In CE NPN transistor `10^(10` electrons enter the emitter in `10^(-6)` s when it is connected to battery. About `5%` electrons recombine with holes in the base. The current gain of the transistor is ……. . `(e=1.6xx10^(-19)C)`

Options

A. 0.98
B. 19
C. 49
D. 0.95

Answer

To solve the problem, we will follow these steps:

### Step 1: Calculate the Emitter Current (I_E)

The emitter current (I_E) can be calculated using the formula:
\[ I_E = \frac{Q}{T} \]
where \( Q \) is the total charge and \( T \) is the time.

The total charge \( Q \) can be calculated as:
\[ Q = n \cdot e \]
where \( n \) is the number of electrons and \( e \) is the charge of an electron.

Given:
- \( n = 10^{10} \) electrons
- \( e = 1.6 \times 10^{-19} \, C \)
- \( T = 10^{-6} \, s \)

Calculating \( Q \):
\[ Q = 10^{10} \cdot 1.6 \times 10^{-19} = 1.6 \times 10^{-9} \, C \]

Now substituting \( Q \) into the formula for \( I_E \):
\[ I_E = \frac{1.6 \times 10^{-9}}{10^{-6}} = 1.6 \, mA \]

### Step 2: Calculate the Base Current (I_B)

Given that 5% of the electrons recombine in the base, we can calculate the base current \( I_B \):
\[ I_B = 0.05 \cdot I_E \]

Substituting the value of \( I_E \):
\[ I_B = 0.05 \cdot 1.6 \, mA = 0.08 \, mA \]

### Step 3: Calculate the Collector Current (I_C)

The collector current \( I_C \) can be calculated using the relationship:
\[ I_E = I_B + I_C \]
Rearranging gives:
\[ I_C = I_E - I_B \]

Substituting the values:
\[ I_C = 1.6 \, mA - 0.08 \, mA = 1.52 \, mA \]

### Step 4: Calculate the Current Gain (β)

The current gain \( \beta \) of the transistor is given by:
\[ \beta = \frac{I_C}{I_B} \]

Substituting the values of \( I_C \) and \( I_B \):
\[ \beta = \frac{1.52 \, mA}{0.08 \, mA} = 19 \]

### Final Answer
The current gain of the transistor is \( \beta = 19 \).

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