
In steady state, the capacitor acts as an open circuit (infinite resistance). Therefore, the 5 $\Omega$ resistor is not part of the circuit. The circuit simplifies to:
The total resistance is 2 $\Omega$ + 3 $\Omega$ = 5 $\Omega$. Using Ohm's law, I = V/R = $\frac{10 \, V}{5 \, \Omega} = 2 \, A$


The output (Y) of the given logic implementation is similar to the output of an/a …………. gate.