In the CB mode of a transistor, when the collector voltage is changed by 0.5 volt. The collector current changes by 0.05 mA The output resistance will be

 

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

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

By using 

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

The charge on hole is positive.

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 $Y=\stackrel{?}{A+B}=\bar{A}\cdot \bar{B}$