Question

In the circuit shown in the figure, the input voltage $V_1$ is $20\text{V},V_{BE}=0\text{and}{V}_{CE}=0$. The values of $I_B,I_C$ and $\beta$ are given by

• A. $I_B=401\mu \text{A},I_C=10\text{mA},\beta =250$
• B. $I_B=20\mu \text{A},I_C=5\text{mA},\beta =250$
• C. $I_B=40\mu \text{A},I_C=5\text{mA},\beta =125$
• D. $I_B=25\mu \text{A},I_C=5\text{mA},\beta =200$

Answer 1

The correct option is C $I_B=40\mu \text{A},I_C=5\text{mA},\beta =125$

For Emitter-Base circuit,
$V_i=I_B{R}_B+V_{BE}$

$20=I_B\times (500\times {10}^3)+0$

$\therefore I_B=\frac{20}{500\times {10}^3}=40\mu A$

For CE circuit
$V_{CC}=I_C{R}_C+V_{CE}$

$I_C=5\times {10}^{-3}=5\text{mA}$

$\beta =\frac{I_C}{I_B}=\frac{5\times {10}^{-3}}{40\times {10}^{-6}}=125$.

Hence $\left(D\right)$ is the correct answer.