A) 0.98
B) 0.97
C) 0.96
D) 0.94
Correct Answer: C
Solution :
No. of electrons reaching the collector,\[{{n}_{C}}=\frac{96}{100}\times {{10}^{10}}=0.96\times {{10}^{10}}\] Emitter current, \[{{I}_{E}}=\frac{{{n}_{E}}\times e}{t}\] Collector current, \[{{I}_{C}}=\frac{{{n}_{C}}\times e}{t}\] \[\therefore \]Current transfer ratio,\[\alpha =\frac{{{I}_{C}}}{{{I}_{E}}}=\frac{{{n}_{C}}}{{{n}_{E}}}=\frac{0.96\times {{10}^{10}}}{{{10}^{10}}}=0.96\]
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