Consider an NPN transistor amplifier in common-emitter configuration. The current gain of the transistor is $100.$ If the collector current changes by $1\ mA$, what will be the change in emitter current
A$1.1 mA$
B$1.01 mA$
C$0.01 mA$
D$10 mA$
[AllMS 2005]
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B$1.01 mA$
Current gain $\beta=\frac{\Delta i_c}{\Delta i_b} \Rightarrow \Delta i_b=\frac{1 \times 10^{-3}}{100}=10^{-5} A=0.01 \mathrm{~mA}$.
By using $\Delta i_e=\Delta i_b+\Delta i_c \Rightarrow \Delta i_e=1.01+1=1.01 \mathrm{~mA}$.
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