Since the circuit voltage is equal to the sum of the drop voltage and the collector voltage, "V_c=V_{cc}-R_c" , where "V_c" is the collector voltage, therefore
"V_c=12-1,2=10.8" V, since the current voltage falls to the base, therefore, according to the law of Oma "I_b=\\frac {R_c}{R_b}" , and since "I_c=I_b\\times \u03b2" ; "I_c=\\frac {R_c}{R_b} \\times \u03b2 =\\frac {1,2}{47\\times10^3} \\times 200 \\approx 0.051" A.
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