In the circuit shown, the capacitor is ini- tially uncharged. At t1 = 0, the switch S is moved to position a. R2 WM C R1 S b Vo a Find VRI, the voltage drop across R1, as a function of time t1. 1. VR = Voe ti/[(Ri+ R2) C'] 2. VRI = Voe-ti/(RIC) 3. VRI = Vo 1-e ti/[(Rui+ R2) C]] 4. VRI = Voe (RitR2) ti/(Ri R2 C) 5. VR = Vo 1 - e (RitR2) ti/(R1 R2 C) 6. VRI = Vo 1-e ti/(Ric)] 7. VRI = Voe -ti/ (R2 C)8. VRI = Vo 1-e ti/(R2 C) 012 (part 2 of 4) 10.0 points Find Vc, the voltage across C, as a function of time t1. 1. Vc = Vo 1-e-ti/(RIC) ] 2. Vo = Voe-ti/(RIC) 3. Vc = Voe-ti/(R2 C) 4. Vc = Vo 1-e-t1/(R2 C) ] 5. Vc = Voe (RitR2) ti/(R1 R2 C) 6. Vc = Vo 1-e (RitR2) t1/( R1 R2 C)] 7. Vc = Voe-ti/[(Ri+ R2) C] 8. Vc = Vo 1-e ti/[(Ri+ R2) C']013 (part 3 of 4) 10.0 points Much later (t1 ~oo), at some time to = 0 (the clock is restarted at t2 = 0), the switch is moved from position a to position b. Find the voltage drop, VR1, across R1, as a function of time t2. R1 - 1. VR1 = Vo - e-t2/(Ri C)] R1 + R2 R1 2. VRI = Vo e -(Ri+ R2) t2/(R1 R2 C) R1 + R2 R1 3. VR = Vo R1+ R2 x 1 - e (Ri+R2)t2/(RiR2C) 4. R1 VR1 = Vo 1 - e-t2/[(R1+R2) C'] R1+ R2 R1 5. VRI = Vo 1 - e-t2/(R2 C) ] R1+ R2 6. VRI = Vo R1 -t2/(R1 C) R1+ R2 R1 7. VRI = Vo e-t2/(R2 C) R1 + R2 R1 8. VRI = Vo e -t2/[(R1+ R2) C'] R1 + R2 014 (part 4 of 4) 10.0 points Find Vc as a function of time t2. 1. Vc = Vo 1 1 - el-tz/(Rui+R2) C'] 2. Vc = Voe-t2/(Ri C) 3. Vc = Voe-t2/[(R1+ R2) C]