How would you set up a circuit to obtain a $NOT$ gate using a transistor?

(N/A) To obtain a $NOT$ gate using an $NPN$ transistor, the transistor is used in the common-emitter configuration. The input $A$ is applied to the base through a resistor $R_B$, and the output $C$ is taken from the collector terminal.
For case $I$:
When $A=0$ (low input), the base current $I_B = 0$. Consequently, the collector current $I_C = \beta I_B = 0$. Using Kirchhoff's voltage law in the output loop, $V_{CC} = I_C R_C + V_{CE}$. Since $I_C = 0$, $V_{CE} = V_{CC} = 5 \text{ V}$. Thus, the output $C = 1$ (high).
For case $II$:
When $A=1$ (high input), the transistor is driven into saturation, making $I_B$ and $I_C$ maximum. In this state, the voltage drop across $R_C$ is nearly equal to $V_{CC}$, so $V_{CE} \approx 0$. Thus, the output $C = 0$ (low).
This confirms the operation $C = \bar{A}$, which is the characteristic of a $NOT$ gate.

Input $A$	Output $C$
$0$	$1$
$1$	$0$