Answer:
[tex]T_2 = 0.592[/tex]
Explanation:
Given
[tex]T_1 = 1.48s[/tex]
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:
[tex]\frac{1}{C_{up}} = \frac{1}{C} + \frac{1}{C}[/tex]
Take LCM
[tex]\frac{1}{C_{up}} = \frac{1+1}{C}[/tex]
[tex]\frac{1}{C_{up}}= \frac{2}{C}[/tex]
Cross Multiply
[tex]C_{up} * 2 = C * 1[/tex]
[tex]C_{up} * 2 = C[/tex]
Make [tex]C_{up}[/tex] the subject
[tex]C_{up} = \frac{1}{2}C[/tex]
The bottom two are also connected serially.
In other words, the upper and the bottom have the same capacitance.
So, the total (C) is:
[tex]C_1 = 2 * C_{up}[/tex]
[tex]C_1 = 2 * \frac{1}{2}C[/tex]
[tex]C_1 = C[/tex]
The total capacitance in (b) is calculated as:
First, we calculate the parallel capacitance (Cp) is:
[tex]C_p = C+C[/tex]
[tex]C_p = 2C[/tex]
So, the total capacitance (C2) is:
[tex]\frac{1}{C_2} = \frac{1}{C_p} + \frac{1}{C} + \frac{1}{C}[/tex]
[tex]\frac{1}{C_2} = \frac{1}{2C} + \frac{1}{C} + \frac{1}{C}[/tex]
Take LCM
[tex]\frac{1}{C_2} = \frac{1 + 2 + 2}{2C}[/tex]
[tex]\frac{1}{C_2} = \frac{5}{2C}[/tex]
Inverse both sides
[tex]C_2 = \frac{2}{5}C[/tex]
Both (a) and (b) have the same resistance.
So:
We have:
Time constant is directional proportional to capacitance:
So:
[tex]T\ \alpha\ C[/tex]
Convert to equation
[tex]T\ =kC[/tex]
Make k the subject
[tex]k = \frac{T}{C}[/tex]
[tex]k = \frac{T_1}{C_1} = \frac{T_2}{C_2}[/tex]
[tex]\frac{T_1}{C_1} = \frac{T_2}{C_2}[/tex]
Make T2 the subject
[tex]T_2 = \frac{T_1 * C_2}{C_1}[/tex]
Substitute values for T1, C1 and C2
[tex]T_2 = \frac{1.48 * \frac{2}{5}C}{C}[/tex]
[tex]T_2 = \frac{1.48 * \frac{2}{5}}{1}[/tex]
[tex]T_2 = \frac{0.592}{1}[/tex]
[tex]T_2 = 0.592[/tex]
Hence, the time constance of (b) is 0.592 s
