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The transistor's α & β are related by
1. β = α/(1 - α)
2. α = β/(1 - β)
3. β = α/(1 + α)
4. α = (1 + β)/β 

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Correct Answer - Option 1 : β = α/(1 - α)

The transistor α and β are related as:

\(α = \frac{β }{{β + 1}}\)

This can also be written as:

\(β = \frac{α }{{1-α}}\)

Derivation:

The common-emitter current gain (β) is the ratio of the transistor's collector current to the transistor's base current, i.e.

\(β = \frac{{{I_C}}}{{{I_B}}}\)

And the common base DC current gain (α) is a ratio of the transistor's collector current to the transistor's emitter current, i.e.

\(α = \frac{{{I_C}}}{{{I_E}}}\)

The transistor currents are related by the relation:

IE = IB + IC

α can now be written as:

\(α = \frac{{{I_C}}}{{{I_B+I_C}}}\)

Dividing both the numerator and denominator by IB, we get:

\(α = \frac{{{I_C/I_B}}}{{{1+I_C/I_B}}}\)

Since \(β = \frac{{{I_C}}}{{{I_B}}}\)

\(α = \frac{β }{{β + 1}}\) 'or'

\(β = \frac{α }{{1-α}}\)

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