The current gain of a common base transistor circuit is 0.96. On changing the emitter current by 10.0 mA, the change in the base current will be

Question

The current gain of a common base transistor circuit is 0.96. On changing the emitter current by 10.0 mA, the change in the base current will be (A) 9.6 mA (B) 0.4 mA (C) 19.6 mA (D) 24 mA

Tardigrade Admin · Accepted Answer

Current gain for common base transistor α=((Δ IC/Δ IE))VC Given, α=0.96 , Δ IE =10.0mA 0.96=(Δ IC/10.0) Δ IC =0.96 × 10.0 =9.6 mA

Q. The current gain of a common base transistor circuit is 0.96. On changing the emitter current by 10.0 mA, the change in the base current will be

9.6 mA

0.4 mA

19.6 mA

24 mA

Current gain for common base transistor
$α = (\frac{Δ I _{C}}{Δ I _{E}})_{V_{C}}$
Given, $α = 0.96, Δ I_{E} = 10.0 m A$
$0.96 = \frac{Δ I _{C}}{10.0}$
$Δ I_{C} = 0.96 \times 10.0 = 9.6 m A$

Q. The current gain of a common base transistor circuit is 0.96. On changing the emitter current by 10.0 mA, the change in the base current will be

9.6 mA

0.4 mA

19.6 mA

24 mA

Current gain for common base transistor α=(ΔIE​ΔIC​​)VC​​ Given, α=0.96,ΔIE​=10.0mA 0.96=10.0ΔIC​​ ΔIC​=0.96×10.0=9.6mA

Current gain for common base transistor
$α = (\frac{Δ I _{C}}{Δ I _{E}})_{V_{C}}$
Given, $α = 0.96, Δ I_{E} = 10.0 m A$
$0.96 = \frac{Δ I _{C}}{10.0}$
$Δ I_{C} = 0.96 \times 10.0 = 9.6 m A$