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

`T_2 = 0.592`

Step-by-step 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:

`(1)/(C_(up)) = (1)/(C) + (1)/(C)`

Take LCM

`(1)/(C_(up)) = (1+1)/(C)`

`(1)/(C_(up))= (2)/(C)`

Cross Multiply

`C_(up) * 2 = C * 1`

`C_(up) * 2 = C`

Make
`C_(up)` the subject

`C_(up) = (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 * (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:

`(1)/(C_2) = (1)/(C_p) + (1)/(C) + (1)/(C)`

`(1)/(C_2) = (1)/(2C) + (1)/(C) + (1)/(C)`

Take LCM

`(1)/(C_2) = (1 + 2 + 2)/(2C)`

`(1)/(C_2) = (5)/(2C)`

Inverse both sides

`C_2 = (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 = (T)/(C)`

`k = (T_1)/(C_1) = (T_2)/(C_2)`

`(T_1)/(C_1) = (T_2)/(C_2)`

Make T2 the subject

`T_2 = (T_1 * C_2)/(C_1)`

Substitute values for T1, C1 and C2

`T_2 = (1.48 * (2)/(5)C)/(C)`

`T_2 = (1.48 * (2)/(5))/(1)`

`T_2 = (0.592)/(1)`

`T_2 = 0.592`

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