If the current gain in common-base configuration ( α ) of a transistor increases from 0.95 to 0.96, the percentage change in current gain in common-emitter configuration ( β ) is approximately

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The relation between common-emitter current gain $β$ and common-base current gain $α$ is given by $β = \frac{α}{1 - α}$ .

For the initial state, $α_{1} = 0.95$ .

$β_{1} = \frac{0.95}{1 - 0.95} = \frac{0.95}{0.05} = 19$

For the final state, $α_{2} = 0.96$ .

$β_{2} = \frac{0.96}{1 - 0.96} = \frac{0.96}{0.04} = 24$

The change in current gain is $Δ β = β_{2} - β_{1} = 24 - 19 = 5$ .

The percentage change in $β$ is $\frac{Δ β}{β _{1}} \times 100 = \frac{5}{19} \times 100 \approx 26.31%$ .

Thus, the percentage change is approximately $26%$ .

Answer: $26%$

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