The potential across a capacitor in an RC circuit is expressed as:

$\overline{){\mathbf{V}}{\mathbf{=}}{{\mathbf{V}}}_{{\mathbf{0}}}{{\mathbf{e}}}^{\mathbf{-}\frac{\mathbf{t}}{{\mathbf{R}}_{\mathbf{e}\mathbf{q}}{\mathbf{C}}_{\mathbf{e}\mathbf{q}}}}}$, where V is the voltage at a time t, V_{0} is the initial voltage, t is the time taken for the voltage to reduce from V_{0} to V, R_{eq} is the equivalent resistance, and C_{eq} is the equivalent capacitance.

Equivalent resistance for resistors in series:

$\overline{){{\mathbf{R}}}_{{\mathbf{eq}}}{\mathbf{=}}{{\mathbf{R}}}_{{\mathbf{1}}}{\mathbf{+}}{{\mathbf{R}}}_{{\mathbf{2}}}{\mathbf{+}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{+}}{{\mathbf{R}}}_{{\mathbf{n}}}}$

For capacitors in parallel, the equivalent capacitance is:

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

In the circuit shown in the figure (Figure 1), both capacitors are initially charged to 50.0 V.

Part A) How long after closing the switch S will the potential across each capacitor be reduced to 10.0 V?

Part B) What will be the current at that time?

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