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An $n-p-n$ transistor is connected in common-emitter configuration as shown in the figure. If the collector current is $5 \mathrm{~mA}$, $V_{B E}=0.6 \mathrm{~V}, V_{C E}=3 \mathrm{~V}$ and common-emitter current amplification factor is 50 , then the values of $R_1$ and $R_2$ are respectively.
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Verified Answer
The correct answer is:
$74 \mathrm{k} \Omega, 1 \mathrm{k} \Omega$
In given circuit, $V_{C C}=i_B R_B+V_{B E}$
$$
\Rightarrow \quad R_B=R_1=\frac{V_{C C}-V_{B E}}{i_B}
$$
As, $i_B=\frac{i_C}{\beta}=\frac{5 \times 10^{-3}}{50}=1 \times 10^{-4} \mathrm{~A}$
$$
\Rightarrow \quad R_1=\frac{8-0.6}{1 \times 10-4}=7.4 \times 10^4=74 \times 10^3 \Omega=74 \mathrm{k} \Omega
$$
and by KVL in closed collector loop, we get
$$
\Rightarrow \quad \begin{aligned}
V_{C C} & =i_C R_L+V_{C E} \\
R_L & =\frac{V_{C C}-V_{C E}}{i_C} \\
& =\frac{8-3}{5 \times 10^{-3}}
\end{aligned}
$$
So,
$$
R_2=R_L=1.0 \mathrm{k} \Omega
$$
$$
\Rightarrow \quad R_B=R_1=\frac{V_{C C}-V_{B E}}{i_B}
$$
As, $i_B=\frac{i_C}{\beta}=\frac{5 \times 10^{-3}}{50}=1 \times 10^{-4} \mathrm{~A}$
$$
\Rightarrow \quad R_1=\frac{8-0.6}{1 \times 10-4}=7.4 \times 10^4=74 \times 10^3 \Omega=74 \mathrm{k} \Omega
$$
and by KVL in closed collector loop, we get
$$
\Rightarrow \quad \begin{aligned}
V_{C C} & =i_C R_L+V_{C E} \\
R_L & =\frac{V_{C C}-V_{C E}}{i_C} \\
& =\frac{8-3}{5 \times 10^{-3}}
\end{aligned}
$$
So,
$$
R_2=R_L=1.0 \mathrm{k} \Omega
$$
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