If \({V}_{{A}}\) and \({V}_{{B}}\) are the input voltages (either \(5\, {V}\) or \(0\, {V}\) ) and \({V}_{0}\) is the output voltage then the two gates represented in the following circuit \((A)\) and \((B)\) are:- \((A)\quad\quad\quad\quad(B)\)

\({V}_{{A}}=0 {V} \Rightarrow {A}=0\)

\({V}_{{B}}=5 {V} \Rightarrow {B}=1\)

\({V}_{{B}}=0 {V} \Rightarrow {B}=0\)

If \({A}={B}=0\), there is no potential anywhere here \({V}_{0}=0\)

If \({A}=1, {B}=0\), Diode \({D}_{1}\) is forward biased, here \({V}_{0}=5 {V}\)

If \({A}=0, {B}=1\), Diode \({D}_{2}\) is forward biased hence \({V}_{0}=5 {V}\)

If \({A}=1, {B}=1\), Both diodes are forward biased hence \({V}_{0}=5 {V}\)

Truth table for \({I}^{{st}}\)

\(\therefore\) Given circuit is \(OR\) gate

For \(\mathbb{I}^{{Id}}\) circuit

\({V}_{{B}}=5 {V}, {A}=1\)

\({V}_{{B}}=0 {V}, {A}=0\)

When \({A}=0, {E}-{B}\) junction is unbiased there is no current through it

\(\therefore {V}_{0}=1\)

When \({A}=1, {E}-{B}\) junction is forward biased

\({V}_{0}=0\)

\(\therefore\) Hence this circuit is not gate.