For a transistor, (1)/(αDC) - (1)/(βDC) is equal to [ αDC and βDC are current amplification factors] (a) three (b) two (c) zero (d) one

Answer: (d). α = (β)/(1+β) ⇒ (1)/(α) = (1+β)/(β) = (1)/(β) + 1 (1)/(α) - (1)/(β) = 1 So the answer is (d) one.

For a transistor, (1)/(αDC) - (1)/(βDC) is equal to [ αDC and βDC are… | MHT CET 2021 (21 Sep, Shift 1)

Q62MHT CET 2021 · 21 SEP · SHIFT 1Easy

Junction Transistor

For a transistor, $\frac{1}{\alpha_\text{DC}} - \frac{1}{\beta_\text{DC}}$ is equal to [ $\alpha_\text{DC}$ and $\beta_\text{DC}$ are current amplification factors]

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Answer: (d)

$\alpha = \frac{\beta}{1+\beta} \Rightarrow \frac{1}{\alpha} = \frac{1+\beta}{\beta} = \frac{1}{\beta} + 1$
$\frac{1}{\alpha} - \frac{1}{\beta} = 1$
So the answer is (d) one.

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Junction Transistor question from MHT CET 2021 (21 Sep, Shift 1)

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