An NPN transistor amplifier has current gain 100. If the collector current changes by 1 mA, The emitter current will change by

Options

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

Answer

To solve the problem, we need to understand the relationship between the collector current (IC), base current (IB), and emitter current (IE) in an NPN transistor. The current gain (β) of the transistor is given, which is defined as the ratio of the change in collector current (ΔIC) to the change in base current (ΔIB).

### Step-by-Step Solution:

1. **Identify Given Values:**
- Current gain (β) = 100
- Change in collector current (ΔIC) = 1 mA = 0.001 A

2. **Calculate Change in Base Current (ΔIB):**
Using the formula for current gain:
\[ β = \frac{ΔIC}{ΔIB} \]
Rearranging gives:
\[ ΔIB = \frac{ΔIC}{β} \]
Substituting the known values:
\[ ΔIB = \frac{1 \text{ mA}}{100} = 0.01 \text{ mA} \]

3. **Relate Emitter Current Change (ΔIE) to Collector and Base Current Changes:**
The emitter current (IE) is related to the collector current and base current by the equation:
\[ IE = IC + IB \]
Therefore, the change in emitter current (ΔIE) can be expressed as:
\[ ΔIE = ΔIC + ΔIB \]

4. **Substitute the Values:**
Now substituting the values we have:
\[ ΔIE = ΔIC + ΔIB = 1 \text{ mA} + 0.01 \text{ mA} = 1.01 \text{ mA} \]

5. **Conclusion:**
The change in emitter current (ΔIE) is 1.01 mA.

### Final Answer:
The emitter current will change by **1.01 mA**.

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