An LC circuit like the one in the figure below contains an 65.0 mH inductor and a 16.0 µF capacitor that initially carries a 185 µC charge. The switch is open for t < 0 and is then thrown closed at t = 0. a)Find the frequency (in hertz) of the resulting oscillations. Hz (b)At t = 1.00 ms, find the charge on the capacitor. µC (c)At t = 1.00 ms, find the current in the circuit. mA (d)What If? What are the first three times (in ms), after t = 0,when thecapacitor is fully charged again? smallest value ms mslargest value ms This image illustrates an LC (inductor-capacitor) circuit. It consists of the following components:1. **Capacitor (C):** On the left side of the circuit, represented by parallel lines, the capacitor holds an electric charge \( Q_{\text{max}} \). The positive and negative terminals are indicated, showing that it is charged.2. **Inductor (L):** On the right side, represented by a coil symbol, the inductor stores energy in a magnetic field when current flows through it.3. **Switch (S):** Located at the bottom, this switch can open or close the circuit.The circuit diagram shows the DC current path when the switch (S) is closed. Initially, when the switch is closed, the capacitor will start discharging through the inductor, which can lead to oscillations of energy between the electric field of the capacitor and the magnetic field of the inductor, characterizing an LC oscillating circuit.

LC Oscillator LC Oscillator is a circuit in which two energy storing elements are connected in a circuit which is the inductor and the capacitor. The circuit energy oscillates between these two elements.Answer: Given: inductance = 65 mH capacitance = 16uF initial charge = 185 uC The switch is open for t < 0 and is then thrown closed at t = 0. Part a) frequency of resulting oscillation: w = 1/ (LC)0.5 w = 1/ (65 * 10-3 * 16 * 10-6) w = 990.09 rad/sec frequency = w/2*pi = 990.09 / 6.28 = 157.65 Hz Part b) At t = 1.00 ms, find the charge on the capacitor. q = qo cos(wt) qo = maximum charge q = 185 * 10-6 cos (990.09 * 1 * 10-3) C q = 101.5 uC Part c) current i is the derivative of charge q I = dq/dt I = -w * qo * sin(wt) i = - 990.09 * 185 * 10-6 sin(990.09 * 1 * 10-3) A i = -0.153 A Part d) for the capacitor to be fully charged, q = qo qo = qo * cos(wt) wt = cos-1(1) = n * pi [here n is an integer] for n = 1, t = 3.14 / w = 3.14 / 990.09 = 3.17 ms --> smallest value for n = 2, t = 3.14 / w = 2 * 3.14 / 990.09 = 6.34 ms for n = 3, t = 3.14 / w = 3 * 3.14 / 990.09 = 9.51 ms --> largest value