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Consider the natural response of a parallel RLC Overdamped circuit, we're required to find the time it takes for the voltage across the Resistor to reach it's first peak.

I got the equation for the natural response and differentiated it to get a maxima, and the time corresponding to it is the time for the first peak.

But my instructors said that there's no peak in case of an Overdamped RLC circuit? Is he right?

Then what about the value I got after differentiating the natural response equation, that should correspond to something, if it's not the first peak?

enter image description here

The equation for the natural response of the Overdamped case when differentiated and equated to 0, to find the maxima/minima, gives a time at which it happens, if it doesn't correspond to the time when it reaches the first peak then what does it correspond to?

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An overdamped RLC when "hit" with a step change input does not produce any overshoot: -

enter image description here

I got the equation for the natural response and differentiated it to get a maxima, and the time corresponding to it is the time for the first peak.

You used the wrong equation then. There are three typical equation examples from this site: -

enter image description here

You need to use the over-damped equation. And note that only the underdamped equation produces ringing.

But my instructors said that there's no peak in case of an Overdamped RLC circuit? Is he right?

Yes, he is correct.

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  • \$\begingroup\$ Thanks. I've done an edit, please have a look. The graph you showed suggests that the slope for the overdamped case will be 0 only at infinity, but I've recalculated it and got the time when the derivative of the voltage across the resistor is 0 to be 1.07s, what does this 0 slope means if it isn't the first peak? \$\endgroup\$ – Adarsh Kumar Mar 21 at 17:12
  • \$\begingroup\$ It cannot be anything else other than infinity for an overdamped case. Try modelling it in a simulator if you don't believe. \$\endgroup\$ – Andy aka Mar 22 at 8:58

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