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

• A

  $I_B=20 \mathrm{\mu A}, I_C=5 \mathrm{mA}, \beta =250$
• B

  $I_B=25 \mathrm{\mu A}, I_C=5 \mathrm{mA}, \beta =200$
• C

  $I_B=40 \mathrm{\mu A}, I_C=10 \mathrm{mA}, \beta =250$
• Correct answer
  D

  $I_B=40 \mathrm{\mu A}, I_C=5 \mathrm{mA}, \beta =125$

Solution

${\text{V}}_{\text{i}}={\text{I}}_{\text{B}}{\text{R}}_{\text{B}}+{\text{V}}_{\text{B}\text{E}}$ (By Kirchoff's Voltage Law)

$20={\text{I}}_{\text{B}}\times \left(500\times {10}^3\right)+0$

${\text{I}}_{\text{B}}=\frac{20}{500\times {10}^3}=40\mathrm{ \mu A}$

${\text{V}}_{\text{C}\text{C}}={\text{I}}_{\text{C}}{\text{R}}_{\text{C}}+{\text{V}}_{\text{C}\mathrm{E }}$(By Kirchoff's Voltage Law)

$20={\text{I}}_{\text{C}}\times \left(4\times {10}^3\right)+0$

${\text{I}}_{\text{C}}=5\times {10}^{-3}=5 \mathrm{mA}$

$\beta =\frac{{\text{I}}_{\text{C}}}{{\text{I}}_{\text{B}}}=\frac{5\times {10}^{-3}}{40\times {10}^{-6}}=125$

Hence, the correct answer is option: 'D'.