Question

(1 point) In the circuit below, R1 = 70 12, R2 = 40 S2, R3 = 16 12, C = 58 mF, V1 = 12 V, and V2 = -44 V. The circuit has been running for a long time before the switch flips to the left at t = 0 Find iz(t) for t > 0 R2 t= 0 R3S 2017 Paul Hummel CCOO BY NC SA iz(t) for t> 0 =

Answer 1

Solution Geiren: R=702 R2 = 402 R2=162 (=58mF V=12V V2 =-440 Refrieving the given Ckt R, 4002tzo 12VY R2 e 7 Figl Briginal Ckt Griveni • Ckt is in steady state before t=0 Q befose vaith ing • Steady stale + Capacitor open att Drawing the che al t= I 700R 6 12 =-440 Fig2. Cke at t= 0 (steady state)

from the above ckt P060)=-44 volt now since the capacitor voltage cannot change instantaneously 100)=() = 27(04) = -44V|---0 •Now at to switching access is dis connected from "Che" and vi gets Gomected • with Connected, the cke once again reaches Steady State at ta we once again draw the che site al tros Steady state → once again capacitor = open Cat R₂-402 vy by ve (W) Figs. Cke allt=00) steady state] By voltage division 29(0) = 1; 23 = 120

of 1026)= 3.43 volt) --- Now the general equation of voltage across Capacitor fortyo is ple) = 1EC2+ (2017-2018 |--- where t is time constant of the che and is given FE Reg C] - - 6 Reg is the equivalent resistance seen by Capacitor after witching calculation of Ree mg Rea 7 Reg= P21/23 = 401/16 = 11-4322 en a gives Te keq c = (1:43 (58x103) seconds or (T= 0.663 Seconds ? ---0

living the values from eq 0, @ and 6 cm eq ® we obtain -t/-663 V(t)= 3-43+ (-44-3-43) e' volts vo(t) = 3-43-47.43 é volts -1.51€alls Finally from rigt -1.56 3-43-47-4B e re PC2 - 3-43 — +1-03e < - 51% 7 live = 0.214375 -2.964e ampere