In an NPN transistor, the collector current is 24 mA. If 80% of electrons reach the collector, what is its base current in mA?

A) 36
B) 26
C) 16
D) 6

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The operation of an NPN transistor involves the flow of electrons from the emitter to the collector, with a certain portion passing through the base. Given that the collector current ($I_C$) is 24 mA and that 80% of the electrons reach the collector, we can determine the base current ($I_B$).

Firstly, let's understand that the emitter current ($I_E$) is the sum of the collector current and the base current:

$$ I_E = I_C + I_B $$

Since 80% of the emitter current reaches the collector, the missing 20% is the base current. Therefore, we can calculate the emitter current as follows, knowing that $I_C$ is 80% of $I_E$:

$$ I_C = 0.8 I_E \ 24 \text{ mA} = 0.8 I_E \ I_E = \frac{24 \text{ mA}}{0.8} $$

Calculating $I_E$:

$$ I_E = 30 \text{ mA} $$

Now, we find $I_B$ by subtracting $I_C$ from $I_E$:

$$ I_B = I_E - I_C \ I_B = 30 \text{ mA} - 24 \text{ mA} $$

Calculating $I_B$:

$$ I_B = 6 \text{ mA} $$

Thus, the base current $I_B$ is 6 mA which corresponds to option D.

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