Question

In an n-p-n transistor 200 electrons enter the emitter in $10^{-8}\ \text{second}$. If $1\%$ electrons are lost in the base, then the current that enters the emitter and the current amplification factor are respectively [$e = 1.6 \times 10^{-19}\ \text{C}$]

• A. $2 \times 10^{-10}\ \text{A}$ and $49$
• B. $3.2 \times 10^{-9}\ \text{A}$ and $99$
• C. $1.6 \times 10^{-19}\ \text{A}$ and $90$
• D. $1.7 \times 10^{-11}\ \text{A}$ and $70$

Hint:

You can find the amplification factor $\beta$ using a quick percentage trick without calculating the currents! If base loss is $1\%$, then the alpha current gain is $\alpha = 99\% = 0.99$. Using the conversion formula: $$\beta = \frac{\alpha}{1 - \alpha} = \frac{0.99}{1 - 0.99} = \frac{0.99}{0.01} = 99$$ This allows you to identify the correct option immediately by matching $\beta = 99$.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
Calculate the current amplification factor β of a transistor given the base current loss percentage.

Step 2: Key Formula or Approach:
If 1% of emitter current is lost to the base, then α = I_c/I_e = 0.99. The conversion to β uses β = α/(1 - α).

Step 3: Detailed Explanation:
Substituting α = 0.99 into the conversion formula: β = 0.99/(1 - 0.99) = 0.99/0.01 = 99. This percentage shortcut directly yields β without computing individual currents. The relation β = α/(1-α) bridges the common-base and common-emitter gains in a single step, making it a powerful rapid-calculation tool.

Step 4: Final Answer:
The amplification factor is β = 99.