Animated Solution for Physics - Current Electricity: Find the time constant for the given RC circuits in correct order (in $\mu \text{s}$). $R_1=1\Omega ,R_2=2\Omega ,C_1=4\mu \text{F},C_2=2\mu \text{F}$.

When dealing with RC circuits, the time constant $\tau$ is the fundamental parameter that dictates how quickly the circuit charges or discharges. For any single-loop circuit, it is elegantly defined as the product of the equivalent resistance and the equivalent capacitance:

To solve this problem, our mission is simple: break down each of the three circuits, find their respective $R_{eq}$ and $C_{eq}$, and multiply them together. Let's dive in!

Series and Parallel

In the first circuit, we observe the left branch containing resistors $R_1$ and $R_2$ connected end-to-end. This is a classic series combination. Therefore, their equivalent resistance is simply their sum:

On the right branch, the capacitors $C_1$ and $C_2$ are connected across the same two nodes, placing them in parallel. Remember, capacitors in parallel add up directly (unlike resistors):

Multiplying these together gives us the first time constant:

The Inverse Setup

Moving to the second circuit, the arrangement is flipped. The resistors $R_1$ and $R_2$ are now in parallel. We use the product-over-sum rule to find their equivalent resistance:

Conversely, the capacitors $C_1$ and $C_2$ are now in series. We apply the product-over-sum rule for them as well:

Multiplying these equivalent values yields the second time constant:

The Hybrid Configuration

In the final circuit, both the resistors and the capacitors are arranged in parallel configurations.

For the resistors, we already calculated the parallel equivalent in the previous step:

For the capacitors, we also calculated their parallel equivalent in the first step:

Multiplying these together gives us the third time constant:

We have successfully calculated the time constants for all three circuits: - ${\tau }_1=18\mu \text{s}$ - ${\tau }_2=8/9\mu \text{s}$ - ${\tau }_3=4\mu \text{s}$

Arranging them in the requested order gives us 18, 8/9, 4, which perfectly matches option (b). By systematically breaking down complex networks into their equivalent components, even the most intimidating circuits become a walk in the park!