# Charging and Discharging of a capacitor in an LC circuit

sophiecentaur
Gold Member
The barrier i find in understanding it is in any case of capacitor discharging, the current is suppoused to decrease,
Why is it "supposed to decrease"? ***You are trying to impose your (inaccurate) intuition. An L and C, together not behave like an R and C. An RC circuit follows an exponential law. An L and C follow a sinusoidal variation in time; the energy, initially stored in the Capacitor does not get dissipated but passes into the Inductor and back again and so on. As the Voltage across the Capacitor approaches Zero, it changes faster and faster and (as it can hardly suddenly change direction or remain at zero) it carries on into the negative region.
I have to point out that an arm waving description of a process like this is really not adequate. There are some things for which a simple verbal description is just not enough - which is why Science can be very difficult. That's something people just need to accept. This link describes what happens and (with Maths, of course) derives an equation that describes the waveforms. Notice, the whole mathematical description takes up very little space because it is concise and cannot be misinterpreted - unlike so many other attempts at descriptions..

*** Explanation of why this idea is wrong: There is a constant amount of energy in the circuit, sloshing back and forth between the L and C.
Energy in a Capacitor is CV2/2
Energy in an Inductor is LI2/2
As Vc reduces, the energy drops faster and faster because of the squared term. This means that the Current IL has to be Increasing to a Maximum as VC passes zero.
With a CR circuit, the current drops as the VC drops; the Energy decays.

Nikhil Rajagopalan