A) \[\frac{1}{\alpha }\]
B) \[\frac{\alpha }{1+\alpha }\]
C) \[\frac{\alpha }{1-\alpha }\]
D) \[\alpha -\frac{1}{\alpha }\]
Correct Answer: C
Solution :
Current gain in common-base configuration is. \[\alpha ={{\left( \frac{\Delta {{i}_{C}}}{\Delta {{i}_{E}}} \right)}_{{{V}_{CB}}}}\] Current gain in common-emitter configuration is, \[\beta ={{\left( \frac{\Delta {{i}_{C}}}{\Delta {{i}_{B}}} \right)}_{{{V}_{CE}}}}\] \[Also{{i}_{B}}={{i}_{E}}-{{i}_{C}}\] \[or\Delta {{i}_{B}}=\Delta {{i}_{E}}-\Delta {{i}_{C}}\] \[\therefore \beta =\frac{\Delta {{i}_{C}}}{\Delta {{i}_{B}}}=\frac{\Delta {{i}_{C}}}{\Delta {{i}_{E}}}\times \frac{\Delta {{i}_{E}}}{\Delta {{i}_{B}}}\] \[or\beta =\alpha \times \frac{\Delta {{i}_{E}}}{\Delta {{i}_{E}}-\Delta {{i}_{C}}}\] \[orB=\alpha \times \frac{1}{1-\frac{\Delta {{i}_{C}}}{\Delta {{i}_{E}}}}\] \[or\beta =\frac{\alpha }{1-\alpha }\] Note: \[\beta \] is always greater than \[\alpha \]. Also \[\alpha <1\] and \[\beta >1\].
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