The current amplification factor of a transistor is 50. The input resistance when used in common emitter mode is 1 k . The peak value for an a.c. input voltage of 0.01 V peak is

Step-by-step solution

aligned & Given: =50, R_i=1 k =10^3 , & V_i=0.01 ~V & = I_C I_B =50 I_C=50 I_B & V_i=I_B R_i & I_B= V_i R_i = 0.01 10^3 =10⁻⁵ & I_C=50 10⁻⁵=500 10⁻⁶ ~A =500 ~A aligned