For the emitter-bias network shown below the collector current is:

Concept:

DC analysis of emitter-bias network;

Base current → \(I_{B} = \frac{V_{CC}-V_{BE}}{R_{B}+(1+β)R_{E}}\)

Collector Current → \(I_{C} = β I_{B}\)

Calculation:

Given;

V_CC = 20 V

R_B = 430 KΩ 

V_BE = 0.7 V

β = 50

R_E = 1 KΩ 

Then;

Base current → \(I_{B} = \frac{V_{CC}-V_{BE}}{R_{B}+(1+β)R_{E}}= \frac{20-0.7}{430+(1+50)1}=0.04\, mA\)

Collector Current → \(I_{C} = β I_{B} = 50 \times 0.04=2.01\, mA\)