Question

solve this question ...​

Answer 1

Hello Mate,

Here is your answer,

We Know,

$i_{e} = i_{c} + i_{b} \\ \\ i_{c} = \beta i_{b} \\ \\ v_{cc} = i_{c} r_{c} + i_{e} r_{e} + v_{ce} \\ \\ v_{cc} = ri_{b} + i_{e} r_{e} + v_{be} \\ \\ r + \beta r_{e} = v_{cc} - v_{be} \\ \\ i_{b} = \frac{ v_{cc} - v_{be} }{r + \beta r_{e}} \\ \\ substituting \: values \: we \: get \\ \\ \frac{11.5}{200} \: ma \\ \\ r_{c} + r_{e} = \frac{ v_{cc} - v_{ce} }{ \beta i_{b} } \\ \\ substituting \: values \: we \: get \\ \\ \frac{2}{11.5} (12 - 3)k\: \: ohms \\ \\ = 1.56k \: \: ohms \\ \\ r_{c} + r_{e} = 1.56 \\ \\ r_{c} = 1.56 - 1 \\ \\ = 0.56k \: \: ohms$

Hope it helps:)

Answer 2

Answer:

We Know,

$\begin{lgathered}i_{e} = i_{c} + i_{b} \\ \\ i_{c} = \beta i_{b} \\ \\ v_{cc} = i_{c} r_{c} + i_{e} r_{e} + v_{ce} \\ \\ v_{cc} = ri_{b} + i_{e} r_{e} + v_{be} \\ \\ r + \beta r_{e} = v_{cc} - v_{be} \\ \\ i_{b} = \frac{ v_{cc} - v_{be} }{r + \beta r_{e}} \\ \\ substituting \: values \: we \: get \\ \\ \frac{11.5}{200} \: ma \\ \\ r_{c} + r_{e} = \frac{ v_{cc} - v_{ce} }{ \beta i_{b} } \\ \\ substituting \: values \: we \: get \\ \\ \frac{2}{11.5} (12 - 3)k\: \: ohms \\ \\ = 1.56k \: \: ohms \\ \\ r_{c} + r_{e} = 1.56 \\ \\ r_{c} = 1.56 - 1 \\ \\ = 0.56k \: \: ohms\end{lgathered} \:$