1. In the circuit below R = 2 kQ, L = 250 mH, and C = 10 nF. The initial current Io = -4 A and the initial voltage on the capacitor is 0 V. Find the following: a. The damped form of the solution b. iR(0+) c. ic (0*) dv(0+) PP d. ic + iR dt e. v(t) for t≥0 C Vo L Io R V - 1. a. Overdamped (a = 25,000 rad/s, wo = 20,000 rad/s) b. iR (0+) = 0 A c. ic(0+) = 4A dv(0+) d. = 4 × 108 V/s dt e. v(t) = 13,333(e−10,000t. - e-40,000+)V

The questions states that at t=0 the switch is closed for the circuit. Initial value of the capacitor and inductor are zero (0). We have to find the expression for capacitor voltage (Vc)

For the circuit in Figure 3 the switch is in the left position for several minutes: (a) Find the Initlal voltage, V, on the capacitor just before the switch is flipped (b) Find an expression v(t) that describes the voltage across the 20 N resistor after the switch has been Figure 3 U 09 flipped to the right NOTE: Remember what we said in class: use a Circuit-Specific Equation to get a value you know. Then solve for whatever else the problem asks for +50 µF 380 0 20 2

An LC circuit like the one in the Figure contains an 25.1-mH inductor and a 224-µF capacitor that initially carries a 212-μC charge. The switch is open for t < 0 and then thrown closed at t = 0. At t = 1.97-ms, find the current in the circuit (unit in mA). t C -ww- L

A series circuit consisting of an inductor, a resistor, and a capaci-tor is open. There is an initial charge of 2 coulombs on the capacitor, and 3 amperes of current is present in the circuit at the instant the circuit is closed. A voltage given by E(t) = 20 cos t is applied. In this circuit the voltage drops are numerically equal to the following: across the resistor to 4 times the instantaneous change in the charge, across the capacitor to 10 times the charge, and across the inductor to 2 times the instantaneous change in the current. Find the charge on the capacitor as a function of time. Determine the charge on the capacitor and the current at time t = 10.

2. Find the differential equation to be solved by the following analog computer circuit. Resistors are in the order of MQ, capacitors are in the order of uf. So, start the solution with RC-1 acceptance. is the voltage at point C. (Make the solution in detail. Explain how the stress values at points A, B, D, E and F are found.) 0,5 ww 0,5 ww 12V Hit www me D

Home Work In the figure below, the switch has been open for a significant period of time. The switch is closed at t3D0. Find the current through the resistor at t=0+, and at t=1.25s. Find the energy in the inductor at t=1.25s. 50V 8H 202

PROBLEM 4 Consider this circuit. Find the value of the voltage across the capacitor vc(t) assuming the value of vi(t) = 10u(t), and assume that at t = 0, -1A flows through the inductor and +5V is across the capacitor. (Hint: Use nodal analysis. Add a current source parallel to the inductor and voltage source in series with the capacitor) V1 vi(t) R1 10/3 Ohms L1 5 H C1 0.1 F vc(t)

For the circuit below with switch U1 closing at time t-0 and the initial current in L1 equal zero, find the transient response voltage across the inductor L1: R1 U1 100 V1 20Vde L1 2mH

An RC circuit has an emf given (in volts) by 400 cos 2t, a resistance of 100 ohms, and a capacitance of 102 farad. Initially there is no charge on the capacitor. Find the current in the circuit at time t = O 2.664 A 0.868 A 2.330 A 3.278 A 0.5 s.