b
(b)Current gain \(\beta = \frac{{\Delta {i_c}}}{{\Delta {i_b}}}\)\( \Rightarrow \Delta {i_b} = \frac{{1 \times {{10}^{ - 3}}}}{{100}} = {10^{ - 5}}A\) \(=0.01\,mA.\)
By using \(\Delta {i_e} = \Delta {i_b} + \Delta {i_c}\)\( \Rightarrow \Delta {i_e}\)
\(= 1.01 + 1 = 1.01\,mA.\)