A) \[\beta =\frac{1+\alpha }{\beta }\]
B) \[\alpha =\frac{\beta }{1+\beta }\]
C) \[\alpha =\frac{\beta }{1-\beta }\]
D) \[\beta =\frac{\alpha }{1+\alpha }\]
Correct Answer: B
Solution :
Current gain in common-base configuration \[\alpha =\frac{\Delta {{I}_{C}}}{\Delta {{I}_{E}}}\] Current gain in common-emitter configuration \[\beta =\frac{\Delta {{I}_{C}}}{\Delta {{I}_{B}}}\] and \[{{I}_{E}}={{I}_{B}}+{{I}_{C}}\] \[\Rightarrow \] \[\Delta \,{{I}_{E}}=\Delta \,{{I}_{B}}\,+\,\Delta \,{{I}_{C}}\] \[\therefore \] \[\alpha =\frac{\Delta \,{{I}_{C}}}{\Delta \,{{I}_{B}}+\Delta \,{{I}_{C}}}\,=\frac{\Delta \,{{I}_{C}}/\Delta \,{{I}_{B}}}{1+\frac{\Delta \,{{I}_{C}}}{\Delta \,{{I}_{B}}}}\,=\,\frac{\beta }{1+\beta }\] or \[\alpha =\frac{\beta }{\beta +1}\] NOTE : is about 1 to 5% of \[{{\text{l}}_{E}}.\alpha \], is about 0.95 to 0.99 and \[\beta \] is about 20 to 100. By simple mathematics we can prove that \[\beta =\frac{\alpha }{1-\alpha }\]
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