A common emitter amplifier is designed with NPN transistor ($\mathrm{\alpha }$ = 0.99). The input impedance is 1 K$\mathrm{\Omega }$ and load is 10 K$\mathrm{\Omega }$. The voltage gain will be:

(1) 9.9                           

(2) 99

(3) 990                         

(4) 9900

The symbol given in figure represents 

(1) NPN transistor

(2) PNP transistor

(3) Forward biased PN junction diode

(4) Reverse biased NP junction diode

The most commonly used material for making transistor is

(1) Copper                 

(2) Silicon

(3) Ebonite                 

(4) Silver

An NPN-transistor circuit is arranged as shown in figure. It is
 

1. A common base amplifier circuit

2. A common emitter amplifier circuit

3. A common collector amplifier circuit

4. Neither of the above

The part of a transistor which is heavily doped to produce a large number of majority carriers is-

1. Base                 

2. Emitter

3. Collector           

4. None of these

For a transistor, the current amplification factor is 0.8 when it is connected in the common base configuration. The transistor is now connected in common emitter configuration. The change in the collector current when the base current changes by 6 mA, is:
1. 6 mA                       

2. 4.8 mA

3. 24 mA                     

4. 8 mA

In a common base amplifier circuit, calculate the change in base current if that in the emitter current is 2 mA and $\mathrm{\alpha }$ = 0.98

(1) 0.04 mA                           

(2) 1.96 mA

(3) 0.98 mA                           

(4) 2 mA

In a transistor circuit shown here the base current is 35 $\mu$A. The value of the resistor Rb is

1. 123.5 k$\mathrm{\Omega }$

2. 257 k$\mathrm{\Omega }$

3. 380.05 k$\mathrm{\Omega }$ 

4. None of these

For a transistor, in a common emitter arrangement, the alternating current gain $\mathrm{\beta }$ is given by

(1) $\mathrm{\beta }={\left(\frac{∆{\mathrm{I}}_{\mathrm{c}}}{∆{\mathrm{I}}_{\mathrm{B}}}\right)}_{{\mathrm{V}}_{\mathrm{C}}}$                       

(2) $\mathrm{\beta }={\left(\frac{∆{\mathrm{I}}_B}{∆{\mathrm{I}}_C}\right)}_{{\mathrm{V}}_{\mathrm{C}}}$

(3) $\mathrm{\beta }={\left(\frac{∆{\mathrm{I}}_{\mathrm{c}}}{∆{\mathrm{I}}_E}\right)}_{{\mathrm{V}}_{\mathrm{C}}}$                       

(4) $\mathrm{\beta }={\left(\frac{∆{\mathrm{I}}_E}{∆{\mathrm{I}}_C}\right)}_{{\mathrm{V}}_{\mathrm{C}}}$

The relation between $\mathrm{\alpha }$ and $\mathrm{\beta }$ parameters of current gains for a transistors is given by

(1) $\mathrm{\alpha }=\frac{\mathrm{\beta }}{1-\mathrm{\beta }}$               

(2) $\mathrm{\alpha }=\frac{\mathrm{\beta }}{1+\mathrm{\beta }}$

(3) $\mathrm{\alpha }=\frac{1-\mathrm{\beta }}{\mathrm{\beta }}$               

(4) $\mathrm{\alpha }=\frac{1+\mathrm{\beta }}{\mathrm{\beta }}$