A) \[\frac{1}{19}\]
B) 19
C) 1.5
D) 0.95
Correct Answer: B
Solution :
From definition \[\alpha =\frac{\Delta {{i}_{C}}}{\Delta {{i}_{E}}}\]and \[\beta =\frac{\Delta {{i}_{C}}}{\Delta {{i}_{B}}}\] where, \[\Delta {{i}_{C}}=\] change in collector current \[\Delta {{i}_{E}}=\]change in emitter current \[\Delta {{i}_{B}}=\]change in base current Also, \[\Delta {{i}_{B}}=\Delta {{i}_{E}}-\Delta {{i}_{C}}\] \[\therefore \] \[\beta =\frac{\Delta {{i}_{C}}}{\Delta {{i}_{E}}-\Delta {{i}_{C}}}=\frac{\alpha }{1-\frac{\Delta {{i}_{C}}}{\Delta {{i}_{E}}}}\] \[\beta =\frac{\alpha }{1-\alpha }\] Putting the numerical values from the question, we have \[\beta =\frac{0.95}{1-0.95}=\frac{0.95}{0.05}\] \[\beta =1.9\] Alternative: a is very nearly unity, so \[(1-\alpha )\] is very small hence, P is very large. Therefore, options (a), (c) and (d) get eliminated. You are left only with option (b).
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