Consider an `n-p-n` transistor amplifer in common-emitter configuration. The current gain of the transistor is `100`. If the collector current changes by `1mA`, what will be the change in emitter current?

Options

A. 1.1 mA
B. 1.01 mA
C. 0.01 mA
D. 10 mA

Explanation

To solve the problem step by step, we will use the relationship between the collector current (Ic), base current (Ib), and emitter current (Ie) in a transistor, along with the current gain (β).

### Step 1: Understand the given data
- Current gain (β) of the transistor = 100
- Change in collector current (ΔIc) = 1 mA

### Step 2: Relate collector current change to base current change
The current gain (β) is defined as:
\[ \beta = \frac{\Delta I_C}{\Delta I_B} \]
From this, we can express the change in base current (ΔIb) as:
\[ \Delta I_B = \frac{\Delta I_C}{\beta} \]

### Step 3: Substitute the known values
Substituting the values we have:
\[ \Delta I_B = \frac{1 \text{ mA}}{100} = 0.01 \text{ mA} \]

### Step 4: Relate emitter current change to collector and base current changes
For an NPN transistor, the change in emitter current (ΔIe) is given by:
\[ \Delta I_E = \Delta I_C + \Delta I_B \]

### Step 5: Substitute the values of ΔIc and ΔIb
Now substituting the values we calculated:
\[ \Delta I_E = 1 \text{ mA} + 0.01 \text{ mA} = 1.01 \text{ mA} \]

### Final Answer
The change in emitter current (ΔIe) is:
\[ \Delta I_E = 1.01 \text{ mA} \]