# how does current go to infinity in an ideal LC circuit at resonance?

Consider an ideal LC series circuit, excited by a sinusoidal source at resonant frequency. Assume zero initial conditions, for capacitor voltage and inductor current.

simulate this circuit – Schematic created using CircuitLab

1) Can anyone explain how does the current amplitude go to infinity, at resonance?

I understand the impedance goes to zero,which explains the steady state current as infinite. But can you explain the same by simple circuit analysis?I mean the voltage across inductor is Vsinωt - Vc = L di/dt, so based on derivative being positive, the current increases.But what makes the current go to infinite? After all for a simple L only circuit, the voltage across inductor is Vsinωt = L di/dt and it is not getting infinite

2)Now at resonance what happens internally, from an energy perspective?Is there any transfer of energy between inductor and capacitor OR Source is throughout supplying energy to inductor and capacitor?

3)Will the nature of current differ if we have initial conditions?

• You recieved some nice answers down below. One thing that might help you is to solve for the voltage at the node between the L and C. Commented Jul 27, 2014 at 14:26

Can anyone explain how does the current amplitude go to infinity, at resonance?

The brief answer is that the AC steady state current isn't infinite but, rather, the circuit has no AC steady state solution.

Recall that one of the assumptions justifying AC (phasor) analysis is that the circuit is in AC steady state, i.e., that all transients have decayed.

For the circuit given, the time domain solution for the current is proportional to

$$i(t) \propto t \cos\left( \frac{t}{\sqrt{LC}}\right),\, t \ge 0$$

The amplitude of the current starts at zero and grows linearly with time once the switch is closed but for any value of time $t$, the current is finite, i.e., the current is never infinite.

Note that this solution has no sinusoidal steady state - the amplitude does not approach a constant as $t \rightarrow \infty$ so this solution has no phasor representation and, thus, we should not be surprised that applying phasor analysis to this problem produces an undefined division by zero result.

Given the solution for the current, one can solve for the voltages across the inductor and capacitor as well as the energy stored in each as a function of time.

Different initial conditions will have different initial energies but will not affect the main result that the amplitude of the current will grow without bound once the switch is closed.

Can you please list the steps that lead to the derivation of the expression i(t)∝ tcos(t/√LC),t≥0 ?

By KVL, we have

$$v_S = v_L + v_C = L\frac{di}{dt} + \frac{1}{C}\int_0^ti(\tau)d\tau$$

(we assume zero initial voltage across the capacitor).

Differentiating both sides with respect to time and dividing through by $L$ yields

$$\frac{d^2i}{dt^2} + \frac{1}{LC}i = \frac{1}{L}\frac{dv_S}{dt}$$

Assuming $v_S = V\cos\omega_0 t$ yields the following non-homogeneous 2nd order ODE:

$$\frac{d^2i}{dt^2} + \frac{1}{LC}i = -\frac{\omega_0 V}{L}\sin \omega_0 t$$

where

$$\omega_0 = \frac{1}{\sqrt{LC}}$$

Assume a solution of the form

$$i(t) = t\left(A \cos \omega_0 t + B \sin \omega_0 t \right)$$

Substitute this $i(t)$ into the ODE to find

$$A = \frac{V}{2L}, B = 0$$

• Can you please list the steps that lead to the derivation of the expression i(t)∝ tcos(t/√LC),t≥0 ? Commented Jul 27, 2014 at 14:29
• @DivyaK.S, see update to my answer Commented Jul 27, 2014 at 15:19
• ,Thanks a lot.It really helped to clarify the concept. Commented Jul 27, 2014 at 16:34
• The above mentioned solution should correspond to the particular integral. After adding the homogeneous solution to it, and applying initial conditions, the total response is found to be $$i(t) =\dfrac{V}{2L}[t\cos(\omega_0t)+\dfrac{1}{\omega_0}\sin(\omega_0t)]$$. Commented Jul 28, 2014 at 7:40

I understand the impedance goes to zero,which explains the steady state current as infinite. But can you explain the same by simple circuit analysis?I mean the voltage across inductor is Vsinωt - Vc = L di/dt, so based on derivative being positive, the current increases.But what makes the current go to infinite?

Yes, the net impedance falls to zero in a series resonant circuit when R is zero and, the voltage across each of L and C also rises to infinity - it couldn't be anything else or you would not get infinite current. That's the simple circuit analysis - trying to equate $V_L$ to the input voltage minus $V_C$ doesn't really help other than remind you that the two voltages (across L and C) are oppositely phased i.e. they are 180 degrees apart.

Should you think of the driving source as a constant current (just to prevent infinities and keep things sensible) that current appears to conduct thru a dead-short-circuit at resonance and that current will produce an equal finite voltage magnitude across each element determined by the components impedance (equal at resonance).

The current will reach infinity non-instantaneously because resonance is at a pure sinewave and a pure sinewave has to have been in existence a very long time ago!

Energy transfer is exactly the same as when not exactly at resonance or with a little bit of resistance except the numbers are infinity (not helpful to analyse in my book).

There are initial conditions - the application of a sinewave via a switch is an initial condition but, if you mean a dc voltage across the capacitor then this will continue to be present (on average) because the energy contained in it has nowhere to be dissipated (neither the perfect voltage source nor inductor) can do this. In a real circuit this DC "energy" will be burnt in the non-zero resistance of the wires.

• "The current will reach infinity non-instantaneously because resonance is at a pure sinewave and a pure sinewave has to have been in existence a very long time ago!"- Correct, But what will be the nature of current if we excite at t=0 (instead of t= -inf)? Commented Jul 27, 2014 at 14:35
• @DivyaK.S If i were you I wouldn't hesitate downloading LTSpice from Linear Technology (it's free) to look at this in detail. Of course there is a small learning curve but you will find that this tool (or any equiv spice simulator) will be invaluable to you on other EE problems. Mathematically I'm not going to try and derive this because it would take me too much time. Suffice to say, the current reaches infinity when t=infinity. Commented Jul 27, 2014 at 15:12
• @ Andy akka Thanks a lot,i did try in Psim but was not clear with the concept.Your explanation really helped to clarify the concept. Commented Jul 27, 2014 at 16:33

I mean the voltage across inductor is Vsinωt - Vc = L di/dt, so based on derivative being positive, the current increases.But what makes the current go to infinite? After all for a simple L only circuit, the voltage across inductor is Vsinωt = L di/dt and it is not getting infinite

In an L only circuit, the voltage across inductor is $$V\sin(\omega t) = L \dfrac{di}{dt}$$This means current is either increasing or decreasing (as per sine wave). In the case of resonance, voltage across inductor is $$V\sin(\omega t) - V_c$$ Here the key point is that the source voltage and capacitor voltage need not be of same polarity. Infact, they will be almost 90 degree out of phase, which makes the difference more than that of a simple 'L' circuit.

This difference becomes more and more as time evolves. It looks like capacitor stores energy from source and adds it with source in next cycle to increase the current more. Increased current means more energy storage in inductor, which later supplies back to capacitor which results in more charging(more voltage) and the process will reinforce each other.