A capacitor C=8.0 μF is connected between nodes p and n, with capacitor voltage…

A capacitor \(C=8.0\,\mu\mathrm{F}\) is connected between nodes \(p\) and \(n\), with capacitor voltage \(v_C=v_{pn}\) polarised positive at \(p\). For all \(t<0\) an external path holds \(v_C=48\,\mathrm{V}\). At \(t=0\) that path is removed, and for \(t\ge 0\) the only remaining element is a resistor \(R=12.5\,\mathrm{k}\Omega\) connected between \(p\) and \(n\). The resistor current \(i_R\) is defined leaving \(p\) through \(R\) and entering \(n\). The capacitor current \(i_C\) obeys the passive sign convention. (a) State the continuity principle for capacitor voltage and determine \(v_C(0^-)\) and \(v_C(0^+)\). Determine the time constant \(\tau\) of the source-free circuit. (b) Write the natural responses \(v_C(t)\) and \(i_R(t)\) for \(t\ge 0\). Determine \(i_C(0^+)\). (c) Determine the energy stored in the capacitor at \(t=0^+\) and as \(t\to\infty\). Determine the energy absorbed by the resistor over \(0\le t<\infty\).

After \(t=0\) the circuit is source-free: the capacitor simply discharges through \(R\), and capacitor voltage cannot jump.