Question

} likr st = 65uf ON ___ory 94+ -°F ATN Pvt

Answer 1

Please find the solution in the images below

Image 1

Initially the Charged fully switch is closed the capacitor and acts as open circuit o Voltage across the capacitor at t w 33K2 Neto) V (01) - Voltage: 5v Q UK across 11 kr Resistor By voltage Liuision rule - 5x11x10 = 1.25v V. (09)- 33 +11) 109 At to switch opens and disconnects the source Now, Capacitor discharges through liks resistor with time constant of ?= RC = 11000x 65x10 - R = 0.715 sec -E/RC -+/0.715 v t>o»Ųc (ot) = V(0-) e . = 1.25e for

for Explanation please refer the below images

Image 2

Discharge Phase : . . upto now we discussed about the charging phase of a Capacitior. At any point the swild hoint the switch is moved to or disconnector from source and conneted to short circuit the through resistor it starts Lischarging, at of the rate of time constant T=Rc. " ang .

Image 3

In this case the capacitor acts as battery and reverse current direction and discharges completty. g dir the current R ww et - VR + Ve=tv 2 As we discussed earlier the capacitor charges to Evolts S By compared with case is ic is reversed so a case is : ic=-ic By kul on look From above fig = c = ir Ve=UR R ( Up-ve) o ne dt = 1 av Apply go on Both sides C de over de ste dve

Image 4

(t-o) = loge Velt) - loge E e che = loge Velt) loga-log 6 = log a -t/RC . EL e = Volt) -t/Rc. [VeLE)= Ee the .. (((t)= c d Velt) at pact) =cd Eet/Re. =&x Ex = Il e-tre 12 indicates that reverse in current direction. UR | VR = į R. - - - - e - t/RC x L - E => Eet/Rc) revese in voltage

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