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I am in my last year of high school. We are studying free RLC circuits but we're not diving too deep into the theory. We know how to derive the differential equations governing charge, current and voltage but not how to solve them. We also know how the curves are supposed to look.

In a circuit with a charged capacitor, a coil and a resistor connected in series without a source, supposing that the circuit is underdamped:

$u_c$ is the voltage in the capacitor which has capacitance $C$.

$u_b$ is the voltage in the coil which has inductance $L$ and internal resistance $r$.

$u_R$ is the voltage in the resistor with resistance $R$.

$i$ is the current and $q$ is the charge in the capacitor.

We are taught that when $u_c$ is maximal or minimal, $i = 0$. This easy to prove since $i = \frac{dq}{dt} = C \frac{du_c}{dt} = 0$. We are also taught that whenever $i$ is maximal or minimal, $u_c = 0$. This looks right if we look at the graph. But I found a reasoning which leads to a contradiction. Can someone tell me what's wrong with it?

According to Kirchoff's law,

\begin{align*} u_c &= -(u_b + u_R) \\ &= -(L \frac{di}{dt} + ri + Ri) \end{align*} When $i$ is maximal, $i = I_{\max}$ and $\frac{di}{dt} = 0$. As such, \begin{align*} u_c &= -(r+R)I_{\max} \neq 0 \end{align*}

Any help would be greatly appreciated.

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...whenever $i$ is maximal or minimal, $u_c=0$.

It is true in a pure LC circuit. Not if there is ohmic resistance.

The mechanic analogy with an harmonic oscillator is useful. If it is not damped, the max. velocity (equivalent to max. current) happens when the spring (equivalent to capacitor) is at its equilibrium position (no force is equivalent to $V=0$).

But if the oscillator is damped, with a drag force (equivalent to resistance) proportional to the velocity, the max. velocity happens when the total force on the mass is zero. But that means spring force minus drag force.

After that moment, the drag force is greater, and the velocity starts to decrease. When the spring force is zero, the drag force is already decelerating the mass for some time.

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  • $\begingroup$ Thank you very much for your answer. It makes a lot of sense. I appreciate your help. Have a good night! $\endgroup$ – Oussema Dec 14 '20 at 21:04

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