(b) In an n-p-n transistor emitter current ($i_{e}$) is the sum of base current ($i_{b}$) and collector current ($i_{c}$).
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Given, $\frac{90}{100}i_{e} = i_{c}$ …(i)
Also, $i_{c} = 10$ mA …(ii)
From Eqs. (i) and (ii), we get
$i_{e} = \frac{10 \times 100}{90} = 11\ mA$