Question

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

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

Answer 1

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

$V_i=V_{BE}+R_B{i}_B$---------(1)

$20=500\times {10}^3\times i_B$

$i_B=40\mu A$

$V_o=V_{CE}-R_C{i}_C$------------(2)

$20=4\times {10}^3{i}_C$

$i_C=5mA$

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