Please help with this problem. The regular physics section said it was too advanced. 2) An inductor L and a capacitor C=1 uF are completely discharged and usedto build up the circuit in figure. The switch is first closed (at t=0) to positionA and kept in position A for enough time to store an amount of energy U0.5 J in the inductor through a resistor R = 10 Q. Then, it is switched toposition B to generate an ac voltage signal at frequency f 1 kHz andmaximum amplitude A, where A LIAB, with IAB current in theinductor when the switch goes from position A to B.How much time does the switch need to be in position B to store a total energyU=0.5 J?%3D%3Di)ii)What is the amplitude A of oscillation?HINT: In your strategy, you do not need to follow the same timing as the switchR=10 ohmt= 0A.C = 1uF100V

For t=0, the switch is connected to position 'A' for a long time. In this position of the switch, the current supplied by the 100 V source will flow through the resistor and inductor. We know that if an inductor is connected to a DC circuit for a long time, it acts as short-circuit and stores the maximum energy of 0.5LImax2 Joule in form of a magnetic field. In this condition, the circuit will look like the figure as shown below: The maximum value of the current in the circuit is Imax=100R=10010=10 A. The energy stored in the inductor is U=0.5 J0.5LImax2=0.5L×102=1L=0.01 Henry.Now, the switch is moved to position 'B'. In this condition, only two opposite energy storage elements (inductor and capacitor) are connected. Inductor possesses the energy of 0.5 J and capacitor has no energy stored in it. We know that an inductor has stored energy, does not change its direction of current when the source is removed but voltage polarity across it reverses. Since the capacitor is initially uncharged, so the initial voltage across it will be zero i.e., v0=0 V and i0=Imax=10 A.Since there is no resistance present in the circuit, no energy will be lost in any form. Thus, the maximum energy stored in the inductor is equal to the maximum energy stored in the capacitor at a later time. From the conservation of energy, 0.5LImax2=0.5qmax2Cqmax2=Cqmax=1×10-6qmax=1 mC. The maximum stored charge on the capacitor will be 1 mC. The natural frequency of the circuit is ω=1LC=10.01×1×10-6=10,000 rad/s. The equation of the charge stored on the capacitor is qt=qmaxcosωt=1×10-3cos10,000t C. The expression of the current is it=dqtdt=ddt1×10-3cos10,000t=-1×10-3×10,000×sin10,000t=-10sin10,000t A. The maximum amplitude of the AC voltage is 'A'=ImaxωL=10×10,000×0.01=1000 V. In the circuit, when the switch moves from position 'A' to position 'B', the total energy stored was 0.5 J and it will remain the same irrespective of time. It may happen that at any particular instant of time, this energy distribution occurs such that the inductor has less energy as compared to that of the capacitor or vice-versa at another instant and or also both may have the same energy stored, keeping the total energy to a value of 0.5 J.