-

-

-

+

+

+

+

+

+

+

+

+ E0

(b) 00

AC 8pF

d

Then 0 0A AAC .k .6 .12

d d/ z d= 96pF.

OR



A(+2 C

30o

30o

B(-3 C C(-3 C M

q

10 cm 10 cm

10 3 cm

20 cm 20 cm

12 6

o

2 20 0

1 3 2 1 1 2q 10cos30 2 .

4 420 10 3

100 100

6 62 3 10 3 / 2 q 10

20 20 100 3

69 3

q 104

= 3.9 coulomb.

8. Draw V – I graph for ohmic and non-ohmic materials. Give an example for each.

Sol.

V

I

Ohmic materials

E.g., metallic conductor

V

I Non-ohmic materials E.g., semiconductor device

p-n junction dide)

9. Define the terms ‘Magnetic Dip’ and ‘Magnetic Declination’ with the help of relevant diagrams.

Sol. Magnetic dip( )

The angle through which the N-pole dips down with reference to horizontal is called the angle of dip.



BV

A

B

B

C D

C

B

BH

Magnetic declination ( )

The angle between the magnetic meridian and the geographic meridian at a place is known as the declination at that place.

10. In the figure given below, a bar magnet moving towards the right or left induces an emf in the coils (1) and (2). Find, giving reason, the directions of the induced currents through the resistors AB and CD when the magnet is moving (a) towards the right, and (b) towards the left.

A B

(1)

N S

C D

(2)

Sol. From Lenz’s law induced emf will oppose the cause, therefore

(a) In coil (1) current will be from A to B

In coil (2) current will be from D to C

(b) In coil (1) current will be from B to A

In coil (2) current will be from C to D.

11. (i) Draw the graphs showing variation of inductive reactance and capacitive reactance with frequency of applied a.c. source.

(ii) Can the voltage drop across the inductor or the capacitor in a series LCR circuit be greater than the applied voltage of the a.c. source? Justify your answer.

Sol.(i)



v

XL

Xi = L = 2 vL

v

Xc

Xc = 1/ C = 1/2 vC

(ii) No, because there must be potential drop across the resister.

12. The image of a candle is formed by a convex lens on a screen. The lower half of the lens is painted black to make it completely opaque. Draw the ray diagram to show the image formation. How this image be different from the one obtained when the lens is not painted black?

Sol.

A

B

B

A

Intensity will become 1

4

th of the intensity when it was not painted black.

13. An electric dipole is held in a uniform electric field. (i) Using suitable diagram, show that it does not undergo any translatory motion, and (ii) derive an expression for the torque acting on it and specify its direction.

Sol. Consider an electric dipole of dipole moment

p 2qa

E -q -qE

2a +q

+qE

Held at an angle in uniform electric field E

as shown. The force experienced by the charges q and –q are

qE

and qE

Hence net force = qE qE 0



However the force qE

and qE

m being equal in magnitude and opposite in direction, constitute a couple.

Torque = either force perpendicular distance between parallel forces

= qE 2asin 2qa Esin

= pE sin

Vectorially, p E

Direction of Torque

p

E

Inwardly perpendicular to the 2D plane of E

and p

14. A galvanometer with a coil of resistance 120 ohm shows full scale deflection for a current of 2.5 mA. How will you convert the galvanometer into an ammeter of range 0 to 7.5 A? Determine the net resistance of the ammeter. When an ammeter is put in a circuit, does it read slightly less or more than the actual current in the original circuit? Justify your answer.

Sol. A galvanometer to converted to an ammeter by connecting a low resistance (shunt) S in parallel with it

Shunt resistance S = gGi

i ig

= 3

3

120 2.5 10

7.5 2.5 10 = 40 m .

Net resistance of the ammeter will be equivalent of resistance of the galvanometer resistance and shunt resistance as shown in figure.

G

120

3

net 3

SG 40 10 120R S11G

S G 40 10 120 = 40 m .

Ammeter is always connected in series in the circuit to measure the current. Since ammeter it self has a resistance (small) therefore it increases the net resistance of the circuit. And there by decreases the current. Therefore it will read less current.

15. Define the term ‘resistivity’ and write its S. I. unit. Derive the expression for the resistivity of a conductor in terms of number density of free electrons and relaxation time.

OR

State the principle of potentiometer. Draw a circuit diagram used to compare the emf of two primary cells. Write the formula used. How can the sensitivity of a potentiometer be increased.

Sol. Resistivity is the resistance of the material of a conductor of unit length and unit area of cross-section. It SI

unit is .

Consider a conductor of length l and area of cross section A. Let a potential V be applied across it. An

electric filed is developed across the conductor that drives the electron in one direction and constitutes current I in it



V

El

We know that dI neAv (1)

Where n is no-density of free electrons and vd is the drift velocity of electrons

Now, current density

dj nev but d

eEv

m =

2ne E

m

conductivity, 2ne

E m

OR

E1

E2

G R.B.

A

B

E

Principle when there is a constant current in a wire of uniform cross-section and composition, the fall of

potential along any length of the wire is directly proportional to the length i.e., V l.

Comparison

Comparison of e.m.f of two primary cells can be done with the help of the circuit shown. If l1 and l2 are the

balancing length obtained for the cells of emf. E and E2 respectively for the same potential gradient along the potentiometer wire then

1 1

2 2

E l

E l

Sensitivity of the potentiometer increases will the increase in the length of the potentiometer wire.

16. Explain with the help of diagram, the principle and working of an a.c. generator.

Write the expression for the emf generated in the coil in terms of its speed of rotation.

Sol. Principle - It is based upon the principle of electromagnetic induction. The direction of the corresponding induced current is given by Fleming’s right hand rule.

Working – Thus in one complete revolution, the current alternates once. If one way flow of current is taken as positive, the reverse flow is obviously negative. The corresponding wave form is sinusoidal

Let be the angular velocity of the coil

At time t = = t



= NBA cos t

d

NBAdt

sin t

Induced e.m.f E = -d

NBA sin tdt

E NBAsin t .

17. Give reasons for the following:

(i) Long distance radio broadcasts use short-wave bands.

(ii) The small ozone layer on top of the stratosphere is crucial for human survival.

(iii) Satellites are used for long distance TV transmission.

Sol.(i) Long distance radio broadcasts are sky waves. Then sky waves suffer total internal reflection from Ionosphere. Thus they have smaller wavelengths and hence short wave band is used.

(ii) The ozone layer blocks the passage of UV radiations and protects human beings from the harmful portions of solar radiations.

(iii) TV transmission requires frequency greater than 40 MHz. For frequencies above 40 MHz, the ionosphere is unable to reflect the signals back to earth. That is why a satellite is used to transmit back the signal to earth.

18. A figure divided into squares, each of size 1 mm2, is being viewed at a distance of 9 cm through a

magnifying lens of focal length 10 cm, held close to the eye.

(a) Draw a ray diagram showing the formation of the image.

(b) What is the magnification produced by the lens?

How much is the area of each square in the virtual image?

(c) What is the angular magnification of the lens?

Sol.(a)

A

A

B

A

B

D

(b) 1 1 1

v u f

1 1 1

v 9 10

v = -90 cm

v 90

M 10u 9

Area of image = 12 10

= 100 mm2

(c) Angular magnification = D 25

2.78u 9

19. Ultraviolet light of wavelength 2271 Å from a 100 W mercury source radiates a photo cell made of molybdenum metal. If the stopping potential is 1.3 V, estimate the work function of the metal. How would the photo cell respond to high intensity (10

5 Wm

-2) red light of wavelength 6328 Å produced by a He – Ne laser?



Plot a graph showing the variation of photoelectric current with anode potential for two light beams of same wavelength but different intensity.

Sol. oV = 1.3 V , = 2271 10-10

m

2o max o o

1h h mv h W eV

2

o o o o

hcW h eV W eV

34 8

19o 1

6.62 10 3 10W 1.3 1.6 10

2271 10

Wo = 66.665 10-19

J 4.2 eV

7o o

o o

hc hcW 2.98 10

W m

2 2980 Å

Photocells will not respond to = 6328 Å because it is greater than

o = 2980 Å

20. (a) Draw a graph showing the variation of potential energy of a pair of nucleons as a function of their separation. Indicate the regions in which nuclear force is (i) attractive and (ii) repulsive.

(b) Write two characteristic features of nuclear force which distinguish it from the coulomb force.

Sol. (a) Potential energy of a pair of nucleons as a function of their separation is plotted as shown. For a separation greater than ro, the force is attractive and for separation less than ro the force is strongly repulsive. The attractive force is strongest when the separation is about 1 fm.

ro 1 2 3

-100

0

100

Po

tenti

al en

erg

y (

MeV

)

r (fm)

(b) Two characteristics are

(i) The nuclear force is very stronger force than coulomb force.



(ii) The nuclear force is chare independent.

21. (a) Show that the decay rate ‘R’ of a sample of a radionuclide is related to the number of radioactive

nuclei ‘N’ at the same instant by the expression R = N.

(b) The half life of 23892 U against -decay is 1.5 10

17 s. What is the activity of a sample of 238

92 U

having 25 1020

atoms?

Sol.(a) The rate of decay is directly proportional to the quantity of material actually present at that instant.

Thus if N is the number of radioactive atoms present at any instant, the rate of decay

dN

dt N

or dN

R Ndt

(b) Activity oA N = 20 3o 17

0.693 0.693N 25 10 11.55 10

T 1.5 10

= 11550 disintegration per second.

22. Explain, with the help of a circuit diagram, how the thickness of depletion layer in a p-n junction diode changes when it is forward biased. In the following circuits which one of the two diodes is forward biased and which is reverse biased?

(i)

–10 V

(ii)

–10 V

Sol.

p n

+

Forward biased p-n junction

Due to concentration difference majority carrier starts crossing the junction and keeps re-combining. In forward biasing, potential barrier is reduced. As a result of this a larger number of majority carriers diffuse through the junction and a larger current flows. Therefore this depletion layer offers lesser resistance.

(i) Forward biased

(ii) Forward biased



23. Distinguish between analog and digital communication. Write any two modulation techniques employed for the digital data. Describe briefly one of the techniques used.

Sol. Analog communication is based upon analog signal. It is a continuous function with amplitude being continuous. Whereas digital communication is based upon digital signal. These are discrete signals which are discontinuous in time, i.e. they are defined only at discrete times.

Two modulation techniques employed for the digital data are

(i) Pulse – amplitude modulation (PAM)

(ii) Pulse – position modulation (PPM)

Pulse – amplitude modulation (PAM)

The modulating signal is sampled at the basic rate, usually fm/2, where fm is the maximum value. The amplitude of the modulation determines the amplitude of the transmitted pulse.

24. Draw a schematic diagram of a single optical fibre structure. Explain briefly how an optical fibre is fabricated. Describe in brief, the mechanism of propagation of light signal through an optical fibre.

Sol. In transmission of light through optical fibre light is incident on the boundary of core and cladding at an angle of incidence which is greater than the critical angle of incidence

i.e. I > C { c = sin-1

(n2/n1)}.

The incident beam undergoes total internal reflection and the light beam remains confirmed within the core as shown in the figure.

Fabrication

If we soften (nearly melt) glass and pull it, then we get the fibres, for obtaining optical fibre, there are two steps:

Step I

Preformed glass rods are first prepared with different inner and outer glass compositions.

Step II

The preformed glass rod is softened in a furnace and fibres are drawn. Finally, the plates buffer coating is done to obtain finished optical fibre.

25. (a) With the help of a labeled diagram, explain the principle and working of a moving coil galvanometer.

(b) Two parallel coaxial circular coils of equal radius R and equal number of turns N carry equal currents I in the same direction and are separated by a distance 2R. Find the magnitude and direction of the net magnetic field produced at the mid-point of the line joining their centers.

OR

(a) State Biot-Savart’s law. Using this law, derive the expression for the magnetic field due to a current carrying circular loop of radius R at a point which is at a distance x from its center along the axis of the loop.

(b) Two small identical circular loops, marked (1) and (2), carrying equal currents, are placed with the geometrical axes perpendicular to each other as shown in the figure. Find the magnitude and direction of the net magnetic field produced at the point O.



i

x

90o

x

i

(1)

(2)

O

Sol. (a) Principle of a Moving Coil Galvanometer

When current is passed through a freely suspended coil in uniform magnetic

Field, the coil is a measure of the torque experienced by the coil which in turn is

a measure of the current flowing in the coil.

Construction.

A moving coil galvanometer consists of a light coil wound over a non-metalic frame with central soft iron core.

The coil is free to rotate about an axis lying the plane of the coil and fixed normal to the direction of a radial magnetic field. At the axis of the coil are attached two spiral springs which produce torsional couple which

holds the coil in the deflected position at an angle which is proportional to the torque on the coil duet to

current in it. Since the torque on the current carrying coil in a radial magnetic field is directly proportional to the current, the deflection shown by the galvanometer coil is directly proportional to the current in the coil.

We know the torque on a current carrying coil in a magnetic field is given by

= INAB sin , As = 90 in a radial magnetic field

= INAB

Here = INAB is the deflecting torque.



As the coil rotates, the spiral spring attached to its axis gets wound, a restoring torque C is developed in

them.

Now is equilibrium, the deflecting couple is equal to the restoring couple.

So NAB = C , OR I = C

NAB ,

OR I ,

Thus the current flowing in the coil is directly proportional to the deflection. This is due to the radial magnetic field in which the plane of the coil remains parallel to the direction of the magnetic field in every orientation.

As a result the deflecting couple on the coil does not depend on the angle of deflection of the soil.

(b)

I I R R

A B

O O P

2R

Applying the formula for 03 / 2

2 21

2IAB

I n a

at point P, due to both the coils, both the fields will be having

equal magnitudes and opposite direction.

Net B = 0.

OR

The magnetic field strength at point P due to a small length all of the conductor carrying

O r

P

current I is given by

0

2

IdlsindB .

4 r

Where 7 10 4 10 TmA is called the permeability of free space or vacuum.

Consider a circular current carrying conductor of radius r.



x

O r

r

dB dB

dB cos dB cos

Idl

According to Biot-Savart’s law,

The magnetic field at P due to current element Id l

is

= o

o o2 2

Idlsin90 IdldB

4 4s s

2 2S r x

o

2 2

IdldB

4 r x

Direction of dB

is perpendicular to the plane containing S

and d l

. We resolve dB

into rectangular

components. dB cos and dB sin . Now the component dB cos gets cancelled by the corresponding component of magnetic field at P produced by current element diametrically opposite to the previous one. It

is only dB sin that is added up for magnetic field of every current element on the loop.

Thus, total magnetic field is

o

2 2

IdlsinB dBsin

4 [ r x

2 2

rsin

x r

o1/ 2

2 2 2 2

IB 2 r

4 r x x r

2

o

3 / 22 2

IrB

2 r x

I

x

x

I B1

45o R

B2 Bnet

2

01 3/ 2

2 2

iRB

2 R x



B2 = 99

2

0

3 / 22 2

2 iRA

2 R x

Direction is as shown in the figure.

26. (a) How is a wavefront different from a ray? Draw the geometrical shape of the wavefronts when (i) light diverges from a point source, and (ii) light emerges out of a convex lens when a point source is placed at its focus.

(b) State Huygen’s principle. With the help of a suitable diagram, prove Snell’s law of refraction using Huygen’s principle.

OR

(a) In Young’s double slit experiment, deduce the conditions for (i) constructive, and (ii) destructive interference at a point on the screen. Draw a graph showing variation of the resultant intensity in the interference pattern against position ‘x’ on the screen.

(b) Compare and contrast the pattern which is seen with two coherently illuminated narrow slits in Young’s experiment with that seen for a coherently illuminated single slit producing diffraction.

Sol.(a) A wavefront is defined as the locus of points having the same phase of oscillation.

Whereas a line perpendicular to a wavefront is called a ray.

Figure shows the wavelengths corresponding to diverging rays.

S

Diverging spherical wavefront.

F

(b) Huygen's Principle: (i) Every point on a given wave front acts as a fresh source of secondary wavelets

which travel in all directions with the speed of light.

(ii) The forward envelope of these secondary wavelets gives the new wave front of any instant.

Let AB the incident wave front incident on a refracting surface XY at an angle of incidence i. Let the

refractive index of medium 2 with respect of medium 2 be . The time in which point A of the wave front sends waves a distance AC in medium 1, point B of the waves front will send wave in medium 2 which will

travel as distance AC/ in the same time. To obtain refracted wave front CD let us draw a semicircle of

radius AC/ in medium 2 with B as its center. A tangent CD from C onto this semicircle give the refracted wave front CD.



A

C 1

2 B

D

r i X Y

OR

(a) Condition for construction and destruction interference.

y1 = a sin t

and y2 = b sin ( t + )

where a and b are the respective amplitudes of the two waves and is the constant phase angle by which second wave leads the first wave.

According to superposition principle, the displacement (y) of the resultant wave at time (t) would be given by

y = y1 + y2 = a sin t + b sin ( t + )

Thus the resultant wave is a harmonic wave of amplitude R.

2 2R a b 2abcos

As resultant intensity I is directly proportional to the square of the amplitude of the resultant wave.

I R2

i.e. I (a2 + b

2 + 2ab cos )

For constructive interference

I should be maximum, for which

cos = max = +1

= 0, 2 , 4 ,………

i.e. = 2n where n = 0, 1, 2…….

If x is the path difference between the two waves reaching point P, corresponding to phase difference , then

x 2n n2 2

i.e. x = n

Hence condition for constructive interference at a point is that phase difference between the two waves

reaching the point should be zero or an even integral multiple of .

Equivalently, path difference between the two waves reaching the point should be zero or an integral multiple of full wavelength.

For destructive interference,

I should be minimum

cos = minimum = -1

, 5 , ….

or = (2n -1) where n = 0, 1, 2,……

i.e. x 2n 12

Hence, the condition for destructive interference at a point is that phase difference between the two waves

reaching the point should be an odd integral multiple of or path difference between the two waves reaching the point should be an odd integral multiple of half the wavelength.

For YDSE



S1

P

S2 d sin

D

For constructive interference

path difference: d sin = n

For destructive interference

path difference: d sin = (2n – 1)

I

4Io

O 2

Interference pattern

Intensity against phase difference at

point of observation

Intensity

a

2

a a

a

3a

2

a

5

27. (a) Distinguish between metals, insulators, and semiconductors on the basis of the energy bands.

(b) Why are photodiodes used preferably in reverse bias condition? A photodiode is fabricated from a semiconductor with band gap of 2.8 eV. Can it detect a wavelength of 6000 nm? Justify.

OR

(a) Explain briefly, with the help of circuit diagram, how V-I characteristics of a p-n junction diode are obtained in (i) forward bias and (ii) reverse bias. Draw the shape of the curves obtained.

(b) A semiconductor has equal electron and hole concentration of 6 108 /m

3. On doping with certain

impurity, electron concentration increases to 9 1012

/m3.

(i) Identify the new semiconductor obtained after doping.

(ii) Calculate the new hole concentration.

Sol. (a) Insulator – The energy band diagram is as shown in figure (a)



The conduction band (CB) is empty but valence band (VB) is full. The forbidden energy gap is = 6 Ev, i.e., an insulator requires 6 eV of energy to become conductor. However this energy cannot be practically possible. Hence an insulator cannot conduct electricity.

Conductor – The energy band diagram is as shown in figure (b). The CB and VB overlap. The forbidden energy gap do not exist. Hence the conductor always conducts electricity.

Semiconductor – The energy band diagram is as shown in figure (c) with energy gap = 1 eV. At 0 K; the semiconductor do not have this much energy and hence do not conduct. But at room temperature some electrons are able to jump to CB and thus semiconductors have poor conductivity at room temperate.

~6eV

CB

VB

Figure (a)

CB

VB

Figure (b)

~1 eV

CB

VB

Figure (c)

(b) In forward bias, we get positive output in the form of conductivity. However in photo diode the electron jump from valence band to the conduction band by absorbing energy from some external source of energy. If the incident visible light is the external source of energy, then the semiconductor is said to be photosensitive. When visible light is incident on a photosensitive semiconductor, more electrons becomes available to participate in conduction. Thus the conductivity of a photo-sensitive conductor increases when light in incident on it.

Band energy E = 2.8 eV

= 2.8 1.6 10-19

J

= 4.48 10-19

J

Energy produced by incident wave length

= 34 8

9

hc 6.6 10 3 10

6000 10 = 3.3 10

-20 J

Which is less than the Band energy so photodiode cannot detect it.

OR

Forward Bias:

Figure shows the circuit diagram with the help of which we can study the forward-bias characteristic of PN junction diode. The battery used is generally of the order of 1.5 V. Different readings are taken b the varying the voltage. Each time the voltage across the diode and the current flowing through the diode is measured by means of a voltmeter and a milli-ammeter respectively. Since current flows easily through a forward-baised diode, a resistance R is included in the circuit so as to limit the current.

+

Current-limiting

resistance

+

mA

+

Experiment arrangement for the study of

forward-bias characteristic

X

Y

Fo

rwa

rd c

urr

en

t (m

A)

I

knee

point

0.7 V

Forward bias V

O

Forward-bias characteristics



Reverse Bias:

Figure shows the circuit diagram with the help of which we an study the reverse-bias characteristics of PN junction diode. Instead of a milliammeter, we use a microammeter. The voltage across the diode is increased gradually in suitable intervals depending upon the maximum reverse voltage. The corresponding values of element are measured with the help of microammeter.

+

-

mA

-

Experiment arrangement for the study of

forward-bias characteristic

I

VZ

Rev

erse Bias (

A)

Reverse-bias characteristics

Reverse-bias Zener

voltae

Cry

stal

Break

dow

n

igure shows the reverse-bias characteristics. In the reverse bias, the diode current (called reverse current) is

very small (only a few A for germanium diodes and only a few nA for silicon diode). It remains small and approximately constant for all voltages less than the break down voltage Vi. At breakdown, the current increases rapidly for small increases in voltage.

(b) (i) n - type

(ii) np = nipi = ni2

282

4 3

12

6 104 10 /

9 10

inP m

n

(delhi region) - nims dubai4. electrons are emitted from a photosensitive surface when it is...

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