In the circuit shown in fig. (a), final the value of `R_(C)`.

Explanation

Consider the fig. (b) to solbe this question.
`I_(E) = I_(C) + I_(B) " and " I_(C) = beta I_(B)`.....(i)
`I_(C) R_(C) + V_(CE) + I_(E) R_(E) = V_(C C)` .....(ii)
`RI_(B) + V_(BE) + I_(E) R_(E) = V_(C C)`....(iii)
`I_(E) ~~ I_(C) = beta I_(B)`

From Eq. (iii),
`(R + beta R_(E)) I_(B) = V_(C C) - V_(BE)`
`rArr I_(B) = (V_(C C) - V_(BE))/(R + beta. R_(E))`
`= (12 - 0.5)/(80 + 1.2 xx 100) = (11.5)/(200) mA`
From Eq. (ii), `(R_(C) + R_(E)) = (V_(CE) - V_(BE))/(I_(C)) = (V_(C C) - V_(CE))/(beta I_(B)) " " ( :. I_(C) = beta I_(B))`
`(R_(C) + R_(E)) = (2)/(11.5) (12 - 3) k Omega = 1.56 k Omega`
`R_(C) + R_(E) = 1.56`
`R_(C) = 1.56 - 1 = 0.56 k Omega`