For a CE-transistor amplifier,

Solution:

Collector resistance, RC = 2 kΩ = 2000 Ω

Audio signal voltage across the collector resistance, V = 2 V

Current amplification factor of the transistor, β = 100

Base resistance, RB = 1 kΩ = 1000 Ω

Input signal voltage = Vi

Base current = IB

We have the amplification relation as:

Voltage amplification $=\frac{V}{V_{i}}=\beta \frac{R_{\mathrm{C}}}{R_{\mathrm{B}}}$

$V_{i}=\frac{V R_{\mathrm{B}}}{\beta R_{\mathrm{C}}}$

$=\frac{2 \times 1000}{100 \times 2000}=0.01 \mathrm{~V}$

Therefore, the input signal voltage of the amplifier is 0.01 V.

Base resistance is given by the relation:

$R_{\mathrm{B}}=\frac{V_{\mathrm{i}}}{I_{\mathrm{B}}}$

$=\frac{0.01}{1000}=10 \times 10 \not \rho \mathrm{A}$

Therefore, the base current of the amplifier is 10 μA.