Answer:
New total resistance = R/2
The new resistor is connected in parallel.
Explanation:
Let [tex]T_1[/tex] be initial time constant and [tex]T_2[/tex] the new time constant.
The 3 capacitors are connected in parallel. The total capacitance, [tex]C_t[/tex], is their sum. Hence
[tex]C_t=C + C + C = 3C[/tex]
[tex]T_1 = R \times 3C = 3RC[/tex]
The new time constant is one-third of the initial time constant.
[tex]T_2 = \dfrac{T_1}{3} = RC[/tex]
If one of the capacitors is removed, then the new total capacitance, [tex]C_{t2}[/tex] is
[tex]C_{t2} = 2C[/tex]
If the new total resistance isc[tex]R_{t2}[/tex], then
[tex]T_2 = R_{t2}C_{t2}[/tex]
[tex]RC = R_{t2}\times2C[/tex]
[tex]R_{t2} = \dfrac{R}{2}[/tex]
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:
[tex]\dfrac{1}{R/2}=\dfrac{1}{R}\dfrac{1}{X}[/tex]
[tex]\dfrac{2}{R}-\dfrac{1}{R}=\dfrac{1}{X}[/tex]
[tex]\dfrac{1}{R}=\dfrac{1}{X}[/tex]
[tex]X = R[/tex]
