Question

You charge an initially uncharged 81.1-mF capacitor through a 39.1-12 resistor by means of a 9.00-V battery having negligible internal resistance. Find the time constant t of the circuit. t= What is the charge Q on the capacitor 1.17 time constants after the circuit is closed? What is the charge Qo after a long amount of time has passed?

Answer 1

0. Capacitance, C=81:1 m F Repistance, R=39.12 Battery voltage, U=90 ③ Time constant of RC circuit is given as, T = RC T-(39.1%81-1*10-3)s īs 3.1715 . Time constant of circuit, t= 3.171S for Let circuit is closed at teos when tto, cabacitor is discharged :. Vc (0-)- Ou Since capacitor does not require sudden change of voltage, Vccot) -OV Od ttoo, capacitom doe gets fully choseged and hence is open com circuited ? d. at tsoo circuit can be brows as, R I - Vega :: V- OxR-VC=Os VeU : Ucco= qu :: ak to the capacitor's voltage can be given ad, Velti: Velo) +[ucios - Vc cool le the

Goli t- 1.175 .. V(14) - 94(0-9) e-tit saCl- e-f1.1787)) =911-e117) Vect) = 6.206 UT Charge at t= 1.170 is given as, Olt)= 62 C Vet) Olt)- 81.1 x6 206 x 10-3 C T044)- 50 3.36 mc :. Charge at t = 1.17t is given as, O=503-36 mc (IT) cct too, Chorg voltage across capacitor is, was to ao c. Charge at top is, Q=&c Vios Q=81.1x9x don3c Q 72499 0 = 729.9mc at tooo, Chaoupe across capacities is 72909 mc