The moment switch is opened, voltage variation takes place across the capacitor as follows-
"V_c(t) = V_0(1-e^{-\\frac{t}{RC}})"
Here, "\\frac{1}{RC}" is the time constant.
On putting limit of (t) tends to zero in the above equation, we get-
= "V_0"
Therefore, after long time, the maximum capacitor voltage will be "V_0"
Since the capacitor is charged with polarity opposite to that of the battery, the current in the circuit will be zero after a very long time. Ideally, it will never be zero, but practically, it will take few seconds to few minutes depending upon the capacitance of the capacitor.
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