In the circuit shown in the figure,when the input voltage $V_i$ is $10\, V$,$V_{BE}$ is zero and $V_{CE}$ is also zero. Find the values of $I_B$,$I_C$,and $\beta$.

(D) In the input section:
$V_i = I_B R_B + V_{BE}$
Given $V_i = 10\, V$,$R_B = 400\, k\Omega = 400 \times 10^3\, \Omega$,and $V_{BE} = 0\, V$.
$10 = I_B (400 \times 10^3) + 0$
$I_B = \frac{10}{400 \times 10^3} = 0.025 \times 10^{-3}\, A = 25 \times 10^{-6}\, A = 25\, \mu A$
In the output section:
$V_{CC} = I_C R_C + V_{CE}$
Given $V_{CC} = 10\, V$,$R_C = 3\, k\Omega = 3 \times 10^3\, \Omega$,and $V_{CE} = 0\, V$.
$10 = I_C (3 \times 10^3) + 0$
$I_C = \frac{10}{3 \times 10^3} = 3.33 \times 10^{-3}\, A = 3.33\, mA = 3333\, \mu A$
Calculating current gain $\beta$:
$\beta = \frac{I_C}{I_B} = \frac{3.33 \times 10^{-3}}{25 \times 10^{-6}} = \frac{3333}{25} = 133.32$