A) 1
B) 0.99
C) 0.1
D) 0.01
Correct Answer: B
Solution :
Current gain a is defined for common-base transistor amplifier while p is defined for common-emitter transistor amplifier. We have \[\alpha =\frac{\Delta {{i}_{C}}}{\Delta {{i}_{E}}}\] ?. (i) and \[\beta =\frac{\Delta {{i}_{C}}}{\Delta {{i}_{B}}}\] ?. (ii) Also, \[{{i}_{E}}={{i}_{C}}+{{i}_{B}}\] \[\Rightarrow \] \[\Delta {{i}_{E}}=\Delta {{i}_{C}}+{{i}_{B}}\] .... (iii) From Eq. (ii), \[\beta =\frac{\Delta {{i}_{C}}}{\Delta {{i}_{B}}}=\frac{\Delta {{i}_{C}}}{\Delta {{i}_{E}}-\Delta {{i}_{C}}}\] or \[\beta =\frac{\Delta {{i}_{C}}/\Delta {{i}_{E}}}{1-\Delta {{i}_{C}}/\Delta {{i}_{E}}}\] or \[\beta =\frac{\alpha }{1-\alpha }\] or \[\alpha =\frac{\beta }{1+\beta }\] ?. (iv) Here, \[\beta =100\] So, From Eq. (iv) \[\alpha =\frac{100}{1+100}=0.99\] Note: For a transistor amplifier, \[\alpha \] is about 0 95 to 0.99 and \[\beta \] is about 20 to 100.
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