Here is a LC circuit with a DC supply.

When the switch is closed at t=0 capacitor behaves as a short circuit while the inductor behaves as an open circuit as the voltage across the inductor immediately jumps to battery voltage.

Now we know that the current in inductor increases while in a capacitor current decreases with respect to time.

I wonder then how the current will overall behave in that circuit as when capacitor is short circuit and inductor is open circuit at t=0 then when current in capacitor decreases then at the same time in inductor it is expected to increase. So what will be the overall shape of the current with respect to time. How will its graph look like and whats the theoratical explaination for that without going into the differential equations ?

enter image description here

  • $\begingroup$ It will behave like a resistor. $\endgroup$ Apr 30, 2018 at 9:16

4 Answers 4


Assuming that there is no resistance in the circuit the current in the circuit will be given by the equation $I(t) = \mathcal E \sqrt {\frac CL} \sin \omega_0 t$ where $\omega_0 = \sqrt\frac {1}{LC}$.

enter image description here

At all times in your circuit the total voltage must add up to zero.

$$\mathcal E + v_{\rm capacitor} + v_{\rm inductor} = 0 \left [ \Rightarrow \mathcal E + \frac Q C + L \frac {dI}{dt} =0 \right]$$

with the differential equation which you do not wish to be used in brackets.

To try to explain what happens I have drawn a series of time sequenced diagrams with $T = 2 \pi \sqrt{LC}$.

enter image description here

Diagram 1
The instant the switch is closed the current is zero and the voltage across the inductor opposes the applied voltage from the cell because although the current is zero there is a rate of change of current.
There is no charge on the capacitor and so the voltage across the capacitor is zero.

Diagram 2
There is now a current, $i$, in the circuit, the capacitor is charging which I have shown with the plus and minus signs between the "plates" of the capacitor and there is a voltage $v_{\rm C}$ across the capacitor.
However the rate of change of current has decreased so there is now a smaller voltage across the inductor v_{\rm L}$.
As there is a current flowing through the inductor it has energy stored in its magnetic field (shown in red) and there is also energy stored in the electric field produced by the capacitor.
All that energy has come from the cell.

Diagram 3
The current in the circuit reaches a maximum value $I$ and the voltage across the capacitor is now equal in magnitude to the voltage across the cell $\mathcal E$.
At this time the instantaneous rate of change of current is zero and so there is no voltage across the inductor and again the total voltage in the circuit is zero.
Both the inductor and the capacitor have more energy stored in their fields.

Diagram 4
This diagram may surprise you because the voltage across the capacitor is now greater than the voltage across the cell.
This happens because the current which is shown to flow in Diagram 3 cannot stop flowing instantaneously and so the capacitor continues to be charged but with a reduced current $i$ in the circuit.
Note that because the current is now decreasing the voltage across the inductor has reversed polarity and again the total voltage in the circuit is zero.
The inductor has given some of its stored energy but the energy stored in the capacitor is still increasing.

Diagram 5
Eventually the voltage across the capacitor reaches twice the voltage of the cell and current ceases to flow.
The charge on the capacitor is a maximum and so is the energy store within it.
Although there is no current flowing through the inductor there is still an instantaneous rate of change of current which produces a voltage across the inductor equal to the voltage of the cell $\mathcal E$ and so the total voltage in the circuit is still zero.
The inductor has no magnetic field associated with it and so is storing no energy.

Hopefully you will now be able to follow the subsequent diagrams and realise that the diagram after Diagram 8 is Diagram 1 as the whole sequence is repeated (for ever).

Overall in one cycle there is no net energy transfer between the cell and the rest of the circuit.

The voltage across the capacitor is $v_{\rm C} = (-) \mathcal E(1-\cos \omega_0 t)$ and the voltage across the inductor is $v_{\rm L} = (-) \mathcal E\, \cos \omega_0 t$.

If there had been resistance in the circuit then the current would tend to zero as the time tended to infinity with the exact form of the variation of current with time depending on the values of the capacitance, inductance and resistance in the circuit.

You will observe that there is a greater deal of similarity between your circuit and the circuit which was dealt with in this question where the capacitor had an initial charge and there was no cell in the circuit.

Update as a result of a comment from @Alex

To illustrate what happens as the resistance changes I have used a copy of MIT's Circuit Sandbox to produce some current against time graphs for a series LCR circuit with a step input and different values of resistance.
(Note that this Circuit Sandbox did not work for me using Firefox so I used Edge instead).

enter image description here

Critical damping for this circuit occurs when the resistance is $2\Omega$ and the system reaches its steady state (current = zero) without overshoot in the shortest time.


Here are the voltage and current graphs when the resistance is $0.2\Omega$.

enter image description here

  • $\begingroup$ Pedantic suggestion: $\omega$ is usually a variable while $\omega_0$ is a parameter, e.g., $\omega_0 = \frac{1}{\sqrt{LC}}$ $\endgroup$ May 1, 2018 at 0:55
  • $\begingroup$ @AlfredCentauri Many thanks for your suggestion which I have acted upon. I would have had no objection to you making such a change. :-) $\endgroup$
    – Farcher
    May 1, 2018 at 4:57
  • $\begingroup$ really thanks alot @Farcher . great answer. Now what happens if we add a resistor in series with the inductor and capacitor ? how the resistor will change the shape of the current ? after solving differential equation we get three conditions, under damped critically damped and over damped depending upon the value of resistor. will you please further tell me that how a resistor or a value of resistor can reshape this sine wave to something else ? $\endgroup$
    – Alex
    May 6, 2018 at 13:43
  • $\begingroup$ underdamped is a graph in which the sine wave's amplitude becomes less and less with respect to time and eventually dies out. that makes sense because in each cycle of current the resistor consumes some of the energy but i do not understand what happens in case of critically(R square = 4L/C) and ovderdamped(R>4L/C) case as in those cases no sine wave is formed. $\endgroup$
    – Alex
    May 6, 2018 at 13:51
  • 1
    $\begingroup$ @Alex I have added an update to my answer as a result of your recent comments. $\endgroup$
    – Farcher
    May 8, 2018 at 8:07

Since you want to avoid differential equations, I will instead consider the so-called Phasor domain, which is actually nothing but the Fourier transform of the original signals.

In phasor domain, we will be basically considering complex values: Complex voltages, complex resistances (which are denoted by $Z$ and called impedance): This is simply the mathematical convenience, and we will always get back the physical real values at the end.

For the time being, forget about the active parts of the circuit and focus on the passive elements, that is forget the switch and the DC source. Now let us check the behavior of remaining elements if we were to give the complex voltage $V(t)=e^{i\omega t}$:

  • For capacitor, we have the equation $C\frac{dV}{dt}=I$, hence we get $$I=i\omega C e^{i\omega t}$$ meaning the impedance is $$Z=\frac{V}{I}=\frac{1}{i\omega C}$$

  • For inductor, we have the equation $L\frac{dI}{dt}=V$, hence we get $$ i\omega L I=e^{i\omega t}$$ meaning the impedance is $$Z=\frac{V}{I}=i\omega L$$

Since the capacitor and the inductor are in series, the total impedance is their sum (just like the case for resistances), hence the impedance of the passive part of the circuit is $$Z=i\omega L+\frac{1}{i\omega C}=\frac{1-\omega^2 LC}{i\omega C}$$

This means that the voltage across the capacitor for a sinusoidal AC signal would be $$V_\text{capacitor}=V_\text{input}\frac{Z_\text{capacitor}}{Z_\text{total}}=V_\text{input}\frac{\frac{1}{i\omega C}}{\frac{1-\omega^2 LC}{i\omega C}}$$ hence the ratio of input voltage to output voltage is $$\frac{V_\text{output}}{V_\text{input}}=\frac{1}{1-\omega^2 LC}$$ where we took the output voltage to be the voltage of the capacitor.

One can immediately notice that the ratio would go to infinity if we were to excite the system with an AC source of the frequency $$\omega=\frac{1}{\sqrt{LC}}$$ This is called the resonance frequency. At this frequency, the system builds up each incoming power cycle and increases its output indefinitely. In real world, the existence of resistors in the systems and breakdown of equipment prevents this divergence to infinity.

If we now go back to our question, we now simply need to know how we can write our excitation in terms of pure sinusoidal waves of constant frequencies. If we know this, then we can examine the output of the system because we are dealing with a linear system: Effect of sum of sinusoidal signals is equal to sum of effects of individual signals.

Our input signal is simply step function, or so called Heaviside-theta function, which is zero before $t=0$ and constant after $t=0$ where we choose $t=0$ as the time the switch is turned on. One now needs to decompose Heaviside-theta function into pure sinusoidals: I'll skip the calculation here; it is basically Fourier transformation of this function. The answer is $$V_\text{input}(f)=E\left(\frac{i}{\sqrt{2\pi}f}+\sqrt{\frac{\pi}{2}}\delta(f)\right)$$ meaning that we are creating an input signal which decomposes of $V_\text{input}(f)$ for all frequencies $f$. We can now find the output signal for individual $f$ values by using the ratio above:

$$V_\text{output}(f)=\frac{E}{1-4 \pi^2 f^2 LC}\left(\frac{i}{\sqrt{2\pi}f}+\sqrt{\frac{\pi}{2}}\delta(f)\right)$$ where we used the fact that natural frequency $\omega$ is $2\pi f$.

We can now get our result taking the inverse Fourier transform, effectively summing over the contribution of all pure sinusoidals with different $\omega$ values: Again, this is due to the fact that the system is linear. I will skip the calculations, but you can easily use Mathematica to do this for you as can be seen at the attached picture. The result is $$V_\text{output}(t)=\left\{\begin{aligned}E\left(1-\frac{1}{2} \cos \left(\frac{t}{2 \pi\sqrt{C L}}\right)\right)\quad t>0\\0\quad t<0\end{aligned}\right.$$ which is mathematically true but with the wrong boundary condition. I comment about this further down. Nonetheless, it shows us the relevant behavior.

You can see in the picture that the voltage simply keeps oscillating. The graphs reflect three facts:

  • Energy keeps shifting from being stored in capacitor and being stored in the inductor.
  • As there is no resistance, there is no damping effect, so the oscillations do not die!
  • The output form is only dependent on the characteristics of the passive circuit; namely the inductance of inductor and the capacitance of capacitor. The input voltage only scales the output form. This is a generic concept: In engineering, this characteristic is called the impulse response whereas it is known to be the Green's function in physics, though impulse response is actually a specific case of the more general Green's functions.

EDITS (Answers to comments):

  • There was a typo earlier where I wrote $\sqrt[4]{LC}$ instead of $\sqrt{LC}$. I corrected it above, everything is now dimensionally consistent.
  • The result respects the boundary value at $t=0$; however, this boundary value is ill-defined. We know that the input voltage is $E$ for $t>0$ and $0$ for $t<0$; however, we do not know what it is exactly at $t=0$ since turning on a switch is a discontinuous jump. The mathematically usual choice is to take it to be the half value, hence $E/2$ at $t=0$, this is indeed in the definition of Heaviside theta in Mathematica as well, and this is why we get $E/2$ for $t=0$ in our output! I was completely incorrect here, I realized it thanks to @Alfred
  • If one wants to define the behavior of the switch differently, one can repeat the calculations accordingly (actually using the differential equations would be a lot easier). The output has nothing to do with switch but with the boundary condition chose. However, with the chosen boundary condition, in the end, the static response $E$ in the output would remain the same whereas the jump response $-\frac{E}{2}\cos(\frac{t}{2\pi\sqrt{L C}})$ would scale to $-E\cos(\frac{t}{2\pi\sqrt{L C}})$, hence we would have $$ V_\text{output}=V_\text{capacitor}=E\left(1-\cos(\frac{t}{2\pi\sqrt{L C}})\right) $$ hence $$I=C\frac{dV_\text{capacitor}}{dt}=\frac{E}{2\pi}\sqrt{\frac{C}{L}}\sin\left(\frac{t}{2\pi\sqrt{L C}}\right)$$

enter image description here

Further Comments

IMO, the simplest approach to the question is to use the differential equations. However, since OP asked us to avoid this, I tried to explain it by using the facts that electrical systems are linear systems and that eingenfunctions of linear systems are exponential functions; that is pure sinusoidals. I believe this is the conceptually closest approximation to using differential equations, though not equally rigorous. I will address some issues in this part regarding the approach above.

First of all, the inverse fourier transform of our output, $$V_\text{output}(f)=\frac{E}{1-4 \pi^2 f^2 LC}\left(\frac{i}{\sqrt{2\pi}f}+\sqrt{\frac{\pi}{2}}\delta(f)\right)$$ is actually $$V_\text{output}=\text{sgn}(t) \sin ^2\left(\frac{t}{4 \pi \sqrt{C L}}\right)+\frac{1}{2}$$ which has the behavior as the plot below: enter image description here

There are a few commands in order:

  • The plot naively looks non-causal, as there is nonzero signal even before the switch is turned on. However that signal is actually the resonant signal we mentioned above. As we discussed there, the effective resistance (impedance) for this signal is zero for the circuit, so that signal can never disappear in a generic solution: It has nothing to do with the switch being turned on or off! The effect of switch is purely shifting the output, in a continuous manner, which is causal.
  • In differential equations language, the resonant signal is the homogeneous solution which is present in the system unless killed off by a boundary/initial condition. The effect of the switch is the particular solution, which is simply the shift of the output signal in a causal manner.
  • In the EDITS part above, I got confused and incorrectly stated that the reason voltage is $1/2$ at $t=0$ is because of the convention of Heaviside-theta. This is not true. I believe it has to do with the fact that signals are assumed to converge to zero at minus infinity in Fourier Transform, hence the mean value of the sine should be 0 for $t<0$, forcing signal to be $1/2$ at $t=0$ to remain continuous with the $t>0$ part. If we instead impose the condition that it should be $0$ at $t=0$, we get the correct result for $t>0$, and irrelevant/unphysical signal for $t<0$.
  • A way better method to do this calculation is to use Laplace domain instead of Fourier domain, where we consider Laplace transformation from $0$ to $\inf$. This would allow us to implement boundary condition directly in the Laplace domain and we would not get the unphysical part for $t<0$. One can also do that; I avoided it as I am not aiming for a rigorous or efficient method but a conceptually easy one, and Phasor space is easier to understand IMO.

To justify the explanations above, let us set up the differential equations: \begin{align} I_c=&I_l \Rightarrow\\ \frac{d I_c}{d t}=&\frac{d I_l}{d t}+\frac{a}{L} \Rightarrow\\ C\frac{d^2 V_c}{d t^2}=&\frac{V_l}{L}+\frac{a}{L} \Rightarrow\\ LC\frac{d^2 V_c}{d t^2}=&V_i-V_c+a \Rightarrow\\ \left(LC\frac{d^2}{d t^2}+1\right)V_c=&V_i+a \end{align} where $V_i$ is the input voltage.

The homogeneous solution is that of $$\left(LC\frac{d^2}{d t^2}+1\right)V_c=0$$ which is simply $$V_c^\text{homogeneous}=b\cos\left(\frac{t+c}{\sqrt{LC}}\right)$$ This is exactly the same signal we observe for $t<0$ in the plot above.

The correct procedure is actually to use Green's function with the input signal with appropriate boundary conditions: This is what EE engineers would call impulse response. Laplace domain is ideally suited for this job for linear systems. My explanation above on the other hand corresponds to finding particular solution for $V_i=E$ instead, combining it with homogeneous solution, imposing $V_E(t=0)=0$ and discarding $t<0$ part.

  • $\begingroup$ The solution you give for $V_{output}(t)$ cannot be correct if this is the voltage across the capacitor. As the OP, the initial condition for the capacitor voltage is $v_C(0+) = 0$. $\endgroup$ Apr 30, 2018 at 14:42
  • $\begingroup$ @Alfred, I edited the post to explain it. $\endgroup$ Apr 30, 2018 at 17:58
  • $\begingroup$ This answer is conceptually flawed. The concept of impedance (the ratio of the voltage and current phasors) assumes sinusoidal excitation. In the context of signals and systems, it is well known that the response to a general excitation is given by the convolution of the input $x(t)$ with the impulse response of the system $h(t)$, i.e., $y(t) = x(t) * h(t)$. For a physical system, $h(t)$ must be causal, i.e., $h(t) = 0$ for $t\lt 0$. By the convolution theorem, $Y(f) = X(f)H(f)$. But, in the above, the $H(f)$ given is non-causal. The Laplace domain is far more appropriate here. $\endgroup$ May 3, 2018 at 0:18
  • $\begingroup$ @Alfred, I totally agree that Laplace transform is conceptually much more appropriate: Not actually because of the causality, but because of the fact that we have an initial condition. With Laplace transformation, we would be able to impose initial condition directly in the transformed domain. Yet, IMO, Fourier transform is conceptually easier to explain, and is sufficient (again IMO) in the current case. $\endgroup$ May 3, 2018 at 1:24
  • $\begingroup$ @Alfred Output signal is naively non-causal, as there is nonzero part for t<0.This follows from that with the Inverse Fourier transform we get the most generic signal which includes the resonant signal for $t<0$: This signal is homogeneous solution of the corresponding DE and is always there unless boundary conditions kill it!So yes, naively there is a nonzero signal for t<0 but this is not the effect of the switch(so no non-causal relation),i.e. it is not particular solution of DE.We can kill it with boundary conditions:In the case above, I simply discarded it by focusing on $t>0$ part only. $\endgroup$ May 3, 2018 at 1:28

Now we know that the current in inductor increases while in a capacitor current decreases with respect to time.

This isn't generally true and it certainly can't be true here since the inductor and capacitor are series connected and so have identical current through.

If you replace the inductor with a resistor (to form a series RC circuit), then you can conclude that the capacitor current decreases with time after the switch closes (assuming zero initial condition).

Similarly, if you replace the capacitor with a resistor (to form a series RL circuit), then you can conclude that the inductor current increases with time after the switch closes (assuming zero initial condition).

But it's an error to apply these conclusions to a series LC circuit. You can however, get an idea of the graph of the current by reasoning carefully.

First, keep in mind that, at all times after the switch is closed, the instantaneous sum of the voltage across the inductor and capacitor must equal the battery voltage $E$.

Second, recognize that the series current must initially be increasing due to the voltage $E$ across the inductor (since the initial voltage across the capacitor is zero). Then, reason as follows:

As the voltage across the capacitor increases (due to the charging current), the voltage across the inductor must decrease (to keep the sum equal to $E$) and so the rate of increase of the series current decreases, i.e., the current is still increasing but not increasing as quickly.

At some time, the voltage across the capacitor reaches $E$ and then the voltage across the inductor is zero and so, the series current has stopped changing - the series current has reached its maximum value and the capacitor is charging at its maximum rate.

The capacitor continues to charge (due to the current) and so the voltage across the capacitor now exceeds $E$ which requires that the inductor voltage becomes negative (to keep the sum equal to $E$). Because the inductor voltage is negative, the series current must now decrease.

The capacitor continues to charge but not as quickly since the current is decreasing and, eventually, the current decreases to zero, the capacitor has reached its maximum voltage and the inductor voltage is now its most negative (to keep the sum equal to $E$). This means that the current is decreasing most rapidly and, in fact, the current decreases through zero to become negative.

The negative current begins the discharge of the capacitor, the capacitor voltage begins to decrease and so the voltage across the inductor becomes less negative. As the capacitor voltage decreases to $E$, the inductor voltage once again is zero and the series current is no longer changing - the series current has reached its most negative value and the capacitor is discharging at its maximum rate.

By now, I suppose, you can easily trace the remainder of the cycle. The capacitor continues to discharge until its voltage is zero, the series current is zero, and the inductor voltage is $E$ at which point the circuit has returned to the state at $t=0+$ and the cycle repeats.


The current through a capacitor is determined (limited) by the external components (and, indirectly, by the voltage on the capacitor) and will be increasing, decreasing or stay at zero, depending on what's going on in the rest of a circuit.

For instance, if a capacitor is connected to a battery, its current would be determined by the voltage difference on the battery and the capacitor and the internal resistance of the battery. As the capacitor charges, the voltage difference decreases and, therefore, the current, (Vbat-Vcap)/r, decreases.

In your circuit, the capacitor is connected in series with the inductor, which, as you've pointed out, will initially behave as an open circuit, so the initial current through the capacitor, determined by the inductor, will be zero.

As the current through the inductor (and the capacitor) increases, initially as di/dt=Vbat/L, the capacitor will start charging. This will reduce the voltage drop on the inductor and, correspondingly, the rate of the current growth, di/dt=(Vbat-Vcap)/L.

So the current in the circuit will start at zero and then grow with a diminishing acceleration. This is actually a beginning of a sine wave with a frequency of 1/2π√(LC).


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