Class 12th

Let $A=\left[\begin{array}{cc}\hfill 2\hfill & \hfill 3\hfill \\ \hfill -4\hfill & \hfill -6\hfill \end{array}\right]\left|A\right|=\left(2*\left(-6\right)-3*\left(-4\right)\right)=-12+12=0$

$\begin{array}{l}{A}_{11}={(-1)}^{1+1}*(-6)=-6\\ A_{12}={(-1)}^{1+2}*(-4)=4\end{array}$

$\begin{array}{l}{A}_{21}={(-1)}^{2+1}*(3)=-3\\ A_{22}={(-1)}^{2+2}*2=2\end{array}$$$

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

Explanation- I_E=I_C+I_B and I_c= $\beta I_B$

I_cRc+V_CE+I_ER_E=V_CC

RI_B+V_BE+I_ER_E=V_cc

I_E=I_c= $\beta I_B$

From above equation

(R+ $\beta R$ _E)I_B=V_CC-V_BE

I_B= $\frac{V_{CC}-V_{BE}}{R+\beta R_E}=\frac{12-0.5}{80+1.2*100}=\frac{11.5}{200}$

( $R_C+R_E$ )= $\frac{V_{CE}-V_{BE}}{I_C}$

( $R_C+R_E$ )=1.56

Rc=1.56-1=0.56kohm

We have,

$A_{11}={(-1)}^{1+1}*\left|\begin{array}{cc}\hfill 3\hfill & \hfill 5\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right|=(3*1-5*0)=3$

$A_{12}={(-1)}^{1+2}*\left|\begin{array}{cc}\hfill 2\hfill & \hfill 5\hfill \\ \hfill -2\hfill & \hfill 1\hfill \end{array}\right|=(-1)(2*1-5*(-2)=-1(2+10)=-12$

$A_{13}={(-1)}^{1+3}*\left|\begin{array}{cc}\hfill 2\hfill & \hfill 3\hfill \\ \hfill -2\hfill & \hfill 0\hfill \end{array}\right|=[2*0-(-2)*3]=6$

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

Explanation-I_c= I_E

R_c= 7.8Kohm

I_c (R_c+R_E)+V_CE= 12

(R_E+R_C) $*1*{10}^{-3}+3=12$

R_E+R_C= 9 $*{10}^3=9kohm$

R_E= 9-7.3= 1.2kohm

V_E= I_{E $*$} R_E

 = 1 $*{10}^{-3}*1.2*{10}^3=1.2V$

Voltage V_B=V_E+V_BE= 1.2+0.5= 1.7V

I= $\frac{1.7}{20*{10}^3}=0.085mA$

Resistance R_B= $\frac{12-1.7}{\frac{I_C}{?}+0.085}=\frac{10.3}{0.01+0.085}=108kohm$

We have,

$A_{11}={(-1)}^{1+1}*4=4$

$A_{12}={(-1)}^{1+2}*3=-3$

$A_{21}={(-1)}^{2+1}*2=-2$

$A_{22}={(-1)}^{2+2}*1=1.$

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

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

whem As is used then it will create n type semiconductor

N_e=N_D= $\frac{1}{{10}^6}*5*{10}^{28}=5*{10}^{22}/m^3$

Number of minority carriers n_h= $\frac{{ni}^2}{ne}=\frac{{(1.5*{10}^{16})}^2}{5*{10}^{22}}$ = 0.45 $*{10}^{10}/m^3$

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

N_h=N_A= $\frac{200}{{10}^6}*\left(5*{10}^{28}\right)=1*{10}^{25}/m^3$

Minority carriers created in p type is

N_e= $\frac{{ni}^2}{ne}=\frac{({1.5*{10}^{16})}^2}{1*{10}^{25}}=2.25*{10}^{27}/m^3$

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

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