# 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

## Question

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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## Answer

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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