Answer :

Given i₀ = 2 A,

R =6 x 10⁵ Ω


C =0.500 μF = 5 x10 F


Hence



So the electric current in a discharging R-C circuit is,



(a) At t = 0.3 s, the current, i = 2 × e-1 =2/e A.


(b) For the rate of change of current, we find out di/dt.






at t = 0.30 s,



(c) Since at t = 0.30, di/dt = -20/3e A/s,


increase in current in 0.01 s A (approx).


Hence the current at 0.31 s = Current at t = 0.30 s + increase in 0.01 s







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