If α and β are the current gain in the CB and CE configurations respectively of the transistor circuit, then (β - α/αβ) =

Question

If α and β are the current gain in the CB and CE configurations respectively of the transistor circuit, then (β - α/αβ) =  (A) zero (B) 1 (C) 2 (D) 0.5

Tardigrade Admin · Accepted Answer

Since β = (α/1-α) ∴ αβ = (α2/1-α) ...(i) β - α = (α/1-α) -α or β - α = (α2/1-α) ...(ii) ∴ (β-α/αβ) = ((α2/1-α))((1-α/α2) ) = 1

Q. If $α$ and $β$ are the current gain in the $CB$ and $CE$ configurations respectively of the transistor circuit, then $\frac{β - α}{α β} =$

zero

$1$

$2$

$0.5$

Since $β = \frac{α}{1 - α}$
$∴ α β = \frac{α ^{2}}{1 - α} ... (i)$
$β - α = \frac{α}{1 - α} - α$ or
$β - α = \frac{α ^{2}}{1 - α} ... (ii)$
$∴ \frac{β - α}{α β} = (\frac{α ^{2}}{1 - α}) (\frac{1 - α}{α ^{2}}) = 1$

Q. If α and β are the current gain in the CB and CE configurations respectively of the transistor circuit, then αββ−α​=

zero

1

2

0.5

Since β=1−αα​ ∴αβ=1−αα2​...(i) β−α=1−αα​−α or β−α=1−αα2​...(ii) ∴αββ−α​=(1−αα2​)(α21−α​)=1

Q. If $α$ and $β$ are the current gain in the $CB$ and $CE$ configurations respectively of the transistor circuit, then $\frac{β - α}{α β} =$

$1$

$2$

$0.5$

Since $β = \frac{α}{1 - α}$
$∴ α β = \frac{α ^{2}}{1 - α} ... (i)$
$β - α = \frac{α}{1 - α} - α$ or
$β - α = \frac{α ^{2}}{1 - α} ... (ii)$
$∴ \frac{β - α}{α β} = (\frac{α ^{2}}{1 - α}) (\frac{1 - α}{α ^{2}}) = 1$