Which one of the following represents the forward bias diode?

1.
2.
3.
4.

For the CE transistor amplifier, the audio signal voltage across the collector resistance of 2k$\Omega$ is 4V. If the current amplification factor of the transistor is 100 and the base resistance is 1k$\Omega$, then the input signal voltage is:

1.  10 mV

2.  20 mV

3.  30 mV

4.  15 mV

The given circuit has two ideal diodes connected as shown in the figure below. The current flowing through the resistance \(R_1\) will be:
   
1. \(2.5~\text{A}\)
2. \(10.0~\text{A}\)
3. \(1.43~\text{A}\)
4. \(3.13~\text{A}\)

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

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. \(0~\text{A}\)

In the given figure, a diode \(D\) is connected to an external resistance \(R = 100~\Omega\)$$ and an EMF of \(3.5~\text{V}\). If the barrier potential developed across the diode is \(0.5~\text{V}\), the current in the circuit will be:
  
1. \(30~\text{mA}\)
2. \(40~\text{mA}\)
3. \(20~\text{mA}\)
4. \(35~\text{mA}\)

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.