e.m.f of the source , E = 1.5 
Voltage drop across the diode , V_d = 0.5 V
Maximum power rating of the diode 
I = (P / Vd)
I = 0.2A
Potential drop across resistance R
V = E - V_d
= 1 V
R = V/1 = 1/0.2 = 5 \(\Omega \)

This is AND logic gate Output wave form

The NOT gate is a device which has only one input and one output i.e. \(\bar { A } \)= Y means Y equals NOT A.
This gate cannot be realised by using diodes. However, it can be realised by making the use of a transistor. This can be seen in the figure given below:

Here, the base B of the transistor is connected to the input A through a resistance R_B and the emitter E is earthed. The collector is connected to 5 V battery. The output Y is the voltage at C with respect to the earth.
The resistors, R_B and R_C are so chosen that, if emitter-base junction is unbiased, the transistor is in cut-off mode and if emitter-base junction is forward biased by 5 V, the transistor is in saturation state.

n-type semiconductor is obtained when pentavalent impurity added to Si or Ge. All these materials are electrically neutral, so n-type semiconductor is also neutral.

(a) 133.3 (b) \(0.2\Omega ^{ -1 }\) (c) 1000
Here, \(R_{ i }=665\Omega , \ R_{ 0 }=5k\Omega =5\times 10^{ 3 } \ \Omega \)
(a) \(\beta _{ ac }=\frac { \Delta I_{ c } }{ \Delta I_{ b } } =\frac { 2mA }{ 15\mu A } =\frac { 2\times 10^{ -3 }A }{ 15\times 10^{ -6 }A } =133.3\)
(b) \(R_{ i }=\frac { \Delta V_{ BE } }{ \Delta I_{ b } } \) or \(\Delta V_{ BE }=R_{ i }\times \Delta I_{ b }\)
 \(=665\times 15\times 10^{ -6 }V\)
\(g_{ m }=\frac { \Delta I_{ c } }{ \Delta V_{ BE } } =\frac { 2\times 10^{ -3 } }{ 665\times 15\times 10^{ -6 } } =0.2\Omega ^{ -1 }\)
\(A_{ V }=\beta _{ a.c. }\times \frac { R_{ 0 } }{ R_{ i } } =133.3\times \frac { 5\times 10^{ 3 } }{ 665 } =10^{ 3 }\)