Step 1: AC (1 + B) = AC.1 [OR law-2]
Step 2: AC . 1 = AC [AND law – 2]
Therefore, AC + ABC = AC
Thus the Boolean identity is proved.
Circuit Description

The output at the 1^st AND gate : A\(\overline { B } \)
The output at the 2^nd AND gate : ĀB
The output at the OR gate: Y = A. \(\overline { B } \) + Ā .B

A	B	C	X = A.B	Y=\(\overline { X.C } \)
0	0	0	0	1
1	1	1	1	0

\({ I }_{ B }=\frac { { V }_{ i } }{ { R }_{ B } } =\frac { 20V }{ 500k\Omega } =40\mu A\)  [∵ V_BE = 0V]
\({ I }_{ C }=\frac { { V }_{ CC } }{ { R }_{ C } } =\frac { 20V }{ 4k\Omega } =5mA\) [∵ V_CE = 0V]
\(\beta =\frac { { I }_{ C } }{ { I }_{ B } } =\frac { 5mA }{ 40\mu A } =125\)

α = \(\frac{I_C}{I_E}\)
I_C = α I_E = 0.95 x 1 = 0.95 mA
I_E = I_B + I_C
∴ I_B = I_E - I_C = 1 - 0.95 = 0.05 mA

\({ E }_{ g }=\frac { hc }{ \lambda } \)
Therefore,
\(\lambda =\frac { hc }{ { E }_{ g } } =\frac { 6.6\times { 10 }^{ -34 }\times 3\times { 10 }^{ 8 } }{ 1.875\times 1.6\times { 10 }^{ -19 } } \)
= 660 nm
The wavelength 660 nm corresponds to red colour light

The P.D. between A and B is given by
\(V =\left[V_{\mathrm{A}}-V_{\mathrm{B}}\right]-V_{\mathrm{b}}(\mathrm{Si}) \)
\(=[3.3-(-7.4)]-0.7 \)
\(=10.7-0.7=10 \mathrm{~V} \)
The value of current flowing through AB can be obtained by using Ohm’s law
\(I=\frac { V }{ R } =\frac { 10}{ 1\times { 10 }^{ 3 } } ={ 10 }^{ -2 }A=10mA\)

The diode is forward biased and it is an ideal one. Hence, it acts like a closed switch with no barrier voltage. Therefore, current that flows through the diode can be calculated using Ohm’s law.
V = IR
\(I=\frac{V}{R}=\frac{15}{5}\) = 3A

V= 10V,  R₁ = 2 Ω, R₃ = 2 Ω
Diode D₁ is reverse biased so it will block the current and diode D₂ is forward biased, so it will pass the current.
\(\mathrm{R} =\mathrm{R}_{1}+\mathrm{R}_{2} \)
\(=2+2=4 \Omega \)
\(\mathrm{I} =\frac{\mathrm{V}}{\mathrm{R}}=\frac{10}{4} \)
I = 2.5 A

(i) De Morgan's First theorem states that the complement of the sum of two inputs is equal to the product of its complements.
\(\overline{\mathrm{A}+\mathrm{B}}=\overline{\mathrm{A}} \cdot \overline{\mathrm{B}}\)
(ii) The second theorem states that the complement of the product of two inputs is equal to the sum of its complements.
\(\overline{\mathrm{A} \cdot \mathrm{B}}=\overline{\mathrm{A}}+\overline{\mathrm{B}}\)

(i) If the portion of the output fed to the input is in phase with the input, then the magnitude of the input signal increases
(ii) It is necessary for sustained oscillations.

(i) The output signal is reversed by 180^o. During the positive half cycle, Input signal (V_s) increases the forward voltage across the emitter-base.
(ii) As a resultthe base current (I_B) increases. Consequently, the collector current (I_C) increases ß times. 
(iii) This increases the voltage drop across R_c (I_C R_C) which in turn decreases the collector-emitter voltage (V_CE).
(iv) Therefore, the input signal in the positive direction produces an amplified signal in the negative direction at the output.

(i) The emitter-base junction is forward biased by a dc power supply Vm and the collector-based junction is reverse biased by the bias power supply V_cB
(ii) The forward bias across the emitter base junction causes the majority charge carriers electrons in the emitter region to flow towards the base region and constitutes the emitter current (I_E).
(iii) Since the base region is very narrow, most of the electrons reach the collector region.
(iv) The electrons that reach the collector region will be attracted by the collector terminal as it has positive potential and flows through the external circuit. This constitutes the collector current(I_c)
(v) The holes that are lost due to recombination in the base region ate replaced by the positive potential of the bias voltage V_EE and constitute the base current (I_B)
\( \mathrm{I}_{\mathrm{E}}=\mathrm{I}_{\mathrm{B}}+\mathrm{I}_{\mathrm{C}} \)
\(\mathrm{I}_{\mathrm{E}} \approx \mathrm{I}_{\mathrm{C}} \)     \((\because \mathrm{I}_{\mathrm{B}} is\ very\ small )\)

The leakage current is the current that the diode will leak when a reverse bias is applied to it.

When a PN junction diode is forward biased, the depletion region decreases and the diode conduct once after the barrier potential is crossed, when it is reverse biased the depletion region increases and the diode does not conduct sci it is called as unidirectional device.

The process of adding impurities to the instrinsic semiconductor is called doping.

(i) To move the hole in a given direction, the valence electrons move in the opposite direction.
(ii) Electrons flow in a N-type semiconductor is similar to electrons moving in a metallic wire.
(iii) The N- type dopant atoms will yield electrons available for conduction.

RADAR stands for RAdio Detection And Ranging.

When the frequency of the baseband signal is zero (no input signal), there is no change in the frequency of the carrier wave. It is at its normal frequency. This is called centre frequency or resting frequency.

i) Noise : Interference of undesirable electrical signal with the transmitted signal.
ii) Attenuation: Loss of strength of a signal.
iii) Medium : It makes some distortion tothe signal.

The total distance d between the transmitting and receiving antennas will be the sum of the individual distances of coverage.
d = d₁ + d₂
\(=\sqrt { 2R{ h }_{ 1 } } +\sqrt { 2{ Rh }_{ 2 } } \)
\(=\sqrt { 2R } \left( \sqrt { { h }_{ 1 } } +\sqrt { { h }_{ 2 } } \right) \)
\(=\sqrt { 2\times 6.4\times { 10 }^{ 6 } } \times (\sqrt { 40 } +\sqrt { 30 } )\)
\(=16\times { 10 }^{ 2 }\sqrt { 5 } \times (6.32+5.48)\)
= 42217 m = 42.217 km