1.

An n-p-n transistor is connected in common - emitter configuration as shown in V_(BE)=0.6V,V_(CE)=3 and common - emitter current amplification factor is 50, then the values of R_(1) and R_(2) are respectively.

Answer»

`1 K OMEGA, 74 kg Omega`
`74 k Omega, 1 k Omega`
`37 K Omega , 2 k Omega`
`2 k Omega, 37 k Omega`

Solution :In given circuit `V_("CC")=i_(B)R_(B)+V_("BE")`
`rArr"" R_(B)=R_(1)=(V_("CC")-V_("BE"))/(i_B)`
As,`i_(B)= (i_c)/(Beta)= (5 XX 10^(-3))/(50)=1 xx 10^(-4)A`
`rArrR_(1)= (8-0.6)/(1 xx 10-4) = 7.4 xx 10^4 = 74 xx 10^3 Omega=74 k Omega`
and by KVL in closed collector loop, we get
`V_("CC")=i_(C)R_(L)+V_(CE)`
`rArr "" R_(L)= V_("CC"-V_("CE"))/(i_C)`
`= (8-3)/(5 xx 10^(-3))`
So,`R_(2)= R_(L)=1.0 k Omega`


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