The Applied Electronics MCQ || Test Revision paper !!!

1) The types of coupling used in the BJT (Bipolar Junction Transistor) amplifier.

a) Resistance capacitance (RC) coupling.

b) Impedance coupling

c) Transformer coupling

d) All of the above

Answer : (d) All of the above.

2) The phase shift of positive feedback circuit.

a) 90° Degree

b) 180° Degree

c) 45° Degree

d) 0° or 360° Degree

Answer : (d) 0° or 360° Degree.

3) The current gain (β) in the power amplifier.

a) Low 5 to 20

b) Low 40 to 65

c) Low 70 to 95

d) Low 25 to 40

Answer : Low 5 to 20

4) What is the angle of conduction of Class A power amplifier ?

a) 90° Degree angle

b) 180° Degree angle

c) 45° Degree angle

d) 360° Degree angle

Answer : 360° Degree angle

5) What is the angle of conduction of Class B power amplifier ?

a) 90° Degree angle

b) 180° Degree angle

c) 45° Degree angle

d) 360° Degree angle

Answer : 180° Degree angle

6) What is the angle of conduction of Class C power amplifier ?

a) 90° Degree angle

b) 100° to 150° Degree angle

c) 180° to 200° Degree angle

d) 360° Degree angle

Answer : 100° to 150° Degree angle

7) What is the angle of conduction of Class AB power amplifier ?

a) 90° Degree angle

b) 100° to 150° Degree angle

c) 180° to 200° Degree angle

d) 360° Degree angle

Answer : 180° to 200° Degree angle

8) What is the application of single tuned amplifier in the below option.

a) T.V Receiver

b) Radio Receiver

c) Tuned Radio Frequency (TRF) Receiver

d) Both a) T.V Receiver & c) 

Answer : d) Both a) T.V Receiver & c) Tuned Radio Frequency (TRF) Receiver

9) What are the application of RC coupled amplifier.

a) In Tape Recorder

b) In stereo amplifier

c) Widely used as voltage amplifier

d) All of the above

Answer : d) All of the above

10) The different types of power amplifier.

a) Class A power amplifier

b) Class B power amplifier

c)Class C power amplifier

d) All of the above

Answer : d) All of the above

The heathy benefits of papaya for healthy body immunity and mental problem

Hello dear all readers,

Welcome to this new concept, so here in this new topic we come to know about the healthy benefits of papaya for healthy body immunity and mental problems.

The papaya is very good for healthy body immunity and mental problems.

The papaya fruit is very soft. The sweet taste of papaya is very tasty.

We must eat papaya at least in a week.

The papaya is one kind of fruit, which we can eat safely in all types of diseases.

Here why we use safely words, means that if you suffer from cold and cough than doctor recommend you to not eat banana fruit and all chill substances.

As we know that banana is not eaten in cold and cough. But we can safely eat papaya in all types of diseases.

Here we know the healthy benefits of eating papaya fruit :

i) The papaya reduced the risk of heart disease.

ii) The papaya reduced the risk of diabetes disease.

iii) The papaya is found to be very helpful in improving blood glucose control in people with diabetes.

iv) The papaya reduced the digestion problems.

v) The papaya is very profitable in lowering the blood pressure.

vi) The papaya reduced the risk of cancer.

Here are the chart table containing by papaya fruit :

• Calories : The papaya contain calories 59

• Carbohydrates : The carbohydrates contain by papaya is 15 gram.

• Fiber : The fibre contain by papaya is 3 gram.

• Protein : The protein contain by papaya is 1 gram.

• Vitamin C : The vitamin C contain by papaya is 157% of the RDI (Recommend Dietary Intake).

• Vitamin A : The vitamin A contain by papaya is 33% of the RDI (Recommend Dietary Intake).

•Vitamin B9 : The vitamin B9 contain by papaya is 14% of the RDI (Recommend Dietary Intake).

• Potassium : The potassium contain by papaya is 11% of the RDI (Recommend Dietary Intake).

And in trace amount the papaya contain calcium, magnesium, vitamin B1, B3, B5, E, and k.

So we hope you feel good to know the knowledge of this papaya fruit.

So we will meet you with next topic.

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Today's 2021 year play Holi festival with corona virus with some taking precaution !!!

Happy Holi - 2021

Hello friends,

Welcome to this new topic.

Here in this topic, we come to know about the celebration of Holi festival

in this year 2021.

So in this year also as like as previous year we have to play Holi with lots of care.

 As you all know that, how fastly corona virus spread every where.

So because of that, we have to take patience and keep calm.

We must have to not get too worried about all that dangerous situation created by this pandemic.

Keep calm, stay at home and celebrate this beautiful and lovely Holi festival with great joy.

So here are the some precaution, if we measures than we can absolutely stay safe, while celebration Holi festival this year.

1) Time to time wash your hand.

2) Keep safe distance with each other.

3) Wear properly mask on your nose and mouth.

4) Avoid to play Holi in public places.

5) Keep sanitizer bottle whenever you go outside from home.

Yeah...!!! It is quite not too much interesting to play Holi festival in this corona time.

We have to keep motivated our inner side.

But still we have to play this Holi festival with lots of care.

By god grace this corona pandemic also go away soon.

We have to keep motivated from our inner side.

And we have to not lost our confidence and motivation from our inner side.

Happy Holi to all...!!!

Keep playing, keep enjoying...!!!

Good bye...!!!

Thank you...!!!

Explain the refraction of light on the basis of wave theory and also prove the laws of refraction of light.

i) The incident rays, refracted rays and normal lie in the equivalent plane.

ii) The incident ray and refracted ray lie on opposite sides of normal. 

iii) The ratio of velocity of light in rarer medium to velocity of light in denser medium is a constant called refractive index of denser medium with respect to rarer medium.

Explanation laws of refraction :

i) Let consider XY be the plane refracting surface separating two media air and glass of refractive indices μ1 and μ2 serially.

ii) A plane wavefront AB is developing at an angle towards XY from the air medium. It is bounded by the rays AA1 and BB1 which are incident rays.

iii) When point 'A' reaches 'A1' then point 'B' will be at 'P'. It even now has to cover distance PB1 to reach XY.

iv) According to Huygens' principle the secondary waves will originate from A1 and will spread over a hemisphere in the glass.

v) All the rays between AA1 and BB1 will reach XY and spread over the hemispheres of increasing radii in the glass. The surface of tangential of all such hemispheres is RB1. This allow to rise refracted wavefront B1R in the glass.

vi)  A1R and B1R1, are refracted rays.

vii) Let consider c1 and c2 be the velocities of light in air and glass .

viii) At any instant of time 't' distance covered by the incident wavefront from point P to B1 = PB1 = c1t.

The distance covered by the secondary wave from point A1 to R = A1R = c2t. 

Proof of laws of refraction:

i)  From figure,

angle AA1M + angle MA1P = 90° .................................(1)

and angle MA1P+ angle PA1B1 = 90° ...........................(2)

From equations (1) and (2),

angle AA1M = angle PA1B1 = i 

ii) Similarly, angle NA1R = angle N1B1R1 = r

We have, angle N1B1R1 + angle N1B1R = 90°.............(3)

 and angle N1B1R + angle A1B1R = 90° ......................(4)

From equations (3) and (4)

angle N1B1R1 = angle A1B1R = r

iii) In triangle A 1PB1, sin i = PB1/A1B1 = c1t/A1B1................(5)

iv) In triangle A1RB1, sin r = A1R/A1B1 = c2t /A1B1..............(6)

v) Dividing equation (5) by (6),

Therefore sin i/ sin r = c1/ c2....................(7)

Also c1/ c2 = μ2/ μ1 = 1μ2 ......................(8)

Where as 1μ2 = R.I. of water with respect to air,

sin i/ sin r = μ2/ μ1

vi) From the explanation, it is clear that incident rays AA, BB, refracted rays A,R, B,R and normal MN and MIN lie on the same plane XY. Also incident ray AA, and refracted ray AR lie on opposite sides of normal MN. Therefore, laws of refraction can be explained.

THE END

Thank you

With the help of a neat diagram | explain the reflection of light from the plane reflecting surface on the basis of wave theory of light.

So here first of all we know that, 

What is the laws of reflection :

i) In the laws of reflection, the incident rays, reflected rays, and normal to the reflecting surface at the point of incidence, all lie in the same level.

ii) In the laws of reflection, the incident rays and the reflected rays lie on the opposite sides of the normal.

iii) The angle of incidence is equal or we can say equivalent to the angle of reflection.

For clear understanding the law of reflection in simple and easy way. So for that lets see the simple explanation about all these points.

Explanation :

i) The plane wavefront AB is developing at an angle towards plane reflecting surface XY.AA1 and BB1 are incident rays.

ii) When point 'A' come to XY at A1, than ray at 'B' reaches at point P and it has to cover distance PB1 to reach the reflecting surface XY.

iii) Let's consider 't' be the time to  required to cover distance PB1. 

In this time meantime, the secondary waves are emitted from the point A1 and it will spread over the hemisphere of the radius A1R in the same medium.

The distance covered by the secondary waves to reach from A1 to R in time 't' is same as the distance covered by the primary waves to reach from point P to B1.

So therefore A1R = PB1 = ct.

iv) All other rays in between AA1 and BB1 will reach XY after A1 and before B1. Therefore they will also emits secondary waves of the decreasing radii.

v) The surface touch all such hemispheres is RB1 which is reflected wavefront, bounded by the reflected rays A1R and B1Q.

vi) Let's draw A1M perpendicular to XY and B1N perpendicular to XY.

Thus,  the angle of incidence is angle AA1M = angle BB1N = i and the angle of reflection is angle MA1R = angle NB1Q = r.

angle RA1B1 = 90 - r

angle PB1A1 = 90 - i

vii) In the triangle A1RB1 and triangle A1PB1

Here angle A1RB1 congruent to angle A1PB1

A1R = PB1,............ (Reflected waves travel in equal in same medium in equal time).

A1B1 = A1B1.........(Common side) 

Therefore triangle A1RB1 congruent to triangle A1PB1

Therefore angle RA1B1 = angle PB1A1

Therefore  90 - r = 90 - i

Therefore  i = r

viii) From the above given figure, it clear that the incident ray, reflected ray and normal lie in the same level or plane.

ix) This laws of reflection of light from the plane reflecting surface on the basis of Huygen's wave theory.

THE END

Thank you

The Huygen's construction of spherical wavefront in simple and easy way !!!

Here we come to know about the Huygen's construction of spherical wavefront.

So let's come to know,

i) The spherical wavefront is formed when the source of light is at the finite distance from the point of observation.

ii) So let's take 'S' be the point source of light in the air.

The point PQR represents spherical wavefront at any instant. 

The wavefront PQR is act as a primary wave, which is propagated through the air.

iii) So according to the Huygens principle, the all points on the PQR will act as a secondary source of light and sending out secondary wavelets with the same velocity 'c' in the air.

iv) To find out new wavefront at a later instant 't',

So for our clear understanding lets draw the hemisphere with the points P, Q, R....as centres and 'ct' as radius in the forward direction.

v) The surface tangential to all such hemisphere is an envelope at that instant 't'.

Such a surface is passing through the points P1Q1R1....so on the hemispheres and touching to all the hemispheres.

This surface is the new wavefront at that instant 't'.

vi) SPN1, SQN2, SRN3 these are the wave normal at points P, Q, R respectively.

vii) These wave normal show the direction of propagation of the spherical wavefront.

viii) The new wavefront P1Q1R1 is parallel to the points P, Q, R at every instant.

THE END

Thank you...!!!

 Explain the Huygen's construction of plane wavefront in easy and simple way !!!

Here we know all about, the construction of the plane wavefront.

So let's come to know,

i) The plane wavefront is formed when the point of observation is very far away or we can say at infinity from the primary source of light.

ii) So here let's PQR represent a plane wavefront at any instant.

So according to the Huygen's principle, all the points on this wavefront will act as a secondary source of light and sending out secondary wavelets in the forward direction only.

iii) For make it to undestand easily, first of all we need to draw hemisphere with Point P,Q,R.... as centres and 'ct' as a radius.

The surface tangential to all such hemisphere is P1Q1R1....at instant 't'.

The new wavefront at time 't'.

iv) The plane wavefront is propagated or travel as a plane waves in the homogeneous isotropic medium. Therefore they are parallel to each other.

v) PP1N1, QQ1N2, RR1N3 these are the wave normal at point P,Q,R respectively.

These wave normal shows the direction of propagation of the plane wavefront.

vi) The new wavefront P1Q1R1 is parallel to the primary wavefront PQR.

THE END

Thank you....!!!!

Difference in between primary source of light and secondary source of light in easy way !!!

Here we know about the distinguish/ difference between primary source of light and the secondary source of light.

Primary source of light

i) The primary source of light is a real source of light.

ii) Primary source of light sends out primary waves in the all possible directions.

iii) The primary source of light is effective at every point on it's surface.

iv) The primary source of light is situated in the air.

The secondary source of light 

i) The secondary source of light is a fictitious source of light.

ii) The secondary source of light sends out secondary waves only in the forward direction.

iii) The secondary waves is effective only at the points where it touches the envelope.

iv) The secondary source of light is situated on the wavefront.

THE END

State what are the Huygen's principle statement !!!

Hello dear all readers,

Welcome to this concept, so here in this concept you come to know about the principle statement of Huygen's. 

So lets start to know,

i) Every point on the primary wavefront it acts as a secondary waves or wavelets in the all possible directions.

ii) The new generated secondary wave is quite more effectiveness in the forward direction only.

iii) The resultant wavefront at any position is given by the tangent to all the secondary waves or wavelets at that instant.

So we hope you understand this concept.

So we will meet you with new concept.

Goodbye !!!

Thank you...!!!