A) \[\beta =\frac{\alpha }{1-\alpha }\]
B) \[\beta =\frac{\alpha }{1+\alpha }\]
C) \[\beta =\frac{1+\alpha }{\alpha }\]
D) \[\beta =1-\alpha \]
Correct Answer: B
Solution :
We know that \[\alpha =\left( \frac{\Delta {{i}_{C}}}{\Delta {{i}_{E}}} \right)\] ... (i) and \[\beta =\left( \frac{\Delta {{i}_{C}}}{\Delta {{i}_{E}}} \right)=\frac{\Delta {{i}_{C}}}{\Delta {{i}_{E}}}\times \frac{\Delta {{i}_{E}}}{\Delta {{i}_{B}}}\] \[=\alpha \frac{\Delta {{i}_{E}}}{\Delta {{i}_{B}}}\] ... (ii) But\[\Delta {{i}_{B}}=\Delta {{i}_{E}}-\Delta {{i}_{C}}\] from (ii) \[\beta =\alpha \frac{\Delta {{i}_{E}}}{\Delta {{i}_{E}}-\Delta {{i}_{C}}}=\frac{\alpha }{1-\frac{\Delta {{i}_{C}}}{\Delta {{i}_{E}}}}\] \[\therefore \] \[\beta =\frac{\alpha }{1-\alpha }\]
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