An n-p-n transistor is connected in common-emitter configuration as shown in the figure. If the collector current is 5 ~mA , V_ B E =0.6 ~V , V_ C E =3 ~V and common-emitter current amplification factor is 50 , then the values of R₁ and R₂ are respectively.

Step-by-step solution

In given circuit, V_ C C =i_B R_B+V_ B E R_B=R₁= V_ C C -V_ B E i_B As, i_B= i_C = 5 10⁻³ 50 =1 10⁻⁴ ~A R₁= 8-0.6 1 10-4 =7.4 10^4=74 10^3 =74 k and by KVL in closed collector loop, we get aligned V_ C C & =i_C R_L+V_ C E R_L & = V_ C C -V_ C E i_C & = 8-3 5 10⁻³ aligned So, R₂=R_L=1.0 k