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Problem 2 HE In the given RC circuit, a capacitor is connected to a resistor in series and is getting charged after closing t

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Problem-2 Vo R = 109 (2) C= P(f) HH AI Switch. V= 1010) Time constant = 10(s) (A) capacitance of the capacitor - Twine constaWe know IR = Ic = c dve dt So, Via Vc t RC duc dt. V = Vct Rc d Vc at (VR= IR) (Vi = v) V-Vc = Ro dve → dvc SV-Vo dt. dt RC VI Cave at Ic = c d f V (1-e-HRc)] dt CV e ARC Utc) V e-t RC Variation of cussent flowing through the Ic X R We know Ic = dac(6) Tinie for the capacitos to the fully charged. (9%) Usung eh Vc = V(l-e-HRc) for fully charge Vc = V (voltage assess capac(C) Charge at the 5th second of switch being closed. Using equation (C) le = CV 1- e-tire] Time constant RC = 10 R = lot 2 tln (0.5) = -t 10 0.6931 10 t = 6.931 seconds of switch habe se) Voltage at 5th second Msing egn (4) closed. being Ve - Vil-e-

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