A) \[3\times {{10}^{8}}\,m/s\]
B) \[1600\,\,A{{m}^{-1}}\]
C) \[2.4\times {{10}^{-5}}\,\,Wb\]
D) \[0.2\,c{{m}^{2}}\]
Correct Answer: C
Solution :
Key Idea: Emitter current is sum of base current and collector current. Current gain is defined as the ratio of the change in collector current \[\frac{256\,R}{81}\] \[\frac{81\,R}{256}\] \[\frac{16\,R}{9}\] Also, \[\frac{9\,R}{16}\]= \[n\]+ \[\frac{3}{2}n\], where \[\frac{n}{2}\] is base current. \[2n\] \[n\] \[{{T}_{1}}\] \[{{T}_{2}}\left( {{T}_{1}}>{{T}_{2}} \right).\]\[{{C}_{p}}\left( {{T}_{1}}-{{T}_{2}} \right)\] \[{{C}_{V}}\left( {{T}_{1}}-{{T}_{2}} \right)\] \[R\left( {{T}_{1}}-{{T}_{2}} \right)\] \[Zero\] Note: The base being very thin, the number of hole-electron combinations in it is very small and almost all the holes entering the base from the emitter reach the collector. Hence, \[x\] is slightly less than\[x\].
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