The transistors provide good power amplification when they are used in 

(1) Common collector configuration

(2) Common emitter configuration

(3) Common base configuration

(4) None of these

The transfer ratio of a transistor is \(50\). The input resistance of the transistor when used in the common-emitter configuration is\(1~\text{K}\Omega\). The peak value for an AC input voltage of \(0.01~\text{V}\) peak is:
1. \(100~\mu\text{A}\)
2. \(0.01~\text{mA}\)
3. \(0.25~\text{mA}\)
4. \(500~\mu\text{A}\)

For a transistor the parameter $\mathrm{\beta }$ = 99. The value of the parameter $\mathrm{\alpha }$ is

(1) 0.9                 

(2) 0.99

(3) 1                   

(4) 9

In a PNP transistor the base is the N-region. Its width relative to the P-region is 

(1) Smaller           

(2) Larger

(3) Same             

(4) Not related

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

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