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Suppose we have a fully charged capacitor and we connect it to an inductor.

Capacitor will discharge. Inductor will get the current from the capacitor. Wave form of the current shows that the current is oscillating.

It means that the current rises intially. But what makes me confuse is that when capacitor will discharge its voltage will become lesser and lesser and it causes decrease in current in the circuit. While at the same time the inductor voltage(opposite to capacitor voltage) also becomes lesser and lesser and it cause more current to flow in the circuit.

So less capacitor voltage makes the current less and less inductor voltage cause more current at the same time. Then what makes sure that current will initially rise and will then fall during the first half cycle of the oscillation ?

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    \$\begingroup\$ Consider this: What happens when the capacitor voltage goes negative? \$\endgroup\$ – Hearth May 10 '18 at 12:11
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The capacitor is initially charged. When the circuit is closed, the capacitor voltage is applied to across the inductor. The nature of a inductor means that the current increases proportional to the applied voltage. With the capacitor voltage applied across the inductor, the inductor current starts going up.

As the inductor current goes up, that current is draining the capacitor, which reduces the voltage across the inductor. This means the current goes up more slowly over time. When the capacitor voltage reaches 0, the current stops going up altogether.

Now there is current but no voltage. That current continues to decrease the capacitor voltage, now making it negative. That negative voltage across the inductor now reduces the current. The current reduces more quickly as the negative voltage builds up on the capacitor. Eventually the current is reduced to 0. At that point, the voltage on the capacitor is the negative of what it started at.

Now we have the same conditions as we started with, except that the capacitor voltage is negative instead of positive. The same thing will happen as happened up until now with all the signs flipped. The result of this is the current is again 0, but the capacitor voltage has flipped again, this time resulting in what it started with when the switch was first closed.

With ideal components, this process repeats indefinitely. The finite energy is constantly sloshed back and forth between being voltage on the capacitor and current thru the inductor. Both the voltage and current have sinusoidal profiles. This is referred to as resonance.

This is also called a tank circuit. Tank circuits are often used as notch filters, since the frequency of oscillation is specific to the particular capacitance and inductance values. It has the interesting property that the impedance across the parallel combination of capacitor and inductor is 0 at 0 and infinite frequency, and infinite right at the resonant frequency. This drastic impedance peak at the resonant frequency is often exploited to either filter in or filter out that frequency.

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  • \$\begingroup\$ thanks for the answer @olin i still do not understand yet that when the potential difference across the capacitor is getting low then why the current rises with time as capacitor is the source which is causing the charge to flow. Although the inductor voltage also offers less and less opposition to current but then current increases ? \$\endgroup\$ – Alex May 15 '18 at 11:23
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what makes sure that current will initially rise and will then fall during the first half cycle of the oscillation

Once the switch closes, the voltage across both components falls (initially more gradually) towards zero volts and when zero volts is reached all that you have for that moment in time is current flow. But that current flow in the inductor is stored energy and it is the same level of energy as stored in the capacitor prior to the switch closing and this further accelerates the decline in voltage from zero to negative values.

As this current gradually reduces to zero, a negative voltage peak is reached where all the energy returns to the capacitor in the form of a negative voltage. The same energy as stored prior to the switch closing.

Then we are back to the start but with the capacitor having a negative voltage of the same magnitude as originally set on it prior to switch closure.

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  • \$\begingroup\$ when "the voltage across both components falls " then inItially decrease in one component voltage cause decrease in current while decrease in others component cause increase in current so what determines that the current will rise initially ? @andy aka \$\endgroup\$ – Alex May 11 '18 at 9:24
  • \$\begingroup\$ @Alex the problem with this "decrease in one component voltage cause decrease in current" statement is that it is wrong. The inductor starts to take more current from the inductor as time passes and this depletes the capacitor of energy and hence the voltage drops and it keeps dropping (whilst current rises) until the shared voltage is zero. At this point the current is maximum. You are trying to analyse it like ohms law but that is not how inductors and capacitors work. For a capacitor, the bigger the rate of change in voltage the higher the current taken from it..... \$\endgroup\$ – Andy aka May 11 '18 at 9:52
  • \$\begingroup\$ Don't think in terms of I = V/R because this does not work with inductors and capacitors when analysing voltage and current changes with time. It's tricky to put across what happens if you are unfamiliar with V = L.di/dt and I = C.dv/dt (equations for an inductor and capacitor). \$\endgroup\$ – Andy aka May 11 '18 at 9:54
  • \$\begingroup\$ i am familiar with V = L.di/dt and I = C.dv/dt but i still do not understand yet that when the potential difference across the capacitor is getting low then why the current rises with time as capacitor is the source which is causing the charge to flow. Although the inductor voltage also offers less and less opposition to current but then what makes the capacitor win over the inductor ? \$\endgroup\$ – Alex May 13 '18 at 16:03

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