The correct relationship between the two current gains `alpha` and `beta` in a transistor is

Options

A. `beta=(alpha)/(1+alpha)`
B. `aplha =(beta)/(1-beta)`
C. `alpha =(beta)/(1+beta)`
D. `alpha=(1+beta)/(beta)`

Answer

To find the correct relationship between the two current gains, alpha (α) and beta (β), in a transistor, we can follow these steps:

### Step-by-Step Solution:

1. **Define Alpha (α)**:
- Alpha (α) is defined as the ratio of the collector current (Ic) to the emitter current (Ie).
- Mathematically, this is expressed as:
\[ \alpha = \frac{I_c}{I_e} \]

2. **Define Beta (β)**:
- Beta (β) is defined as the ratio of the collector current (Ic) to the base current (Ib).
- This can be expressed mathematically as:
\[ \beta = \frac{I_c}{I_b} \]

3. **Relate Emitter Current (Ie)**:
- The emitter current (Ie) can be expressed in terms of the base current (Ib) and the collector current (Ic) using the equation:
\[ I_e = I_b + I_c \]

4. **Express Ic in terms of Ib**:
- From the definition of beta, we can rearrange it to express Ic:
\[ I_c = \beta \cdot I_b \]

5. **Substitute Ic in the equation for Ie**:
- Substitute the expression for Ic into the equation for Ie:
\[ I_e = I_b + \beta \cdot I_b = I_b (1 + \beta) \]

6. **Substitute Ie in the equation for Alpha (α)**:
- Now, substitute the expression for Ie back into the equation for α:
\[ \alpha = \frac{I_c}{I_e} = \frac{I_c}{I_b (1 + \beta)} \]
- Substitute Ic from the beta definition:
\[ \alpha = \frac{\beta \cdot I_b}{I_b (1 + \beta)} = \frac{\beta}{1 + \beta} \]

7. **Final Relationship**:
- Thus, the relationship between α and β is:
\[ \alpha = \frac{\beta}{1 + \beta} \]

### Conclusion:
The correct relationship between the two current gains α and β in a transistor is:
\[ \alpha = \frac{\beta}{1 + \beta} \]