Answer:
New total resistance = R/2
The new resistor is connected in parallel.
Step-by-step explanation:
Let
be initial time constant and
the new time constant.
The 3 capacitors are connected in parallel. The total capacitance,
, is their sum. Hence
![C_t=C + C + C = 3C](https://img.qammunity.org/2021/formulas/physics/college/ymbau72z3wcrbs8w6be9329binsq2iokob.png)
![T_1 = R * 3C = 3RC](https://img.qammunity.org/2021/formulas/physics/college/d93q8pp88clh0vaiqgriz2shyos4fa4ed5.png)
The new time constant is one-third of the initial time constant.
![T_2 = (T_1)/(3) = RC](https://img.qammunity.org/2021/formulas/physics/college/ov0u3op17ili1tc29n8pzjzicp13qzg1vw.png)
If one of the capacitors is removed, then the new total capacitance,
is
![C_(t2) = 2C](https://img.qammunity.org/2021/formulas/physics/college/75ycsbmy90itd1wt9unxtwcylmnpavkab9.png)
If the new total resistance isc
, then
![T_2 = R_(t2)C_(t2)](https://img.qammunity.org/2021/formulas/physics/college/417pca0cdfxg90bj9ywanxkp31rtmnqelo.png)
![RC = R_(t2)*2C](https://img.qammunity.org/2021/formulas/physics/college/8zj8z4s3fdi7khh8r4th7gsgy7upchbmfl.png)
![R_(t2) = (R)/(2)](https://img.qammunity.org/2021/formulas/physics/college/6u8mcg2n5qr5mz4dfw5k3xpwjaflbcevqy.png)
Since this is less than the old total resistance, the new resistor, with resistance, X, must be connected in parallel. Its value will be R, as below:
![(1)/(R/2)=(1)/(R)(1)/(X)](https://img.qammunity.org/2021/formulas/physics/college/q4242881j3zk0dpmb32guhbxk2verjo47x.png)
![(2)/(R)-(1)/(R)=(1)/(X)](https://img.qammunity.org/2021/formulas/physics/college/mxcpdgyq8bpstsi5hjsdaepodfbpekylij.png)
![(1)/(R)=(1)/(X)](https://img.qammunity.org/2021/formulas/physics/college/cvos033izmvdb518jxzu25hxuv8gbsuw46.png)
![X = R](https://img.qammunity.org/2021/formulas/physics/college/nx6pjjocx36hsymdei6j2kavflc0gl82a4.png)