If `alpha=0.98` and current through emitter `i_(e)=20 mA`, the value of `beta` is

Options

A. 4.9
B. 49
C. 96
D. 9.6

Explanation

To solve the problem, we need to find the value of beta (β) given that alpha (α) is 0.98 and the emitter current (i_e) is 20 mA.

### Step-by-Step Solution:

1. **Understand the relationship between alpha and beta:**
\[ \alpha = \frac{I_C}{I_E} \]
where \(I_C\) is the collector current and \(I_E\) is the emitter current. We also know that:
\[ \beta = \frac{I_C}{I_B} \]
where \(I_B\) is the base current.

2. **Calculate the collector current (I_C):**
We can rearrange the formula for alpha to find \(I_C\):
\[ I_C = \alpha \times I_E \]
Substituting the values:
\[ I_C = 0.98 \times 20 \, \text{mA} = 19.6 \, \text{mA} \]

3. **Use the relationship between emitter current, collector current, and base current:**
The emitter current is the sum of the collector current and the base current:
\[ I_E = I_C + I_B \]
Rearranging gives:
\[ I_B = I_E - I_C \]
Substituting the known values:
\[ I_B = 20 \, \text{mA} - 19.6 \, \text{mA} = 0.4 \, \text{mA} \]

4. **Calculate beta (β):**
Now we can calculate beta using the collector current and base current:
\[ \beta = \frac{I_C}{I_B} \]
Substituting the values:
\[ \beta = \frac{19.6 \, \text{mA}}{0.4 \, \text{mA}} = 49 \]

### Final Answer:
The value of beta (β) is 49.