Correct Answer - Option 3 : 1.98 mA and 2 mA

**Concept:**

Collector current can be calculated as:

Ic = βIb ----(1)

where β is common emitter DC gain and

\(β=\frac{\alpha}{1-\alpha}\) -----(2)

Emitter current can be calculated by:

Ie = Ib + Ic -----(3)

**Calculation**:

**Given**:

α = 0.99

I_{b} = 20 μA

From equation (1) and (2),

I_{c} can be calculated:

\(\beta=\frac{0.99}{1-0.99}=99\)

I_{c} = 99 × 20

**I**_{c} = 1.98 mA

From equation (3)

I_{e} can be calculated as:

**I**_{e} = 1.98 + 0.02 = 2 mA.