Four identical capacitors are connected with a resistor in two different ways. When they are connected as in part a of the drawing, the time constant to charge up this circuit is 1.48 s. What is the time constant when they are connected with the same resistor, as in part b

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

terd muffin is the correct answer