A 40 μF capacitor is connected, for all t<0 and for a duration sufficient to reach DC…

A \(40\,\mu\mathrm{F}\) capacitor is connected, for all \(t<0\) and for a duration sufficient to reach DC steady state, directly across a \(25\,\mathrm{V}\) independent DC voltage source, so that \(v_C(0^-)=25\,\mathrm{V}\). At \(t=0\) a switch disconnects the source and simultaneously connects the capacitor across the series combination of a \(6.0\,\mathrm{k}\Omega\) resistor and a \(10\,\mathrm{k}\Omega\) resistor. The definition of \(v_C\) is unchanged. Let \(i\) be the loop current that leaves the positive capacitor plate and then passes first through the \(6.0\,\mathrm{k}\Omega\) resistor and then through the \(10\,\mathrm{k}\Omega\) resistor. (a) State \(v_C(0^+)\) and \(i(0^+)\). Determine the time constant. (b) Write \(v_C(t)\) and \(i(t)\) for \(t>0\). Evaluate both at \(t=4\tau\). (c) Compute the energy stored in the capacitor at \(t=0^+\) and as \(t\to\infty\). Compute the energy absorbed by each of the two resistors over \(0<t<\infty\), and confirm that those two energies sum to the energy lost by the capacitor.

Disconnecting the source leaves a source-free series RC loop; the two resistors never share voltage equally, but they do share a single current.