# How does current increases even if the capacitor discharges in LC oscillation?

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Consider an LC oscillator, where at time t=0 the capacitor is fully charged with charge Q. There are two equivalent explanations to your question: one physical and one mathematical.

The physical explanation is that any inductor is opposing the flow of the current through it. At the first moment (all the charge on capacitor) when the current should have been maximum, the coil is generating an auto-induced voltage on it equal and opposite to that of the capacitor (VL =-VC) . The result is a total null voltage on the inductor, which gives a total current equal to zero. With time, this auto-induced voltage on the coil is becoming smaller and smaller, and hence the total voltage (VL + VC) is becoming higher and higher. This is the physical reason why the current increases even if the capacitor discharges.

The mathematical explanation is a bit more arid and follows.

Faraday law states that the auto-induced voltage on an inductor is

`V_L = -L*(dI)/dt`

where the sign minus shows that is opposite to the external changes.

The voltage on a capacitor is (see capacitance definition)

`V_C =Q/C = (int(I*dt))/C`

By making `V_L =V_C` we obtain the differential equation

`L*(dI)/dt + (int(I*dt))/C =0`

and by differentiating once again and rearranging

`(d^2I)/(dt^2) + 1/(L*C)* I =0` where I=I(t) is a function of time.

The solution of the above equation is well known in mathematics

`I(t) =I_0*sin(omega_0*t)`

where `omega_0 =1/sqrt(LC) ` is the resonance frequency and I0 is the maximum value of the current in the circuit.

If at t=0, (capacitor fully charged) it can be observed that the current in the circuit is zero and is increasing with time t>0 (up to a maximum value when all the capacitor charge has been depleted), even if the capacitor discharges.