An npn transistor operates as a common emitter amplifier with a power gain of 10?. The input circuit resistance is 100Ω and the output load resistance is 10kΩ. The common emitter current gain 'β' will be -----------. (Round off to the Nearest Integer)

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 

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

The charge on hole is positive.

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