CURRENT equations are related as:IC = β × IBIC = α × IEGains are related as:\(\alpha = \frac{\beta }{{1 + \beta }}\)\(\beta = \frac{\alpha }{{1 - \alpha }}\)Voltage Divider Bias or Self BiasThevenin’s EQUIVALENT will be\({V_{th}} = \frac{{{V_{CC}}{R_2}}}{{{R_1} + {R_2}}}\)\({R_{th}} = \frac{{{R_1}{R_2}}}{{{R_1} + {R_2}}}\)Calculation:Given structure is Voltage divider bias circuitVCC = 9 V, R1 = 10 KΩ , R2 = 2 KΩ , RE = 0.2 KΩ , RC = 1 KΩThevenin’s equivalent will beGiven that the base current is negligible. So, β will be of high value.Generally, β varies from 50 to 150Here we assume β = 100Calculating Thevenin’s voltage and Resistance\({V_{th}} = \frac{{9 \times 2K}}{{12K}}\)Vth = 1.5 V\({R_{th}} = \frac{{10K \times 2K}}{{12K}}\)\({R_{th}} = \frac{5}{3}K{\rm{\Omega }}\)Applying the KVL from Base to the emitter- Vth + IB(RTH) + VBE + IE(RE) = 0\({I_B}\left( {\frac{5}{3}k + 0.2K \times 101} \right) = 1.5 - 0.7\)\({I_B} = \frac{{0.8}}{{1.67K + 20.2K}}\)IB = 0.036 mAIC = β × IBIC = 100 × 0.036 mAIC = 3.6 mABased on the options given it is nearer to 4 mA. But to get that β should be 50.If β = 50 then that will be very low. So, assume β = 100