I have this circuit:


simulate this circuit – Schematic created using CircuitLab

If the capacitor is precharged this circuit has an underdamped response and for VC1 = 10V the equation for current I(t)=1.05e^(-5t)sin(95t)A

How do find the equation for voltage across C1? Can I use this formula? \$Vc1 = Vo+\frac{1}{C}*\int(I(t)dt)\$?

  • \$\begingroup\$ An underdamped oscillator will have an exponential decay term. \$\endgroup\$
    – ErikR
    Jun 2, 2021 at 20:22
  • \$\begingroup\$ Yes sorry I only wrote part of the expression now it is corrected. \$\endgroup\$
    – Jun Seo-He
    Jun 2, 2021 at 20:24
  • \$\begingroup\$ btw - I don't think your value of 95 is correct. It should be \$\sqrt{10000-25} = 99.87...\$ \$\endgroup\$
    – ErikR
    Jun 2, 2021 at 22:13
  • \$\begingroup\$ @ErikR the neper frequeny is equal to 5.So the frequency of the underdamped oscillation is natural frequency-neper frequency = 100-5=95 rad/s. \$\endgroup\$
    – Jun Seo-He
    Jun 2, 2021 at 23:31
  • \$\begingroup\$ I'd double check that... everything I've seen says \$\omega_d^2 = \omega_0^2 - \alpha^2\$ where \$\omega_d\$ is the damped frequency, \$\omega_0\$ is the natural frequency and \$\alpha\$ is the neper frequency. \$\endgroup\$
    – ErikR
    Jun 3, 2021 at 0:23

2 Answers 2


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Now you can look up the exponential waveform for this initial condition that includes Q or ζ=1/2Q . this is your homework.

This is how I started doing it in 1975. It still works.


Since you already have the equation for \$I(t)\$, if you want to solve it algebraically I would start with this KVL equation:

$$ L\frac{d}{dt}I(t) + RI(t) + V_C(t) = 0 $$ or $$ V_C(t) = - L\frac{d}{dt}I(t) - RI(t) $$

You'll get a combination of \$\sin\$ and \$\cos\$ terms, e.g.

$$ A\cos 95t + B\sin 95t $$

You can transform that into the form \$C\sin(95t + \delta)\$ by solving:

$$ C = \sqrt{A^2+B^2}, \sin\delta = \frac{A}{C}, \cos\delta = \frac{B}{C} $$

Update: If you follow through with this approach you should be able to derive this:

$$ V_C(t) = -I_0\sqrt{\frac{L}{C}}e^{-\gamma t}\cos(\omega t - \delta) $$ where \$\sin \delta = \frac{RC}{2\sqrt{LC}}\$ and the current is given by \$I(t) = I_0e^{-\gamma t}\sin \omega t \$. And don't forget about the approximation \$\sin \delta = \delta\$ for small values of \$\delta\$.


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