Does an ideal LC circuit oscillate when a battery connects to it?

[k_rlc]

Hi All ,

I was thinking about the time domain unit step response of ideal series LC crcuit. If both cap and inductor are ideal and unchrged initially and I am looking at voltage across capacitor.
Connecting a battery to LC circuit at t=t1 , would the circuit oscillate ?
I am not able to imagine this situation in terms of current and voltages of inductor /capacitor.I would be greatful if someone could intutively expain (without math and analogies) what exactly would happen.
My another question is, in case of underdamped response of RLC . I am again not able to think what causes the ringing behaviour with respect to inductor / capacitor current and voltage.

 

[zoki85]

If an ideal DC battery gets connected to ideal de-energized LC circuit, transient is an ideal sine current. Sinnce there's no damping, it will oscillate forever with amplitude I=V/√(L/C) and angular frequency ω=1/√(LC). Voltage on the capacitor will oscillate between 0 and 2V. The process is hard to explain without any math becouse this is the result obtained from solving differential equation.

 

[Baluncore]

A voltage step applied to an LC series pair will initially appear across the inductor. As current starts to build up through the inductor, (because V_L = L * di/dt), the voltage across the capacitor will begin to ramp up. The inductor current will reach a maximum when the capacitor voltage reaches the step voltage, where the inductor voltage passes through zero. The capacitor voltage will overshoot and the cycle will reverse and repeat. The LC pair will ring at the resonant frequency, (at which X_L + X_C = zero), and because the circuit resistance is zero, it will continue ring forever.

 

[k_rlc]

If an ideal DC battery gets connected to ideal de-energized LC circuit, transient is an ideal sine current. Sinnce there's no damping, it will oscillate forever with amplitude I=V/√(L/C) and angular frequency ω=1/√(LC). Voltage on the capacitor will oscillate between 0 and 2V. The process is hard to explain without any math becouse this is the result obtained from solving differential equation.

A voltage step applied to an LC series pair will initially appear across the inductor. As current starts to build up through the inductor, (because V_L = L * di/dt), the voltage across the capacitor will begin to ramp up. The inductor current will reach a maximum when the capacitor voltage reaches the step voltage, where the inductor voltage passes through zero. The capacitor voltage will overshoot and the cycle will reverse and repeat. The LC pair will ring at the resonant frequency, (at which X_L + X_C = zero), and because the circuit resistance is zero, it will continue ring forever.

Thanks Zoki85

Thanks Baluncore. I had also thought on the same lines. However , one question that was challenging my thought process was that when capacitor charges to step voltage then there should will not be any current flowing in inductor so there should not be any overshoot. Hence , I was confused . Let me know if you have any explanation.

 

[Baluncore]

I am not able to imagine this situation in terms of current and voltages of inductor /capacitor.I would be greatful if someone could intutively expain (without math and analogies) what exactly would happen.

Why do you insist on tying your hands behind your back?

When voltage appears across the capacitor it stores energy, simply E_C = ½ C V²
When current flows through the inductor it stores energy, simply E_L = ½ L I²

Conservation of energy requires that when the capacitor voltage is zero, the inductor current must be at a maximum. Likewise, when the inductor current is zero the capacitor voltage must be at a maximum.

Rather than considering a voltage step applied to an LC series pair, consider the parallel connection of an open circuit inductor with a charged capacitor. The energy flows from the capacitor to the inductor until the capacitor reaches zero voltage, when all the energy is stored in the inductor at maximum current. The energy then passes from the inductor to the capacitor until the capacitor voltage is the negative of it's starting voltage, when the inductor current has fallen again zero. At that point half a cycle has occurred. In effect the capacitor voltage has been reversed. It will reverse again to return the initial situation.

 

[k_rlc]

Thanks a lot baluncore for all the explanations