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?

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Solution

The correct option is B $1.01$ mA
We know that in $N P N$ transistor,
Current gain $β = \frac{Δ i_{c}}{Δ i_{b}} \Rightarrow Δ i_{b} = \frac{1 \times 10^{- 3}}{100} = 10^{- 5} A = 0.01 m A .$
By using $Δ i_{e} = Δ i_{b} + Δ i_{c} \Rightarrow Δ i_{e} = 1.01 + 1 = 1.01 m A .$

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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?

The correct option is B 1.01 mAWe know that in NPN transistor, Current gain β=ΔicΔib⇒Δib=1×10−3100=10−5A=0.01 mA.By using Δie=Δib+Δic⇒Δie=1.01+1=1.01 mA.

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?

The correct option is B $1.01$ mA
We know that in $N P N$ transistor,
Current gain $β = \frac{Δ i_{c}}{Δ i_{b}} \Rightarrow Δ i_{b} = \frac{1 \times 10^{- 3}}{100} = 10^{- 5} A = 0.01 m A .$
By using $Δ i_{e} = Δ i_{b} + Δ i_{c} \Rightarrow Δ i_{e} = 1.01 + 1 = 1.01 m A .$