Physics NCERT Exemplar Solutions Class 12th Chapter Fourteen

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A
alok kumar singh

Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation- a) In V- graph of condition (i), a reverse characteristics is shown in fig. (c). Here A is connected to n - side of p-n junction and B is connected to p -side of p-n junction I with a resistance in series.

(b) In V- graph of condition (ii), a forward characteristics is shown in fig. (d), where 0.7 V is

the knee voltage of p-n junction I 1/slope= (1/1000)? It means A is connected to n -side of p n- junction and B is connected to p-side of p n- junction and resistance R is in series of p n- junction between A and B.

(c) In V- graph of condition (iii), a

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alok kumar singh

Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar

Y= A'B+A.B'=Y1+Y2

Y1= A.B and Y2= A.B'

Y1 can be obtained as output of AND GATE I for which one Input is of A through NOT GATE and

another input is of B. Y2 can be obtained as output of AND GATE II for which one input is of A

and other input is of B through NOT gate.

Now Y2 can be obtained as output from or gate, where, Y1 and Y2 are input of or gate.

Thus, the given table can be obtained from the logic circuit given below

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alok kumar singh

Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar

whem As is used then it will create n type semiconductor

Ne=ND= 1 10 6 * 5 * 10 28 = 5 * 10 22 / m 3

Number of minority carriers nh= n i 2 n e = ( 1.5 * 10 16 ) 2 5 * 10 22 = 0.45 * 10 10 / m 3

But when B is implanted in Si crystal then p type semiconductor is formed

Nh=NA= 200 10 6 * 5 * 10 28 = 1 * 10 25 / m 3

Minority carriers created in p type is

Ne= n i 2 n e = ( 1.5 * 10 16 ) 2 1 * 10 25 = 2.25 * 10 27 / m 3

Minority charge carriers holes in p type would contribute more in reverse saturatiom current.

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A
alok kumar singh

Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar

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A
alok kumar singh

Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation- When the input voltage is equal to or less than 5 V, diode will be revers biased. It will offer high resistance in comparison to resistance ( ) R in series. Now, diode appears in open circuit. The input waveform is then passed to the output terminals. The result with sin wave input is to dip off all positive going portion above 5 V.

If input voltage is more than + 5 V, diode will be conducting as if forward biased offering low resistance in comparison to R. But there will be no voltage in output beyond 5 V as the

voltagebeyond+5Vwillappearacross R.

When

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alok kumar singh

Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar

A         B

C          D

E           F

   G

    H

I

C1

0         0

0          0

1          1

   1

    0

     0

1

1         0

1          0

0          1

   0

    1

     1

0

0         1

0          1

1          0

   0

    1

     1

0

1         1

1          1

0          0

   0

    1

     1

0

A         B

C          D

E           F

   G

C2

0         0

0          0

1          1

   1

0

1         0

1          0

0          1

   1

0

0         1

0          1

1          0

   1

0

1         1

1          1

0          0

   0

1

New answer posted

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A
alok kumar singh

Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation- voltage across RB= 10V

Resistance RB= 400kohm

VBE= 0, VCE= 0, Rc=3kohm

IB= voltage /RB= 10/400 *  103= 25 * 10-6A

Voltage across Rc= 10V

Ic=voltage/Rc= 10/3 * 103= 3.33 * 10-3

β = I c I B = 3.33 * 10 - 3 25 * 10 - 6 = 133

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alok kumar singh

Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar

forward biased resistance = 25ohm

Reverse biased resistance = infinity

As CD branch is reverse biased having infinite resistance.

So I3= 0

Resistance in branch AB= 25+125= 150 ohm (R1)

Resistance in branch EF= 25+125= 150 ohm (R2)

They both are in parallel

 So 1 R = 1 R 1 + 1 R 2 = 1 150 + 1 150 = 2 150

R= 75ohm

So total resistance = 25+75= 100ohm

Current = V/R= 5/100=0.05A

I1=I2+I3+I4

I1=I4+I2

Here the resistance R1 and R2 is same

I4=I2

I1= 2I2

I2= I1/2=0.05/2=0.025A

I4= 0.025A

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