Question

In the circuit shown below, the capacitor was fully charged then the battery was removed Find the time at which the energy in the capacitor decreases to 3% of its maximum (R1=60, R2=20, E-20 and C=1.OUF) R1 2.0 с R2 1.0 22 ti Select one: a. 1.18 us b. 11.8 us OC. 0.471 us d. 23.5 us e. 2.35 PS The electric potential in a region between x=0 m and x = 6:00 misgeby va

Answer 1

For t=i (Just before battery removed) » circuit is in steady state Under steady state steady state condition, Condition capacitor behaves as open circuit for de supply nul 22 mul or iz Ve (6) + 1 ur 22 20v -> current i 10 le v R 2+1 :i. 20 20 3 current i = 20 6+2 : iz = 5A - Vc (0) Apply EVL for 612 -Vc (0-) -aii =0

Velo) 6(3/2) -2(200 Velo) 1.6667 v - Capacitor doesnot allow sudden change in voltage - Velo) = Vc (0) = Vc (ot) 1.6667 v For t>o or unt 22 / — VLO) 22 122 an > capacitor voltage -te Velt) Vcco) e てた where Req С the equivalent resistance where Req is across Capacitor

62 22 M Req un w iz 212 : Req (6+2)(2+1) 8(3) 6+2+2+1 i Ree > 2= Req C 24 or (169) > Velt) – 1.667 844)105) المها ) 17 - ve (t) - 1.6667 e 24 V Energy Ez = 0.34 Emax Emax El Lell(t)) Ź ( Cveco)? 0.34

(Velt))? = 0.34 (veco)? (1417 2*) * = 034 (147) => -22/10634 e 24 = 0.34 (= -22 Cheb) + en (0-34) = -1,07881 + = (-1-07881 (24) (154) ta 1.18 us Coption