The electrical circuit described has the given features:
Capacitance C = 5 * 10^{-3 }Farad
Resistance R = 25 ohm
Excitation 50 volt DC Supply (v)
Initially Q on the capacitor in VC = 0.25 coulomb
At t = 0 when S is switched from node ‘a’ to node ‘b’ and excitation falls instantly to zero
Q1. prepare down the differential equation which includes rate of change on capacitor as a function of time in the range D < t
Q2. Solve the differential equation by using Q_{0} = 0.25 coulomb to find out the constant of integration.
Q3. Use your outcome to show that the change on the capacitor is decreased to Q_{0}/2 after 1.3863 * 10^{-4 }S