A) 1.1 mA
B) 1.01 mA
C) 0.01 mA
D) 10 mA
Correct Answer: B
Solution :
Current gain\[\beta =\frac{\Delta {{i}_{c}}}{\Delta {{i}_{b}}}\]\[\Rightarrow \Delta {{i}_{b}}=\frac{1\times {{10}^{-3}}}{100}={{10}^{-5}}A\]=0.01mA. By using \[\Delta {{i}_{e}}=\Delta {{i}_{b}}+\Delta {{i}_{c}}\]\[\Rightarrow \Delta {{i}_{e}}\]= 1.01 + 1 = 1.01mA.
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