If $\mathrm{\alpha }$ = 0.98 and current through emitter ${\mathrm{i}}_{\mathrm{e}}$ = 20 mA, the value of $\mathrm{\beta }$ is 
(a) 4.9                  (b) 49
(c) 96                   (d) 9.6

In a PNP transistor working as a common-base amplifier, current gain is 0.96 and emitter current is 7.2 mA. The base current is:

(1) 0.4 mA                   

(2) 0.2 mA

(3) 0.29 mA                   

(4) 0.35 mA

 If ${\mathrm{l}}_1$, ${\mathrm{l}}_2$, ${\mathrm{l}}_3$ are the lengths of the emitter, base and collector of a transistor then 
(1)${\mathrm{l}}_1$= ${\mathrm{l}}_2$ = ${\mathrm{l}}_3$                   

(2) ${\mathrm{l}}_3$ <${\mathrm{l}}_2$> ${\mathrm{l}}_1$

(3) ${\mathrm{l}}_3$<  ${\mathrm{l}}_1$ < ${\mathrm{l}}_2$                 

(4) ${\mathrm{l}}_3$> ${\mathrm{l}}_1$>${\mathrm{l}}_2$

In an NPN transistor circuit, the collector current is 10 mA. If 90% of the electrons emitted reach the collector, the emitter current (${\mathrm{i}}_{\mathrm{E}}$) and base current (${\mathrm{i}}_{\mathrm{B}}$) are given by 
(1) ${\mathrm{i}}_{\mathrm{E}}$ = –1 mA, ${\mathrm{i}}_{\mathrm{B}}$ = 9 mA

(2) ${\mathrm{i}}_{\mathrm{E}}$ = 9 mA, ${\mathrm{i}}_{\mathrm{B}}$ = – 1 mA

(3) ${\mathrm{i}}_{\mathrm{E}}$ = 1 mA, ${\mathrm{i}}_{\mathrm{B}}$ = 11 mA

(4) ${\mathrm{i}}_{\mathrm{E}}$ = 11 mA, ${\mathrm{i}}_{\mathrm{B}}$ = 1 mA

In a common emitter transistor, the current gain is 80. What is the change in collector current, when the change in base current is 250 $\mathrm{\mu }$A 
(1) 80 $\times$ 250 $\mathrm{\mu }$A

(2) (250 – 80) $\mathrm{\mu }$A

(3) (250 + 80) $\mathrm{\mu }$A

(4) 250/80 $\mathrm{\mu }$A

Least doped region in a transistor 

(1) Either emitter or collector

(2) Base

(3) Emitter

(4) Collector

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