The charge on a hole is equal to the charge of

Here  $\Delta V_c=0.5V,\Delta i_c=0.05mA=0.05\times 10^{-3}A$

Output resistance is given by

$R_{out}=\frac{\Delta V_c}{\Delta i_c}=\frac{0.5}{0.5\times 10^{-3}}=10^4\Omega =10k\Omega .$

$i_b=\frac{5-0.7}{8.6}=0.5mA$

$\Rightarrow I_c=\beta I_b=100\times 0.5 mA$

By using 

$V_{CE}=V_{CC}-I_C{R}_L=18-50\times 10^{-3}\times 100=13 V$

Power gain = (i_c² R_c) / (i_b² R_B) = (i_c/i_b)² (R_c/R_B) = (10²)² (10? /10³) = 10?
i_c/i_b = 100 ⇒ β = i_c/i_b = 100

Photodiode operate in reverse bias. The photocurrent increases initially and saturates fi

First part of figure shown is or GATE and Second part of figure shown is NOT GATE So, $Y_P=OR+NOT=$  NOR GATE $Y=\stackrel{?}{A+B}=\bar{A}\cdot \bar{B}$