If `alpha and beta` are the current gain in the CB and CE configurations respectively of the transistor circuit, then `(beta-alpha)/(alphabeta)=`

Options

A. zero
B. 1
C. 2
D. 5

Explanation

To solve the problem, we need to derive the expression \((\beta - \alpha) / (\alpha \beta)\) using the definitions of current gains \(\alpha\) and \(\beta\) in a transistor circuit.

### Step-by-Step Solution:

1. **Define Current Gains**:
- The current gain in the Common Base (CB) configuration is given by:
\[ \alpha = \frac{I_C}{I_E} \]
- The current gain in the Common Emitter (CE) configuration is given by:
\[ \beta = \frac{I_C}{I_B} \]

2. **Use the Relationship Between Currents**:
- For a transistor, the relationship between the emitter current \(I_E\), base current \(I_B\), and collector current \(I_C\) is:
\[ I_E = I_B + I_C \]

3. **Express \(I_E\) in Terms of \(I_C\)**:
- Dividing the entire equation by \(I_C\):
\[ \frac{I_E}{I_C} = \frac{I_B}{I_C} + 1 \]

4. **Substitute Current Gains**:
- We know that:
\[ \frac{I_E}{I_C} = \frac{1}{\alpha} \quad \text{and} \quad \frac{I_B}{I_C} = \frac{1}{\beta} \]
- Substituting these into the equation gives:
\[ \frac{1}{\alpha} = \frac{1}{\beta} + 1 \]

5. **Rearranging the Equation**:
- Rearranging gives:
\[ \frac{1}{\alpha} - \frac{1}{\beta} = 1 \]

6. **Finding a Common Denominator**:
- The left-hand side can be expressed with a common denominator:
\[ \frac{\beta - \alpha}{\alpha \beta} = 1 \]

7. **Final Result**:
- Therefore, we can conclude:
\[ \frac{\beta - \alpha}{\alpha \beta} = 1 \]

### Conclusion:
The final answer is:
\[ \frac{\beta - \alpha}{\alpha \beta} = 1 \]