Question

In the circuit shown in the figure, the value of $R_c$ is

• A. $2\text{k}\Omega$
• B. $4\text{k}\Omega$
• C. $6\text{k}\Omega$
• D. $8\text{k}\Omega$

Answer 1

The correct option is C $6\text{k}\Omega$


We know, $I_E=I_B+I_c$ and $I_c=\beta I_B$,

As $\beta >>1$

so, $I_E=(1+\beta )I_B\approx \beta I_B$

From figure, using Kirchhoff's law for collector-emitter loop

$I_c{R}_C+V_{CE}+I_E{R}_E=V_{CC}$

$\Rightarrow \beta I_B{R}_C+V_{CE}+\beta I_B{R}_E=V_{CC}$

$\Rightarrow \beta I_B(R_C+R_E)=V_{CC}-V_{CE}$

$\Rightarrow R_C+R_E=\frac{V_{CC}-V_{CE}}{\beta I_B}---\left(1\right)$

Also, $V_{CC}-R_B(I_B+I)-IR=0$

$\Rightarrow 12-{10}^5(I_B+I)-2\times {10}^{4}I=0$

$\Rightarrow 12-{10}^5{I}_B-12\times {10}^{4}I=0---\left(2\right)$

And applying Kirchhoff's law for base emitter loop,
$-V_{BE}-I_E{R}_E+IR=0$

$\Rightarrow -0.5-\beta I_B\times 1000+I\times 2\times {10}^4=0$

$\Rightarrow -0.5-{10}^5{I}_B+2\times {10}^{4}I=0$

$\Rightarrow {10}^5{I}_B=-0.5+2\times {10}^{4}I---\left(3\right)$

From $\left(2\right)\text{and}\left(3\right)$

$12+0.5-2\times {10}^{4}I-12\times {10}^{4}I=0$

$\Rightarrow I=8.92\times {10}^{-5}---4$

From $\left(3\right)\text{and}\left(4\right)$

$I_B=1.28\times {10}^{-5}\to 5$

From $\left(1\right)\text{and}\left(5\right)$

$R_C+R_E=\frac{12-3}{100\times 1.28\times {10}^{-5}}$

$\Rightarrow R_C+R_E=7031.25\approx 7\text{k}\Omega$

$\Rightarrow R_C=7-1=6\text{k}\Omega$

Hence, option $\left(C\right)$ is correct.