If `alpha` and `beta` current gains in common-base and common-emitter configuration of a transistor, then `beta` is equal to

Options

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

Answer

To find the relationship between the current gains \( \alpha \) (common-base current gain) and \( \beta \) (common-emitter current gain) in a transistor, we can follow these steps:

### Step 1: Understand the definitions
- \( \alpha \) is defined as the ratio of the collector current (\( I_C \)) to the emitter current (\( I_E \)) in a common-base configuration:
\[ \alpha = \frac{I_C}{I_E} \]
- \( \beta \) is defined as the ratio of the collector current (\( I_C \)) to the base current (\( I_B \)) in a common-emitter configuration:
\[ \beta = \frac{I_C}{I_B} \]

### Step 2: Relate the currents
In a transistor, the emitter current (\( I_E \)) is the sum of the collector current (\( I_C \)) and the base current (\( I_B \)):
\[ I_E = I_C + I_B \]

### Step 3: Substitute for \( I_B \)
From the definition of \( \beta \), we can express \( I_B \) in terms of \( I_C \):
\[ I_B = \frac{I_C}{\beta} \]

### Step 4: Substitute \( I_B \) into the equation for \( I_E \)
Substituting \( I_B \) into the equation for \( I_E \):
\[ I_E = I_C + \frac{I_C}{\beta} \]
Factoring out \( I_C \):
\[ I_E = I_C \left(1 + \frac{1}{\beta}\right) \]

### Step 5: Express \( I_C \) in terms of \( I_E \)
Now, we can express \( I_C \) in terms of \( I_E \):
\[ I_C = \frac{I_E}{1 + \frac{1}{\beta}} = \frac{I_E \beta}{\beta + 1} \]

### Step 6: Substitute into the equation for \( \alpha \)
Now, substituting \( I_C \) into the equation for \( \alpha \):
\[ \alpha = \frac{I_C}{I_E} = \frac{\frac{I_E \beta}{\beta + 1}}{I_E} \]
This simplifies to:
\[ \alpha = \frac{\beta}{\beta + 1} \]

### Step 7: Solve for \( \beta \)
To find \( \beta \) in terms of \( \alpha \), we can rearrange the equation:
\[ \alpha (\beta + 1) = \beta \]
Expanding and rearranging gives:
\[ \alpha \beta + \alpha = \beta \]
\[ \beta - \alpha \beta = \alpha \]
Factoring out \( \beta \):
\[ \beta (1 - \alpha) = \alpha \]
Finally, solving for \( \beta \):
\[ \beta = \frac{\alpha}{1 - \alpha} \]

### Final Answer
Thus, the relationship between \( \beta \) and \( \alpha \) is:
\[ \beta = \frac{\alpha}{1 - \alpha} \]