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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

1207_resistance and capacitance.jpg

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 Q0 = 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 Q0/2 after 1.3863 * 10-4 S

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