Question

The network shown is assembled with uncharged capacitors X, Y, and Z, with C_x = 4.00 mF C_Y = 3.00 mF and C_Z = 5.00 mF. The switches S_1 and S_2 are initially open and a potential difference V_ab = 120V is applied between points a and b. In this problem we will assume that there is a resistance, R_s, = 0.800k Ohm, in the switch S_1. (I) The switch S_1 is closed, for a total of 9.40s, and then it is openned again. How much energy is stored in the three capacitors at this time? (II) The network is reset to the original state (uncharged capacitors) and then switch S_2 is closed (no resistance here) and afterwards switch S_1 is closed, for a total of 9.40s, and then opened again. How much energy is stored in the three capacitors at this time? (III) Explain which of the two settings (S_2 open or S_2 closed) is able to store more energy, you may use different arguments but your choice must be supported by theory and/or numerical results.

Answer 1

So emengy stored in X xcx hnf 193.5 mc opened theny.zui" be in sen'esord.+equ! gne,gderrel en 't ba ,xds ㅁ 19,184丁 Cy (พิ) s total emery in (I) 3.741+ 6.2404 2).3 6、246丁 31,284丁 3,744丁 n) when e2.ir coled,-,ne opacs toryusul tPr kl gon-ho 휵 cue 1, Be caule in cue Tr Yonz 1my 12ev ane poraiel while y- not working am xo 0.8 So yaltay e om xondz, will be come Vs 086X120 103.2V So this wi be at badh x ord cyz -0.79% 120