# Why does the current increase in an LC circuit while the capacitor is being discharged?

From the maths governing an LC circuit (eg. $E_B=\frac{LI^2}{2}$) we can deduce, that the current through the inductor will have a maximum value, when there's no energy stored in the capacitor, or else $E_B = max \iff I=max$. But how to explain it intuitively? Suppose we have a fully charged capacitor and we connect it to a inductor. As the voltage between the capacitor's plates decreases, so should the current flowing through the circuit. Yet, we observe the opposite, as the current increases. I suppose it's due to the EMF induced in the inductor, but shouldn't it merely decrease the rate with which the current is decreasing?

As the voltage between the capacitor's plates decreases, so should the current flowing through the circuit.

I don't follow your reasoning here. Recall that, for an ideal capacitor, we have:

$$i_C = C\frac{dv_C}{dt}$$

In words, the current through the capacitor is proportional to the rate of change of the voltage across, not the instantaneous value of the voltage.

So, for example, if the voltage across the capacitor is sinusoidal

$$v_C = V \sin\omega t$$

the current is

$$i_C = \omega CV \cos \omega t$$

which means (1) that the maximum current (magnitude) occurs when the voltage is zero and (2) that the maximum voltage (magnitude) occurs when the current is zero.

Now, for this simple LC circuit, the voltage across the capacitor is identical to the voltage across the inductor:

$$v_C = v_L$$

thus,

$$i_C = C\frac{dv_L}{dt}$$

For an ideal inductor, we have:

$$v_L = L\frac{di_L}{dt}$$

But, the inductor current is

$$i_L = - i_C$$

thus,

$$i_C = -LC\dfrac{d^2i_C}{dt^2}$$

which means that the current is sinusoidal

$$i_C = A \sin \omega t + B \cos \omega t$$

where

$$\omega = \frac{1}{\sqrt{LC}}$$

Since, in your example, the initial current is zero and the initial voltage is $V$, we have

$$i_C(t) = -\frac{V}{\omega L} \sin \omega t$$

• Good, I didn't want to write all the equation for them. – George Herold Jul 20 '14 at 13:15
• @Alfred Shouldn't$$i_C = \omega C V \cos \omega t$$? – JM97 Aug 6 '16 at 10:58

You should write down all the equations of you LC circuit and that may help. The short answer is that when you close the switch and let current flow out of the capacitor, it can't flow right away because the rapidly changing current sets up an opposing voltage in the inductor.

V = L di/dt

And so the current increases slowly, reaching a maximum (as you say) when the capacitor is fully discharged.

Now you tell me what happens after that.

• After that, I'd say, the current starts decreasing, which induces an opposite current in the inductor, thus, in result, the capacitor gets charged conversely. – neverneve Jul 20 '14 at 15:45