Question:
For a common emitter transistor, if \(\frac{I_C}{I_E} = 0.95\), then the current gain is
For a common emitter transistor, if \(\frac{I_C}{I_E} = 0.95\), then the current gain is
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Relation: \(\beta = \frac{\alpha}{1-\alpha}\)
Updated On: Apr 26, 2026
- \(47.5\)
- \(44\)
- \(19\)
- \(15\)
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The Correct Option is C
Solution and Explanation
Concept:
For transistor: \[ \alpha = \frac{I_C}{I_E}, \quad \beta = \frac{\alpha}{1-\alpha} \] Step 1: Given value. \[ \alpha = 0.95 \]
Step 2: Calculate \(\beta\). \[ \beta = \frac{0.95}{1 - 0.95} = \frac{0.95}{0.05} \] \[ \beta = 19 \]
Step 3: Conclusion. Current gain = \(19\)
For transistor: \[ \alpha = \frac{I_C}{I_E}, \quad \beta = \frac{\alpha}{1-\alpha} \] Step 1: Given value. \[ \alpha = 0.95 \]
Step 2: Calculate \(\beta\). \[ \beta = \frac{0.95}{1 - 0.95} = \frac{0.95}{0.05} \] \[ \beta = 19 \]
Step 3: Conclusion. Current gain = \(19\)
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