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

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

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

β = 120
Base current, \({ I }_{ B }=\frac { 25V }{ 1M\Omega } =\frac { 25 }{ 1\times { 10 }^{ 6 } } =25\mu A\)
We know that
\(\beta =\frac { { I }_{ C } }{ { I }_{ B } } \) (or) 
I_C = β I_B = 120 x 25 μA
= 3000 μA = 3 mA
V_CE = V_CC - I_CR_C
= 25 - (3 mA x 5k) = 10 V 

\(\beta =50 \)
\(V_{\beta E} =600 \mathrm{mV} \)
\(=0.6 \mathrm{~V} \)
\(\mathrm{V}_{\mathrm{B}} =\mathrm{V}_{\mathrm{E}}-\mathrm{V}_{\mathrm{EB}} \)
\(\mathrm{V}_{\mathrm{B}} =3-0.6 \)
\(=2.4 \mathrm{~V} \)
\(\mathrm{I}_{\mathrm{B}} =\frac{\mathrm{V}_{\mathrm{B}}}{R_B}=\frac{2.4}{60 \times 10^3}=40 \mu \mathrm{A} \)
\(\mathrm{I}_{\mathrm{C}} =\beta \mathrm{I}_{\mathrm{B}}=50 \times 40 \mu \mathrm{A} =2 \mathrm{~mA} \)
\(V_C=R_FI_C=500 \times 2 \times10^{-3}=1 V\)
\(V_{EC}=V_E-V_C\)
\(V_{EC}=V_E-V_C\)
\(V_{EC}=3-1=2V\)

\(\mathrm{V}_{\mathrm{cc}}=12 \mathrm{~V}, \mathrm{R}_{\mathrm{B}}=10 \mathrm{k} \Omega, \mathrm{R}_{\mathrm{E}}=1 \mathrm{k} \Omega, \mathrm{R}_{\mathrm{c}}=1+1=2 \mathrm{k} \Omega, \alpha=0.99, \mathrm{~V}_{\mathrm{BE}}=0.7 \mathrm{~V}, \mathrm{I}_{\mathrm{c}}=?\)
\(\beta=\alpha /(1-\alpha)=0.99 /(1-0.99)=99\)
\(\mathrm{I}_{\mathrm{B}}=\mathrm{I}_{\mathrm{C}} / \beta=\mathrm{I}_{\mathrm{c}} / 99\)
Applying Kirchoff's Voltage law,
\(I_C R_C+I_n R_n+I_E R_E+V_{u t}=V\)
\(2 \times 10^3 \mathrm{I}_{\mathrm{C}}+10 \times 10^3\left(\mathrm{I}_{\mathrm{C}} / 99\right)+1 \times 10^3\left(\mathrm{I}_{\mathrm{C}}+\mathrm{I}_{\mathrm{C}} / 99\right)+0.7=12 \quad\left(\because \mathrm{I}_{\mathrm{E}}=\mathrm{I}_{\mathrm{n}}+\mathrm{I}_{\mathrm{C}}\right)\)
\(\therefore \mathrm{I}_{\mathrm{C}}=\frac{11.3 \times 10^{-3} \times 99}{298}\)
\(\mathrm{I}_{\mathrm{C}}=3.7 \times 10^{-3} \mathrm{~A}=3.7 \mathrm{~mA}\)

Diode D₁ and D₄ is reverse biased [open]
Diode D₁ and D₃ are forward biased.
The resistances are in series
R = 1 + 10 + 1 - 12 Ω
Barier Potential, V = 0.7 + 0.7 = 1.4 V (Silicon diode)
Applying Kirchhoff's voltage Law,
0.7 + I(1) + I(10) + 0.7 + I(1) = 3V
12 I = 3 - 1.4
12 I = 1.6
\(I=\frac{1.6}{12}=\mathbf{0 . 1 3 3 A}\)