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If the energy is dissipated more quickly in overdamped circuit then why it does have longest settling time?

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This answer uses the example of a series capacitor, inductor and resistor and assumes the capacitor is previously charged to some voltage level.

If the over-damping is massive i.e. a massive series resistor then it dominates the inductance and the circuit becomes a resistor discharging a capacitor hence, the discharge time can be very long if R is very large.

As R reduces, the discharge time reduces. But as you approach critical damping (Q = 0.5) from the over-damped case, the decay changes from being a simple exponential situation to a decaying ripple where the frequency of the ripple is a decaying sinewave having a frequency of \$\omega_n\sqrt{1-\zeta^2}\$, where \$\zeta = \frac{1}{2Q}\$ and \$\omega_n\$ is the natural resonant frequency.

At Q = 0.6 (slightly less than critical damping), the decaying sinewave is very limited in its extent. The overshoot of the first cycle is about 0.9% and, to most folk would hardly count. As the sinewave decays, the first undershoot is about 0.008% so it's pretty much over as soon as it began.

The ratio of the magnitude of the 1st positive peak to the first negative peak is approximately 0.9/0.008 = ~112.

However, as Q rises more and more, the overshoot and undershoot become more dominant. For instance at Q = 1, the first overshoot peak is about 16% and the first undershoot peak is about 2.7%. You can see that these are closer in magnitude than was the case when Q was only 0.6. Ratio is about 6.

As Q rises even more you eventually reach the situation where the overshoot and undershoot continue for a very long time. This means that energy is kept in the reactive components for a very long time.

Conclusion: The energy is dissipated more quickly in a series RLC circuit at around critical damping. Either side of critical damping it takes longer and, at extremes (R = infinity, R = 0), it takes an infinite length of time.

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  • \$\begingroup\$ What about in parallel circuit how does smaller value of resistance dominates over capacitance. If the capacitor is initially charged then energy will be dissipated faster for small R (p=v²/R) \$\endgroup\$ – Aaqib Jun 20 '19 at 9:30
  • \$\begingroup\$ @Aaqib and, after I give details about this you might ask about series-parallel arrangements and there are two of them so, I'm declining to pursue this further. I will say this though: you should get hold of a free simulation tool and perform the experiment by yourself. I use micro-cap student version and find it to be very good and accurate. If you've never used a sim tool before then there is a steep learning curve however, if you are planning a career in EE then you'll have to learn sims sooner rather than later. \$\endgroup\$ – Andy aka Jun 20 '19 at 9:42
  • \$\begingroup\$ I will also add this: If the initial energy was stored in the inductor in a parallel circuit (i.e. current was flowing in the inductor) and you reduced the parallel resistor to zero ohms near instantly then the inductor will continue to circulate the current indefinitely i.e.a very low Q factor will still produce a very long decay. So, this brings in the concept of initial conditions and, those initial conditions can lead to different conclusions. \$\endgroup\$ – Andy aka Jun 20 '19 at 9:51

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