Question

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

Answer 1

Answer:

$T_2 = 0.592$

Explanation:

Given

$T_1 = 1.48s$

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:

$\frac{1}{C_{up}} = \frac{1}{C} + \frac{1}{C}$

Take LCM

$\frac{1}{C_{up}} = \frac{1+1}{C}$

$\frac{1}{C_{up}}= \frac{2}{C}$

Cross Multiply

$C_{up} * 2 = C * 1$

$C_{up} * 2 = C$

Make $C_{up}$ the subject

$C_{up} = \frac{1}{2}C$

The bottom two are also connected serially.

In other words, the upper and the bottom have the same capacitance.

So, the total (C) is:

$C_1 = 2 * C_{up}$

$C_1 = 2 * \frac{1}{2}C$

$C_1 = C$

The total capacitance in (b) is calculated as:

First, we calculate the parallel capacitance (Cp) is:

$C_p = C+C$

$C_p = 2C$

So, the total capacitance (C2) is:

$\frac{1}{C_2} = \frac{1}{C_p} + \frac{1}{C} + \frac{1}{C}$

$\frac{1}{C_2} = \frac{1}{2C} + \frac{1}{C} + \frac{1}{C}$

Take LCM

$\frac{1}{C_2} = \frac{1 + 2 + 2}{2C}$

$\frac{1}{C_2} = \frac{5}{2C}$

Inverse both sides

$C_2 = \frac{2}{5}C$

Both (a) and (b) have the same resistance.

So:

We have:

Time constant is directional proportional to capacitance:

So:

$T\ \alpha\ C$

Convert to equation

$T\ =kC$

Make k the subject

$k = \frac{T}{C}$

$k = \frac{T_1}{C_1} = \frac{T_2}{C_2}$

$\frac{T_1}{C_1} = \frac{T_2}{C_2}$

Make T2 the subject

$T_2 = \frac{T_1 * C_2}{C_1}$

Substitute values for T1, C1 and C2

$T_2 = \frac{1.48 * \frac{2}{5}C}{C}$

$T_2 = \frac{1.48 * \frac{2}{5}}{1}$

$T_2 = \frac{0.592}{1}$

$T_2 = 0.592$

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