The time constant in RC circuit:

$\overline{){\mathbf{\tau}}{\mathbf{=}}{{\mathbf{R}}}_{\mathbf{e}\mathbf{q}}{{\mathbf{C}}}_{\mathbf{e}\mathbf{q}}}$

For capacitors in parallel, the equivalent capacitance is:

$\overline{){{\mathbf{C}}}_{\mathbf{e}\mathbf{q}}{\mathbf{=}}{{\mathbf{C}}}_{{\mathbf{1}}}{\mathbf{+}}{{\mathbf{C}}}_{{\mathbf{2}}}{\mathbf{+}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{+}}{{\mathbf{C}}}_{{\mathbf{n}}}}$

Initially, the capacitors are in parallel.

Equivalent capacitance is:

C_{eq} = C_{1} + C_{2} + C_{3} = C + C + C = 3C

You have an RC timing circuit made from three capacitors, each C and in parallel with each other, and one resistor R. You want to change the RC time constant to one-third what it was by removing one of the capacitors and adding one resistor of the appropriate value. What does the new total resistance of the circuit need to be and how should you add the new resistor (series or parallel)?

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