For a transistor , `alpha_(dc)` and `beta_(dc)` are the current ratios, then the value of `(beta_(dc) - alpha_(dc))/(alpha_(dc) . beta_(dc))`

Options

A. (a)1
B. (b)1.5
C. (c)2
D. (d)2.5

Answer

To solve the problem, we need to find the value of the expression \((\beta_{dc} - \alpha_{dc}) / (\alpha_{dc} \cdot \beta_{dc})\) using the relationship between \(\alpha_{dc}\) and \(\beta_{dc}\).

### Step-by-Step Solution:

1. **Understand the Relationships**:
We know that:
\[ \beta_{dc} = \frac{\alpha_{dc}}{1 - \alpha_{dc}} \]
This relationship helps us connect \(\alpha_{dc}\) and \(\beta_{dc}\).

2. **Rearranging the Relationship**:
From the relationship above, we can express \(1 - \alpha_{dc}\) in terms of \(\beta_{dc}\):
\[ 1 - \alpha_{dc} = \frac{\alpha_{dc}}{\beta_{dc}} \]

3. **Substituting into the Expression**:
Now, we substitute \(1 - \alpha_{dc}\) into our original expression:
\[ \frac{\beta_{dc} - \alpha_{dc}}{\alpha_{dc} \cdot \beta_{dc}} = \frac{\frac{\alpha_{dc}}{1 - \alpha_{dc}} - \alpha_{dc}}{\alpha_{dc} \cdot \frac{\alpha_{dc}}{1 - \alpha_{dc}}} \]

4. **Simplifying the Numerator**:
The numerator becomes:
\[ \frac{\alpha_{dc}}{1 - \alpha_{dc}} - \alpha_{dc} = \frac{\alpha_{dc} - \alpha_{dc}(1 - \alpha_{dc})}{1 - \alpha_{dc}} = \frac{\alpha_{dc} - \alpha_{dc} + \alpha_{dc}^2}{1 - \alpha_{dc}} = \frac{\alpha_{dc}^2}{1 - \alpha_{dc}} \]

5. **Substituting Back**:
Now substituting this back into the expression:
\[ \frac{\frac{\alpha_{dc}^2}{1 - \alpha_{dc}}}{\alpha_{dc} \cdot \frac{\alpha_{dc}}{1 - \alpha_{dc}}} \]

6. **Canceling Terms**:
This simplifies to:
\[ \frac{\alpha_{dc}^2}{\alpha_{dc}^2} = 1 \]

7. **Final Result**:
Therefore, the value of the expression \((\beta_{dc} - \alpha_{dc}) / (\alpha_{dc} \cdot \beta_{dc})\) is:
\[ \boxed{1} \]