Physics

CBSE Class 12

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31.

Draw a simple circuit of a CE transistor amplifier. Explain its working. Show that the voltage gain A_{V}, of the amplifier is given by , where is the current gain, R_{L} is the load resistance and r_{i} 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 V_{I} (to be amplified) is superimposed on the bias V_{BB}, the output, which is measured between collector and ground, increases.

We first assume that V_{i} = 0.

Applying Kirchoff’s law to the output loop,

V_{CC} = V_{CE }+ I_{C} R_{L}

Similarly, the input loop gives,

V_{BB} = V_{BE} + I_{B }R_{B}

When V_{i} is not zero, we have

V_{BE} + V_{i} = V_{BE} + I_{B} R_{B} +

_{}Change in I_{B} causes changes in I_{c}

Hence,

As,

The change in V_{CE }is the output voltage V_{o }is,

The voltage gain of the amplifier is,

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

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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

where, f_{o} is the focal length of the objective, f_{e} 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,

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

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

Therefore, magnification due to eyepiece m_{e = Hence, total magnification, So, size of final image = }

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33.

ii) NAND gate

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 gateii) NAND gate

a)

Working: The AC input voltage across secondary s_{1} and s_{2} changes polarity after each half cycle. Suppose during the first half cycle of input AC signal, the terminal s_{1} is positive relative to centre tap O and s_{2} is negative relative to O. Then diode D_{1} is forward biased and diode D_{2 }is reverse biased. Therefore, diode D_{1} conducts while diode D_{2} does not. The direction of current (i_{1}) due to diode D_{1} in load resistance R_{L} is directed from A to B. In next half cycle, the terminal s_{1} is negative and s_{2} is positive relative to centre tap O. The diode D_{1} is reverse biased and diode D_{2} is forward biased. Therefore, diode D_{2} conducts while D_{1}does not. The direction of current (i_{2} ) due to diode D_{2} in load resistance R_{L} is still from A to B. Thus the current in load resistance R_{L} 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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