2. The voltage drop across a capacitor with capacitance C is given by q(t)/C, where q is the charge on the capacitor. Hence, for the RC series circuit, Kirchhop's second law gives Ri + a = E(t) But current i and charge q are related byi= dq/dt. So, the above equation becomes dq 1 R dt 9 = E(t) Use this differential equation to solve the following: (a) A 100 volt electromotive force is applied an RC series circuit in which the resistance is 200 ohms and the capacitance is 10-+ farad. Find the charge q(t) on the capacitor if q(0) = 0. Find the current i(f). Ans(t) = r (1 – e-50*); i(t) = ļe¬50t (b) Let R = 10 ohms and C = 0.1 farad. Let E(t) be exponentially decaying, say, E(E) = 30e-3t volts. Assuming q(0) = 0, find and graph q(t). At what time does q(t) reach a maximum? What is that maximum charge? Ans: q(t) = 1.5(e-- e-3); q'(f) = 0 gives tm= 0.549; qm= 0.577 coulomb %3D

An open circuit is composed of a capacitor with capacitance C and an inductorwith inductance L. The capacitor is charged to a voltage V and at time t = 0 the circuit isclosed. Find the current as a function of time, I(t).

1. The capacitor charges toward the source voltage through the resistor. The flash turns on when the capacitor voltage reaches 8 V. If C = 10 μF, find R such that the flash will turn on once per second.   2. Find the differential equation for the current in the circuit shown below.

3. In an LC circuit, when the capacitor is maximum, the inductor energy is ........

! Required information Problem 06.001 - DEPENDENT MULTI-PART PROBLEM - ASSIGN ALL PARTS The voltage across a 10-F capacitor is 2te-3t V. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. Problem 06.001.a - Current through a capacitor Find the current through the capacitor. The current through the capacitor is -3t A. |(1-3 t)e¯

A voltage v(t) = 100 cos(60t + 20°) V is applied to a parallel combination of a 40-k resistor and a 50-μF capacitor. Find the steady-state currents through the resistor and the capacitor.