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kvl around the loop - v + iR + V_c = 0 , v_c = voltage across capacitor i = c dV_c/dt c dV_c/dt R + Vertical-bar V_C = v d V_C/dt + V_C/R-c = V/Rc solving the differential equation v_c = V (1 - e^- t/Rc) As the voltage finally settle to 5 v V_C = 5 (1 - e^- t/Rc) V_C = 64% 0 s, = 3.2 v at t = Rc V_c = 5 (1 - e^- 1) = 3.16 v Thus , when V_c = 3.2 v (64 % of 5v) , then t = Rc (time constant) from graph , when V_c almostequalto 3.2 v, t tildeequal 0.08 sec 0.08 = R times 10^- 6 , as c = 14 F R = 80 k ohm
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