In the shown common emitter amplifier circuit,  $\beta =80,V_{BE}=0.7\text{\hspace{0.17em}volt}$. The value of $R_C$  is _____ $\text{k\hspace{0.17em}}\Omega$ [Given $V_{CE}=3\text{\hspace{0.17em}V}$ ]

 

Given  $\beta =\frac{I_c}{I_B}=80$
$V_{BE}=0.7\text{\hspace{0.17em}V}$ 
Applying KVL in loop (1)
$V_{BE}=0.7\text{\hspace{0.17em}V}=8.7-I_B\left(200\times {10}^3\right)$

$\Rightarrow I_B=\frac{8}{200\times {10}^3}=4\times {10}^{-5}\text{\hspace{0.17em}A}$

$\Rightarrow \beta =80=\frac{I_C}{I_B}\Rightarrow I_C=32\times {10}^{-4}\text{\hspace{0.17em}A}$

Applying KVL in loop (2)

$\begin{array}{l}{V}_{CE}=3\mathrm{V}=11-I_C{R}_C\\ 3=11-32\times {10}^{-4}{R}_C\\ \Rightarrow R_C=\frac{8}{32\times {10}^{-4}}\\ \therefore R_C=2.5\mathrm{k\Omega }\end{array}$

Therefore, the correct answer is 2.50