To solve the problem, we need to find the change in collector current when the base current of a transistor changes, given that the transistor is in a common-emitter configuration.
The relationship between the base current (\(I_B\)), emitter current (\(I_E\)), and collector current (\(I_C\)) in a transistor is defined using the parameters \(\alpha\) (common base current gain) and \(\beta\) (common-emitter current gain).
For a transistor, \(\alpha\) and \(\beta\) are related by the equation:
\(\alpha = \frac{\beta}{\beta + 1}\)
We know \(\alpha = 0.9\). Solve for \(\beta\):
\(0.9 = \frac{\beta}{\beta + 1}\)
\(0.9(\beta + 1) = \beta\)
\(0.9\beta + 0.9 = \beta\)
\(0.1\beta = 0.9\)
\(\beta = \frac{0.9}{0.1} = 9\)
Thus, the common-emitter current gain, \(\beta = 9\).
In a common-emitter configuration, the change in collector current \(\Delta I_C\) is given by:
\(\Delta I_C = \beta \times \Delta I_B\)
Given that the change in base current \(\Delta I_B = 2 \mu A\), substitute the values:
\(\Delta I_C = 9 \times 2 \mu A = 18 \mu A\)
Thus, the change in collector current when the base current changes by \(2 \mu A\) is \(18 \mu A\).
Therefore, the correct answer is:
$18\mu \mathrm{A}$