In the circuit given below, if $V_{CE}$ is 5 V and $R=4\times {10}^5\Omega$, then current gain is:
(Neglect \(V_{BE}\))
           

1. 97

2. 98

3. 99

4. 100

A Zener diode is shown in the following circuit diagram. When the source voltage fluctuates such that \(V>V_z\) then:
   

When the emitter current increases by 10 mA, the base current increases by 0.4 mA. The value of voltage gain in the common emitter configuration of the amplifier will be:

(output and input resistance ratio is 50)

1. 1200

2. 1250

3. 150

4. 120

If the current gain in the transistor is 50, then $V_{CE}$ is equal to:
                  
1. 2 V

2. 3 V

3. 4 V

4. 6 V

The net charges on $\mathrm{p}$-type semiconductor and $\mathrm{n}$-type semiconductor are, respectively:

1. \(2\) A and zero
2. \(3\) A and \(2\) A
3. \(2\) A and \(3\) A
4. zero and \(2\) A

If in a reverse-biased \(\mathrm{p\text-n}\) junction, an increase in carrier concentration takes place due to the creation of new hole-electron pairs by the light of wavelength less than or equal to \(620\) nm, then the bandgap is:
1. \(1\) eV
2. \(2\) eV
3. \(20\) eV
4. \(0.2\) eV

A n-p-n transistor operates in a common emitter mode as shown. The value of \(R_{L}\) is:
\(\small\left ( \text{Given,}I_{C}=4 mA,~V_{CE}=4~V,V_{BE}=0.6~V~\text{and}~\beta _{dc}=100 \right )\\\)
        

1. 1 k$\Omega$

2. 2 k$\Omega$

3. 3 k$\Omega$

4. 4 k$\Omega$