In the combination of the following gates, the output \(Y\) can be written in terms of inputs \(A\) and \(B\) as:

    
1. \(\overline {A\cdot B}\)
2. \(A\cdot \overline{B}+ B\cdot \overline{A}\)
3. \(\overline {A\cdot B}+ A\cdot B\)
4. \(\overline {A+ B}\)

In the circuit shown in the figure, the input voltage V_i is 20 V, V_BE = 0, and V_CE = 0. The values of I_B, I_C and $\beta$ are given by:
           

1. ${\mathrm{I}}_{\mathrm{B}}$ $=$ $40$ $\mathrm{\mu A},$ ${\mathrm{I}}_{\mathrm{c}}$ $=$ $10$ $\mathrm{mA},$ $\mathrm{\beta }$ $=$ $250$

2. ${\mathrm{I}}_{\mathrm{B}}$ $=$ $25$ $\mathrm{\mu A},$ ${\mathrm{I}}_{\mathrm{c}}$ $=$ $5$ $\mathrm{mA},$ $\mathrm{\beta }$ $=$ $200$

3. ${\mathrm{I}}_{\mathrm{B}}$ $=$ $20$ $\mathrm{\mu A},$ ${\mathrm{I}}_{\mathrm{c}}$ $=$ $5$ $\mathrm{mA},$ $\mathrm{\beta }$ $=$ $250$

4. ${\mathrm{I}}_{\mathrm{B}}$ $=$ $40$ $\mathrm{\mu A},$ ${\mathrm{I}}_{\mathrm{c}}$ $=$ $5$ $\mathrm{mA},$ $\mathrm{\beta }$ $=$ $125$

In a \(\mathrm{p\text{-}n}\) junction diode, the change in temperature due to heating:

The given electrical network is equivalent to:
   

Which one of the following represents the forward bias diode?

1.
2.
3.
4.

What is the output \(Y\) in the following circuit, when all the three inputs \(A\), \(B\), and \(C\) are first \(0\) and then \(1\)? 
                          
1. \(0,1\)
2. \(0,0\)
3. \(1,0\)
4. \(1,1\)

To get output \(Y=1\) for the following circuit, the correct choice for the input is:
  

Consider the junction diode as an ideal. The value of current flowing through \(AB\) is: 
   
1. \(10^{-2}~\text{A}\)
2. \(10^{-1}~\text{A}\)
3. \(10^{-3}~\text{A}\)
4. \(0~\text{A}\)

The input signal given to a CE amplifier having a voltage gain of 150 is ${\mathrm{V}}_{\mathrm{i}}=2\cos \left[15\mathrm{t}+\frac{{\displaystyle \mathrm{\pi }}}{{\displaystyle 3}}\right]$. The corresponding output signal will be:

1. $30<mspace\; width="1em"></mspace>\cos <mspace\; width="1em"></mspace>\left[15\mathrm{t}+\frac{\mathrm{\pi }}{3}\right]$

2. $75<mspace\; width="1em"></mspace>\cos <mspace\; width="1em"></mspace>\left[15\mathrm{t}+\frac{2\mathrm{\pi }}{3}\right]$

3. $2<mspace\; width="1em"></mspace>\cos <mspace\; width="1em"></mspace>\left[15\mathrm{t}+\frac{5\mathrm{\pi }}{3}\right]$

4. $300<mspace\; width="1em"></mspace>\cos <mspace\; width="1em"></mspace>\left[15\mathrm{t}+\frac{4\mathrm{\pi }}{3}\right]$

If in a \(\mathrm{p\text{-}n}\) junction, a square input signal of \(10~\text{V}\) is applied as shown, 

 
then the output across \(R_L\) will be:

1.		2.
3.		4.