Step 1: Understanding the Question:
The goal of this question is to find the overall time constant ($\tau$) of the given RC network.
The time constant is a characteristic parameter of first-order networks that indicates how quickly the circuit responds to a transient input.
Step 2: Key Formula or Approach:
The time constant of an RC network is generally given by:
\[ \tau = R_{eq} C_{eq} \]
where $R_{eq}$ is the equivalent resistance seen by the capacitor(s) with independent voltage sources short-circuited, and $C_{eq}$ is the equivalent capacitance.
Alternatively, we can derive the transfer function $H(s) = \frac{V_{out}(s)}{V_{in}(s)}$ and find the pole of the system.
Step 3: Detailed Explanation:
• Let us deactivate the independent voltage source $V_{in}(t)$ by replacing it with a short circuit.
• Shorting $V_{in}(t)$ directly shorts the capacitor $C_1$ to ground, which removes $C_1$ from contributing to the dynamic behavior of the output node.
• The node to the left of the parallel combination of $R$ and $C_2$ is now connected to ground.
• Thus, the resistor $R$ and capacitor $C_2$ are connected in parallel between the output node and ground.
• The capacitor $C_3$ is also connected between the output node and ground.
• Therefore, looking from the output terminal to ground, the resistor $R$ is in parallel with the parallel combination of $C_2$ and $C_3$.
• The equivalent capacitance seen by the resistor $R$ is:
\[ C_{eq} = C_2 + C_3 \]
• The equivalent resistance is:
\[ R_{eq} = R \]
• The time constant is then calculated as:
\[ \tau = R_{eq} C_{eq} = R(C_2 + C_3) \]
Step 4: Final Answer:
The time constant of the network is $R[C_2+C_3]$, which matches Option (D).