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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 ?

enter image description here

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  • $\begingroup$ Have you looked at this answer to a related question? physics.stackexchange.com/a/370910/104696 $\endgroup$ – Farcher May 15 '18 at 13:21
  • $\begingroup$ yes i looked at it @Farcher . when we connect a battery with only an inductor then voltage source is assumed to be constant and the inductor determines the amount of the current that flows through the circuit. But when we connect a charged capacitor with an inductor then in this case the voltage source (the capacitor) is not supplying the constant voltage. source voltage is decreasing and it should decrease the current while at the same time inductor voltage is becoming lower and it should increase the current. the link you provided doesnt answer this question i think. $\endgroup$ – Alex May 15 '18 at 15:23
  • $\begingroup$ please tell me the answer if you know . I shall be grateful to you much. $\endgroup$ – Alex May 15 '18 at 15:24
  • $\begingroup$ What is wrong with accepting the @HalHollis answer? $\endgroup$ – Farcher May 15 '18 at 15:27
  • $\begingroup$ it just tells that because voltage across the inductor is positive so the current might be increasing but it doesnt tell that how current is increasing when voltage source that causing the current is decreasing. $\endgroup$ – Alex May 15 '18 at 15:41
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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.

It's true that the initially clock-wise current that discharges the capacitor is increasing at first and the reason for this is that the voltage across the inductor is initially positive (the top terminal of the inductor is more positive than the bottom terminal).

However, and in contrast to your statement quoted above, the current will continue to increase even as the voltage across the capacitor (and the inductor in parallel) decreases towards zero. As long as the voltage is positive, the current must be increasing. Why? It follows from the fundamental inductor equation:

$$v_L(t) = L \frac{di_L}{dt}$$

Since $L$ is a positive constant, a positive inductor voltage requires a positive rate of change of inductor current - positive voltage across implies increasing current through.

When the capacitor is fully discharged, the voltage across is zero and the current is maximum (the rate of change of current is zero at the maximum) but now the current begins to charge the capacitor 'in the opposite direction' and the voltage across becomes negative which implies a negative rate of change of inductor current.

In short, the current doesn't begin to decrease until the voltage across passes through zero and becomes negative.

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It may be informative to apply Kirchoff's law to the circuit. As there is no external voltage source, $Q/C + L \ dI/dt = 0$ where $Q/C$ is the voltage on the capacitor and $L \ d I/dt$ is the voltage on the inductor. So as they sum to zero, the voltages on the capacitor and inductor must always be of the same magnitude and opposite in sign.

Now, since the current $I = dQ/dt$, I can rewrite this equation as:

$\frac{d^2 Q}{d t^2} = - \frac{1}{L C} Q$.

You might recognise this equation as being that of simple harmonic motion. We can then simply write down the solution as $Q(t) = Q_0 \cos \omega t$, and $I(t) = - Q_0 \omega\sin \omega t$, where the frequency of oscillation is given by $\omega^2 = 1/L C$. From this you can immediately see that the capacitor voltage (which is proportional to $Q(t)$) immediately starts to drop, while the current starts to rise from zero. Eventually when the capacitor is empty the current will instead start to reduce with time, and the capacitor will charge up in the opposite polarity. These processes continue to repeat, giving the oscillations in charge and current.

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