A transistor is connected in common emitter circuit configuration, the collector supply voltage is 10 V and the voltage drop across a resistor of 1000 `Omega` in the collector circuit is 0.6 V. If the current gain factor `(beta)` is 24, then the base current is _______ `muA`. (Round off to the Nearest Integer)

Answer

To find the base current (Ib) in a common emitter transistor configuration, we can follow these steps:

### Step 1: Determine the Collector Current (Ic)

The voltage drop across the collector resistor (Rc) is given as 0.6 V, and the resistance is 1000 Ω. We can calculate the collector current (Ic) using Ohm's law:

\[ Ic = \frac{V_{drop}}{R} = \frac{0.6 \, V}{1000 \, \Omega} = 0.0006 \, A = 6 \, mA \]

### Step 2: Use the Current Gain Factor (β)

The current gain factor (β) is given as 24. The relationship between the collector current (Ic) and the base current (Ib) is given by:

\[ Ic = \beta \cdot Ib \]

We can rearrange this formula to find the base current (Ib):

\[ Ib = \frac{Ic}{\beta} \]

### Step 3: Substitute the Values

Now we can substitute the values we have:

\[ Ib = \frac{6 \, mA}{24} = \frac{6 \times 10^{-3} \, A}{24} \]

Calculating this gives:

\[ Ib = 0.00025 \, A = 250 \, \mu A \]

### Step 4: Round Off to the Nearest Integer

Since the question asks for the base current rounded off to the nearest integer, we have:

\[ Ib \approx 250 \, \mu A \]

### Final Answer

The base current is approximately **250 µA**.

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