Truth table of given circuit is given below:

This circuit out the logic operation of AND gate which can be verified also by De-Morgan's theorem
\(Z=\overline { \left( X+Y \right) } =\overline { \overline { A } +\overline { B } } \)
\(Z=\overline { \overline { A } } .\overline { \overline { B } } =A.B\)
This circuit corresponds to AND gate logic symbol is

Given voltage across R_B = 10 V
\(Resistance,{ R }_{ B }=400\kappa \Omega\)
\( { V }_{ BE }=0,{ V }_{ CE }-0 \ { R }_{ C }=3\kappa \Omega\)
\( { I }_{ B }=\frac { Voltage \ across \ { R }_{ B } }{ { R }_{ B } } =\frac { 10 }{ 400\times { 10 }^{ 3 } } \)
\(25\times { 10 }^{ -6 }A=25\mu A\)
\( Voltage \ across{ R }_{ C }=10 \ V\)
\({ I }_{ C }=\frac { Voltage \ across \ { R }_{ c } }{ { R }_{ C } } =\frac { 10 }{ 3\times { 10 }^{ 3 } }\)
\( \beta =\frac { { I }_{ C } }{ { I }_{ B } } =\frac { 3.33\times { 10 }^{ -3 } }{ 25\times { 10 }^{ -6\\ } }\)
\( \beta =1.33\times { 10 }^{ 2 } \)
\(=133\)

As, Concentration of Si atom \(=5\times { 10 }^{ 28 }\) /m³
The doping of indium is 1 atom 10⁶ atoms of Si. But indium has three valence electrons and each doped indium atom creates one hole in Si crystal. Hence, it acts as an acceptor atom.
Concentration of acceptor atoms,
\({ n }_{ h }=5\times { 10 }^{ 28 }\times { 10 }^{ -6 }=5\times { 10 }^{ 22 }/{ m }^{ 3 }\)
Intrinsic carrier concentration,
\({ n }_{ i }=1.5\times { 10 }^{ 16 }/{ m }^{ 3 }\)
Hole concentration is increased
\(=\frac { { n }_{ h } }{ { n }_{ i } } =\frac { 5\times { 10 }^{ 22 } }{ 1.5\times { 10 }^{ 16 } } =3.33\times { 10 }^{ 6 }\)
New electron concentration,
\({ n }_{ e }=\frac { { n }_{ i }^{ 2 } }{ { n }_{ b } } =\frac { { \left( 1.5\times { 10 }^{ 16 } \right) }^{ 2 } }{ 5\times { 10 }^{ 22 } } =0.45\times { 10 }^{ 10 }/{ m }^{ 3 }\)
Electron concentration has been reduced
\(=\frac { { n }_{ i } }{ { n }_{ e } } =\frac { 1.5\times { 10 }^{ 16 } }{ 0.45\times { 10 }^{ 10 } } =3.33\times { 10 }^{ 6 }\)
This means that the hole concentration has been increased over its intrinsic concentration by the same amount with which the electron concentration has been decreased.
The conductivity of doped silicon is given by
\(\sigma =e\left( { n }_{ e }H_{ e }+{ n }_{ h }H_{ h } \right) \)
\(\\ =\quad 1.6\times { 10 }^{ -19 }\left( 0.45\times { 10 }^{ 10 }\times 0.135+5\times { 10 }^{ 22 }\times 0.048 \right)\)
\( \\ =384s/m\)
Resistivity,  \(\rho =\frac { 1 }{ \sigma } =\frac { 1 }{ 384 } =0.0026\Omega -m\)
Conductivity of pure Si crystal,
\(\sigma ={ en }_{ i }\left( { H }_{ e }+{ H }_{ h } \right) \)
\(\\ =1.6\times { 10 }^{ -19 }\times 1.5\times { 10 }^{ 16 }\left( 0.135+0.048 \right) \)
\(\\ =0.4392\times { 10 }^{ -3 }S/m\)
Resistivity, \(\rho =\frac { 1 }{ \sigma } =\frac { 1 }{ 0.4392\times { 10 }^{ -3 } } =2276.\Omega -m\)
Thus, we see that the conductivity of Si-doped within becoming much greater than its intrinsic conductivity and the resistivity has become much smaller than the intrinsic resistivity.

Given, forward biased resistance = \(25\Omega \)
Reverse biased resistance = \(\infty \)
As the diode in branch CD is in reverse biased which having resistance infinite.
so  \(I_{ 3 }\)= 0
Resistance in branch AB = 25 + 125 = 150\(\Omega \)(say R₁)
Resistance in branch EF = 25 + 125 = 150\(\Omega \)(say R₂)
AB is parallel to EF.
so, resultant resistance, \(\frac { 1 }{ { R }^{ \prime } } =\frac { 1 }{ { R }_{ 1 } } +\frac { 1 }{ { R }_{ 2 } } =\frac { 1 }{ 150 } +\frac { 1 }{ 150 } =\frac { 2 }{ 150 } \)
\(\Rightarrow { R }^{ \prime }=75\Omega \)
Total resistance, \(R={ R }^{ \prime }+25=75+25=100\Omega \)
Current, \({ I }_{ 1 }=\frac { V }{ R } =\frac { 5 }{ 100 } =0.05A\)
\({ I }_{ 1 }={ I }_{ 4 }+{ I }_{ 2 }+{ I }_{ 3 }\quad [Here,\quad { I }_{ 3 }=0]\)
\(\\ { I }_{ 1 }={ I }_{ 4 }+{ I }_{ 2 }\)
Here, the resistances R₁ and R₂ are same
\({ I }_{ 4 }={ I }_{ 2 }\)
\(\\ { I }_{ 1 }=2{ I }_{ 2 }\)
\(\\ \Rightarrow { I }_{ 2 }=\frac { { I }_{ 1 } }{ 2 } =\frac { 0.05 }{ 2 } =0.025A\)
\(\\ { I }_{ 4 }=0.025A\)
\(\\ { I }_{ 1 }=0.05A\)
\(\\ { I }_{ 2 }=0.025A\)
\(\\ { I }_{ 3 }=0\)
\(\\ { I }_{ 4 }=0.025A\) 

When the input voltage is equal to or less than 5 V, diode will be reverse biased. It will offer high resistance in comparison to resistance (R) in series. now, diode appears in open circuit. The input waveform is then passed to the output terminals. The result with sine wave input is to dip off all positive going portion above 5 V.
If the input voltage is more than +5V, the diode will be conducting as if forward biased offering low resistance in comparison to R. But there will be no voltage in output beyond 5 V as the voltage beyond +5V appear across R.
When the input voltage is negative, there will be opposition to 5V battery. In p-n junction, input voltage becomes more than -5V, the diode will be reverse biased. It will offer high resistance in comparison to resistance R in series. Now, junction diode appears in open circuit. The input waveform is then passed on to the output terminals.
The output waveform is shown here in the figure: