In a common -emitter transistor amplifier, an increase of `50muA` in the base current causes an increase of `1.0mA` in the collector current . Calculat gain `beta`. What will be the change in emitter current? Also calculate current gain `alpha`

Options

A. 1050 mA
B. 1050 `mu`A
C. 5025 mA
D. 5025 `mu`A

Explanation

To solve the problem step by step, we will calculate the gain \( \beta \), the change in emitter current \( \Delta I_E \), and the current gain \( \alpha \).

### Step 1: Calculate \( \beta \)

The gain \( \beta \) (also known as the current gain) of a transistor is defined as the ratio of the change in collector current \( \Delta I_C \) to the change in base current \( \Delta I_B \).

\[ \beta = \frac{\Delta I_C}{\Delta I_B} \]

Given:
- \( \Delta I_B = 50 \, \mu A = 0.050 \, mA \)
- \( \Delta I_C = 1.0 \, mA \)

Substituting the values into the formula:

\[ \beta = \frac{1.0 \, mA}{0.050 \, mA} = \frac{1.0}{0.050} = 20 \]

### Step 2: Calculate the change in emitter current \( \Delta I_E \)

The change in emitter current \( \Delta I_E \) can be calculated using the relationship:

\[ \Delta I_E = \Delta I_B + \Delta I_C \]

Substituting the known values:

\[ \Delta I_E = 0.050 \, mA + 1.0 \, mA = 1.050 \, mA \]

### Step 3: Calculate \( \alpha \)

The current gain \( \alpha \) is defined as the ratio of the change in collector current \( \Delta I_C \) to the change in emitter current \( \Delta I_E \):

\[ \alpha = \frac{\Delta I_C}{\Delta I_E} \]

Substituting the values we have:

\[ \alpha = \frac{1.0 \, mA}{1.050 \, mA} = \frac{1.0}{1.050} \approx 0.952 \]

### Summary of Results

- \( \beta = 20 \)
- \( \Delta I_E = 1.050 \, mA \) or \( 1050 \, \mu A \)
- \( \alpha \approx 0.952 \)