Question:
If the current gain in common-base configuration (\(\alpha\)) of a transistor increases from 0.95 to 0.96, the percentage change in current gain in common-emitter configuration (\(\beta\)) is approximately
If the current gain in common-base configuration (\(\alpha\)) of a transistor increases from 0.95 to 0.96, the percentage change in current gain in common-emitter configuration (\(\beta\)) is approximately
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Use the relation between beta and alpha, calculate beta for both alpha values, and find the percentage change.
Updated On: Nov 12, 2026
- \(1\%\)
- \(10\%\)
- \(26\%\)
- \(50\%\)
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept
The two current gains are related by \(\beta = \dfrac{\alpha}{1-\alpha}\).
Step 2: Calculation of beta
For \(\alpha = 0.95\): \(\beta_{1} = \dfrac{0.95}{0.05} = 19\).
For \(\alpha = 0.96\): \(\beta_{2} = \dfrac{0.96}{0.04} = 24\).
Step 3: Percentage change
\(\dfrac{\beta_{2}-\beta_{1}}{\beta_{1}}\times 100 = \dfrac{5}{19}\times 100 \approx 26\%\).
Final Answer:
The common-emitter gain beta increases by about 26 percent. \[ \boxed{\text{(C) }26\%} \]
The two current gains are related by \(\beta = \dfrac{\alpha}{1-\alpha}\).
Step 2: Calculation of beta
For \(\alpha = 0.95\): \(\beta_{1} = \dfrac{0.95}{0.05} = 19\).
For \(\alpha = 0.96\): \(\beta_{2} = \dfrac{0.96}{0.04} = 24\).
Step 3: Percentage change
\(\dfrac{\beta_{2}-\beta_{1}}{\beta_{1}}\times 100 = \dfrac{5}{19}\times 100 \approx 26\%\).
Final Answer:
The common-emitter gain beta increases by about 26 percent. \[ \boxed{\text{(C) }26\%} \]
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