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In the circuit shown in fig. (a), final the value of R_(C). |
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Answer» Solution :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` |
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