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For a common emitter transistor, input current is 5μA and β = 100. If circuit is operated at load resistance of 10 k Ω then voltage across load resistance will be :

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5 V

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7.5 V

answer is A.

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Detailed Solution

$\begin{array}{l} I_{B} = 5 μΛ \\ \frac{I_{C}}{I_{B}} = β \\ \frac{I_{C}}{5 μA} = 100 \\ I_{C} = 500 μA \\ V_{load} = I_{C} R_{L} \\ = 500 μA \times 10 kΩ \\ V_{load} = 5 volt \end{array}$

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