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The electrical circuit illustrated has the following features:

Capacitance “C” = 5 * 10-3 Farad

Resistance “R” = 25?

Exitation 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” & excitation falls immediately to zero

(A) Write the differential equation which invoves rate of change of change on capacitor as a function of time in the range D < t

(B) Use your result to show that the change on the capacitor is reduced to Q0 / 2 after 1.3863 * 10-4 S

(C) Solve the differential equation. Using Q0 = 0.25 coulomb to determine the constant of integration

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