Answer:

Explanation:
Given

See attachment for connection
Required
Determine the time constant in (b)
First, we calculate the total capacitance (C1) in (a):
The upper two connections are connected serially:
So, we have:

Take LCM


Cross Multiply


Make
the subject

The bottom two are also connected serially.
In other words, the upper and the bottom have the same capacitance.
So, the total (C) is:



The total capacitance in (b) is calculated as:
First, we calculate the parallel capacitance (Cp) is:


So, the total capacitance (C2) is:


Take LCM


Inverse both sides

Both (a) and (b) have the same resistance.
So:
We have:
Time constant is directional proportional to capacitance:
So:

Convert to equation

Make k the subject



Make T2 the subject

Substitute values for T1, C1 and C2




Hence, the time constance of (b) is 0.592 s