Voltage of capacitor in underdamped RLC circuit

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)\$$?

• An underdamped oscillator will have an exponential decay term. Jun 2, 2021 at 20:22
• Yes sorry I only wrote part of the expression now it is corrected. Jun 2, 2021 at 20:24
• btw - I don't think your value of 95 is correct. It should be $\sqrt{10000-25} = 99.87...$ Jun 2, 2021 at 22:13
• @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. Jun 2, 2021 at 23:31
• 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. Jun 3, 2021 at 0:23

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.

• Cant I solve it algebraically? Jun 2, 2021 at 21:04
• Jun the impedances are on the graph, yes you can solve that in time or frequency domain Jun 2, 2021 at 21:28
• electronics.stackexchange.com/search?q=rlc Jun 2, 2021 at 21:33

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\$$.