Physics NCERT Exemplar Solutions Class 12th Chapter Fourteen
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New answer posted
4 months agoContributor-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
New answer posted
4 months agoContributor-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

New answer posted
4 months agoContributor-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=
Number of minority carriers nh= = 0.45
But when B is implanted in Si crystal then p type semiconductor is formed
Nh=NA=
Minority carriers created in p type is
Ne=
Minority charge carriers holes in p type would contribute more in reverse saturatiom current.
New answer posted
4 months agoContributor-Level 10
This is a Long Answer Type Questions as classified in NCERT Exemplar

New answer posted
4 months agoContributor-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
New answer posted
4 months agoContributor-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
4 months agoContributor-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
New answer posted
4 months agoContributor-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
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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