Question

Problem 1. 15 points) switch The figure above shows a circuit containing an electromotive force in battery), a capacitor with a capacitance of forede (F), and a resistor with a resistance of Rohms (1) The voltage drop across the capacitor in Olc, where is the charge in coulombs, so in this case Kirchhoff's Law gives RI + Since the current is do di R0 = EU). Suppose the resistance is 309, the capacitance is 0.15F, a battery gives a constant voltage of E() SOV, and the initial charge is (0) 0C. Find the charge and the current at time 00- (1)

Answer 1

Solution of Differential Equations: Given, R dQ 1 +9=E(t) dt R=30 12, C = 0.15 F, E (t)= 50 V,(0= 0 We have to find the charge and current at timet. Substitute all the given values in the differential equation and solve as follows: 1 dQ 30 + Q = 50 dt 0.15 dQ 9 +2Q =15 dt This is first order differential equation whose solution is given as follows:

dQ - 15-20 9 dt dQ 1 =- 15-299 Integrate on both sides we get -2dQ 15-20 = 1 do In(15-20) - ;--+Inc 1,(15-20) =--t 9 15-20=CB 2= (15-ccm) Since (0)= 0, therefore

010)-{{15-cento) - 15 == C 2 15 C= 2 Therefore, 15 g 1- 2 Since,

I= dQ dt To d 15 dt 2 1 15 2 2 — X 2 5 =-e 3 Hence, 15 1-e 2 5 I(t)= 3 --e