Question

For a transistor \( \dfrac{1}{\alpha_{dc}} - \dfrac{1}{\beta_{dc}} \) is (\(\alpha_{dc}\) and \(\beta_{dc}\) are current gains)

• A. Zero
• B. \(-1\)
• C. \(1\)
• D. \(2\)

Hint:

Always remember the basic transistor relation: \[ \alpha = \frac{\beta}{1 + \beta} \] This helps in quickly solving numerical and conceptual questions involving current gains.

Answer 1

The Correct Option is C

Step 1: Recall the relation between current gains.
For a transistor, the relation between the common-base current gain \(\alpha_{dc}\) and the common-emitter current gain \(\beta_{dc}\) is:
\[ \alpha_{dc} = \frac{\beta_{dc}}{1 + \beta_{dc}} \]
Step 2: Find the reciprocal of \(\alpha_{dc}\).
\[ \frac{1}{\alpha_{dc}} = \frac{1 + \beta_{dc}}{\beta_{dc}} = \frac{1}{\beta_{dc}} + 1 \]
Step 3: Substitute in the given expression.
\[ \frac{1}{\alpha_{dc}} - \frac{1}{\beta_{dc}} = \left(\frac{1}{\beta_{dc}} + 1\right) - \frac{1}{\beta_{dc}} \]
Step 4: Simplify the expression.
\[ \frac{1}{\alpha_{dc}} - \frac{1}{\beta_{dc}} = 1 \]
Step 5: Conclusion.
Hence, the correct value of the given expression is \(1\).