Subject

Physics

Class

CBSE Class 12

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Sample Papers

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 Multiple Choice QuestionsLong Answer Type

31.

Draw a simple circuit of a CE transistor amplifier. Explain its working. Show that the voltage gain AV, of the amplifier is given by A subscript v space equals space minus fraction numerator beta subscript a c end subscript R subscript L over denominator r subscript i end fraction, where bold italic beta subscript bold italic a bold italic c end subscript is the current gain, RL is the load resistance and ri is the input resistance of the transistor. What is the significance of the negative sign in the expression for the voltage gain?


Circuit diagram of CE transistor amplifier, 

 

When an ac input signal VI (to be amplified) is superimposed on the bias VBB, the output, which is measured between collector and ground, increases.

We first assume that Vi = 0.

Applying Kirchoff’s law to the output loop,

       VCC = VCE + IC RL

Similarly, the input loop gives, 

        VBB = VBE + IB RB

When Vi is not zero, we have

VBE + Vi = VBE + IB RBincrement I subscript B space left parenthesis R subscript B space plus space R subscript i right parenthesis

rightwards double arrow straight space straight V subscript straight i straight space equals straight space increment straight I subscript straight B left parenthesis straight space straight R subscript straight B plus straight R subscript straight i right parenthesis

rightwards double arrow straight space straight V subscript straight i straight space equals straight space straight r increment straight I subscript straight B
Change in IB causes changes in Ic

Hence, straight beta subscript ac straight space equals straight space fraction numerator increment straight I subscript straight c over denominator increment straight I subscript straight B end fraction straight space equals straight space straight I subscript straight C over straight I subscript straight B 

As, increment straight V subscript CC straight space end subscript equals increment straight space straight V subscript CE straight space plus straight space straight R subscript straight L increment straight I subscript straight C straight space equals straight space 0

     rightwards double arrow straight space increment straight V subscript CE straight space equals straight space minus straight R subscript straight L increment straight I subscript straight C

The change in VCE is the output voltage Vis, 

rightwards double arrow straight space straight V subscript straight o straight space end subscript equals straight space minus straight R subscript straight L increment straight I subscript straight C

space space space space space space space space equals space straight beta subscript ac increment straight I subscript straight B straight R subscript straight L

The voltage gain of the amplifier is,


straight A subscript straight V straight space equals straight space straight V subscript straight o over straight V subscript straight i equals fraction numerator increment straight V subscript CE over denominator straight r increment straight I subscript straight B end fraction equals straight space fraction numerator negative straight beta subscript ac increment straight I subscript straight B straight R subscript straight L over denominator straight r increment straight I subscript straight B end fraction equals negative straight beta subscript ac straight R subscript straight L over straight r

Negative sign in the expression shows the output voltage and input voltage have phase difference of straight pi.

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32.

Define magnifying power of a telescope. Write its expression.

A small telescope has an objective lens of focal length 150 cm and an eye piece of focal length 5 cm. If this telescope is used to view a 100 m high tower 3 km away, find the height of the final image when it is formed 25 cm away from the eye piece.


a) Magnifying power of telescope is the ratio of the angle subtended at the eye by the image to the angle subtended at the unaided eye by the object.

Mathematically, we can write

Error converting from MathML to accessible text.
where, fo is the focal length of the objective, fe is the focal length of the eye-piece and D is the least distance of distinct vision.

b) Using, the lens equation for objective lens,

1 over straight f subscript straight o equals 1 over straight v subscript straight o minus 1 over straight u subscript straight o space

rightwards double arrow 1 over 150 equals straight space 1 over straight v subscript straight o minus fraction numerator 1 over denominator negative 3 cross times 10 to the power of 5 end fraction space

rightwards double arrow 1 over 150 minus fraction numerator 1 over denominator 3 cross times 10 to the power of 5 end fraction equals straight space 1 over straight v subscript straight o space

rightwards double arrow straight v subscript straight o straight space equals fraction numerator 3 cross times 10 to the power of 5 over denominator 1999 end fraction equals straight space 150 straight space cm

Hence, magnification due to the objective lens is given by,

bold m subscript bold o straight space equals straight space bold v subscript bold o over bold u subscript bold o equals fraction numerator 150 cross times 10 to the power of negative 2 end exponent over denominator 3000 end fraction equals straight space 10 to the power of negative 2 end exponent over 20
italic rightwards double arrow italic space m subscript o space italic equals space italic 0 italic. italic 05 italic cross times italic 10 to the power of italic minus italic 2 end exponent

Now, using lens formula for eye-piece, we get

space space space space space space space space 1 over bold italic f subscript bold italic e equals 1 over bold italic v subscript bold italic e minus 1 over bold italic u subscript bold italic e

rightwards double arrow space space space space space space 1 fifth equals fraction numerator 1 over denominator negative 25 end fraction minus 1 over bold italic u subscript bold italic e
rightwards double arrow space space space space space bold italic u subscript bold italic e space equals space fraction numerator negative 25 over denominator 6 end fraction bold italic c bold italic m

Therefore, magnification due to eyepiece me = fraction numerator negative 25 over denominator negative 25 over 6 end fraction equals straight space 6 straight space cm
Hence, total magnification, straight m straight space equals straight space straight m subscript straight e cross times straight m subscript straight o

bold italic m equals space 6 space cross times 5 cross times 10 to the power of negative 4 end exponent

space space space equals space 30 cross times 10 to the power of negative 4 end exponent

So, size of final image = 30 cross times 10 to the power of negative 4 end exponent cross times 100 space straight m space equals space 30 space cm

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33.

Draw the circuit diagram of a full wave rectifier using p-n junction diode. Explain its working and show the output, input waveforms.

(b) Show the output waveforms (Y) for the following inputs A and B of

(i) OR gate  

ii) NAND gate




a)
                     

Working: The AC input voltage across secondary s1 and s2 changes polarity after each half cycle. Suppose during the first half cycle of input AC signal, the terminal s1 is positive relative to centre tap O and s2 is negative relative to O. Then diode D1 is forward biased and diode D2 is reverse biased. Therefore, diode D1 conducts while diode D2 does not. The direction of current (i1) due to diode D1 in load resistance RL is directed from A to B. In next half cycle, the terminal s1 is negative and s2 is positive relative to centre tap O. The diode D1 is reverse biased and diode D2 is forward biased. Therefore, diode D2 conducts while D1does not. The direction of current (i2 ) due to diode D2 in load resistance RL is still from A to B. Thus the current in load resistance RL is in the same direction for both half cycles of input AC voltage. Thus for input AC signal the output current is a continuous series of unidirectional pulses.



In a full wave rectifier, if input frequency is f hertz, then output frequency will be 2f hertz because for each cycle of input, two positive half cycles of output are obtained.

b) Output waveforms for the following inputs A and B of OR gate and NAND gate.

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