For the transistor circuit shown in figure , evaluate `V_E , R_B and R_E`. Given `I_C = 1 mA`,
` V_(CE) = 3V , V_(BE) = 0.5 V , V_(CC )= 12 V and beta = 1000`

Explanation

Consider the fig. (b) given here to solve this problem
`I_(C) = I_(E)` [As base current is very small]
From the figure, `I_(C) (R_(C) + R_(E)) + V_(CE) = 12`
`(R_(E) + R_(C)) xx 1 xx 10^(-3) + 3 = 12`
`R_(E) + R_(C) = 9 xx 10^(3) = 9 Omega`
`R_(E) = 9 - 7.8 = 1.2 k Omega`
`V_(E) = I_(E) xx R_(E)`
`= 1 xx 10^(-3) xx 1.2 xx 10^(3) = 1.2 V`
Voltage `V_(B) = V_(E) + V_(BE) = 1.2 + 0.5 = 1.7 V`
Current `I = (V_(B))/(20 xx 10^(3)) = (1.7)/(20 xx 10^(3)) = 0.085 mA`
Resistance `R_(B) = (12 - 1.7)/(I_(C) /(beta) + 0.085) = (10.3)/(0.01 + 0.085) ["Given", Beta = 100]`
`= 108 k Omega`