Explanation

(d) The ratio of collector current $(I_{c}$) to emitter current $(I_{e}$) is known as current gain ($\alpha$) of a transistor. Therefore,
$\alpha = \frac{\mathrm{\Delta}I_{c}}{\mathrm{\Delta}I_{e}}$ …(i)
Also, emitter current is equal to sum of change of base current and collector current. Therefore,
$\mathrm{\Delta}I_{e} = \mathrm{\Delta}I_{b} + \mathrm{\Delta}I_{c}\ \ \ \ \ \ \ \ \ \ \ $…(ii)
From Eqs. (i) and (ii), we get
$\alpha = \frac{I_{c}}{\mathrm{\Delta}I_{b} + \mathrm{\Delta}I_{c}}$
Given, $\alpha =$ 0.9, $\mathrm{\Delta}I_{b} = 4\ mA$
0.9 = $\frac{I_{c}}{4 + \mathrm{\Delta}I_{c}}$
$\Rightarrow \ \ \ \ $0.9 (4 + $I_{c}$) = $I_{c}$
$\Rightarrow$ 3.6 + 0.9 $I_{c} = I_{c}$
$\Rightarrow$ 3.6 = 0.1$\ I_{c}$
$\Rightarrow \ \ \ \ I_{c} = 36$ mA