I think I'm suffering from brain-drain. I'm trying to solve a 2nd order differential equation for the natural response of an under-damped parallel RLC circuit with initial conditions of \$I_L\$ and \$V_C\$ but I'm falling (apparently) at the final hurdle.

All the solutions on the web that I have found seem to skirt over the solution to finding B1 and B2 in this type of formula. Most (if not all the solutions) seem to solve for output voltage but I also want to know how the inductor current decays.

The best single picture I have found shows this: -

enter image description here

And, I am happy that I can use this to solve for the voltage waveform if my initial condition was just the capacitor voltage BUT, I need to know what B1 and B2 are for the intial conditions of BOTH capacitor voltage and inductor current.

So, in short: -

  • What is the formula for \$i_L(t)\$ for a parallel RLC
  • How do I formulate the B1 and B2 values taking into account BOTH initial conditions
  • Ditto for \$v(t)\$
  • \$\begingroup\$ Are you only looking at the underdamped case? Or all three cases? I followed your picture's underdamped case easily and got the entire v(t) equation using both initial conditions pretty easily, too. The v(t) equation in the underdamped case includes an interesting factor, \$\sqrt{\frac{\omega_0^2}{\alpha^2}-1}\$, which I don't see here, yet. But I can't tell if there's further interest. \$\endgroup\$
    – jonk
    May 9, 2018 at 5:35
  • \$\begingroup\$ @jonk I'm only looking at the underdamped case but because it's only the initial conditions that I need, I believe that these apply to over and critically damped? Yes please do make an answer if you have the time but be aware of the aggro that has happened with the other answer and him wasting my time with him misinterpreting the question. That may understandably put you off. Also be aware that my own answer to the question solves for v(t) - I finally got my head around the initial conditions that determine B2 so, I'm just looking for the \$i_L(t)\$ scenario now. \$\endgroup\$
    – Andy aka
    May 9, 2018 at 7:32
  • \$\begingroup\$ @jonk However, if you can confirm that or add to it that would be good. \$\endgroup\$
    – Andy aka
    May 9, 2018 at 7:32
  • \$\begingroup\$ I'm off to sleep (geologist coming over to check out some issues with the land here in the AM.. less than 8 hrs from now.) I'll see what I can do later. In the meantime, if you have V(t) already you also have d I(t) then, since that's just V(t)/L. You also have IL(0) as the initial condition to solve the constant of integration. I haven't taken this particular step yet. But it seems straight forward. I'll look at it after the geologist thing in the morning. I stopped at V(t), myself. \$\endgroup\$
    – jonk
    May 9, 2018 at 8:10
  • \$\begingroup\$ @jonk maybe you can also confirm that my value for B2 makes sense when solving for v(t)? \$\endgroup\$
    – Andy aka
    May 9, 2018 at 9:54

4 Answers 4


Just to be complete:


simulate this circuit – Schematic created using CircuitLab

I gather that's the circuit. And you'd like an expression for \$V_{\left(t\right)}\$ and \$I_{L\left(t\right)}\$ that handles the initial conditions where \$V_{\left(0\right)}\$ and \$I_{L\left(0\right)}\$ may be non-zero and are otherwise known at \$t=0\$.

You also want to focus on the case that is underdamped.

I hope you won't mind if I start at the basics and take this through, carefully. A lot of it will just be repetition of the obvious. I apologize for that.

The basic nodal equation is (treating all currents pointing downwards as positive):

$$\frac{V_{\left(t\right)}}{R}+\frac{1}{L}\int V_{\left(t\right)}\:\text{d}t+C\frac{\text{d}V_{\left(t\right)}}{\text{d}t}=0\:\text{A}$$

Taking the derivative and dividing through by \$C\$ (had I added \$I_{L\left(0\right)}\$ above, it would disappear now, anyway):

$$\begin{align*}\frac{\text{d}^2V_{\left(t\right)}}{\text{d}t^2}+\frac{1}{R\:C}\frac{\text{d} V_{\left(t\right)}}{\text{d}t}+\frac{V_{\left(t\right)}}{L\:C}&=0\:\text{A}\end{align*}$$

A standard 2nd order ODE where the obvious proposed solution is \$V_{\left(t\right)}=A e^{s t}\$. By substitution, we find the characteristic equation is:


Solving with the standard quadratic solution equation, \$\frac{-b\pm\sqrt{b^2-4\:a\:c}}{2\:a}\$, we find that it is convenient (from a cursory examination of the results) to define:


The solutions are then:


At this point, there are three possibilities to consider. One is the critically damped case where \$\alpha = \omega_0\$ and therefore \$s_1=s_2\$ and they are both real-valued. Another is the overdamped case where \$\alpha > \omega_0\$ and therefore \$s_1\ne s_2\$ but they are both real-valued, again. The final case is the underdamped case where \$\alpha < \omega_0\$ and \$s_1\$ and \$s_1\$ are both complex-valued and are conjugates of each other.

This last case is the one you want to address.

In the underdamped case it is again convenient to flip the sign inside the square-root and define another real-valued variable, the damped frequency:


Now the general solution (in the underdamped case there are two conjugate answers, so both are valid and should be included) is:

$$\begin{align*}V_{\left(t\right)}&=A_1\:e^{s_1\:t}+A_2\:e^{s_2\:t}\\\\ &=e^{-\alpha\:t}\left[\:\left(A_1+A_2\right)\operatorname{cos}\left(\omega_d\:t\right)+j\:\left(A_1-A_2\right)\operatorname{sin}\left(\omega_d\:t\right)\:\right]\\\\ \text{setting: } B_1&=A_1+A_2\\B_2&=j\left(A_1-A_2\right)\\\\ &=e^{-\alpha\:t}\left[\:B_1\operatorname{cos}\left(\omega_d\:t\right)+B_2\operatorname{sin}\left(\omega_d\:t\right)\:\right]\end{align*}$$



The derivative is,

$$\begin{align*} \frac{\text{d}V_{\left(t\right)}}{\text{d}t}&=e^{-\alpha\:t}\left[\:\left(\omega_d\:B_2-\alpha\:B_1\right)\operatorname{cos}\left(\omega_d\:t\right)-\left(\omega_d\:B_1+\alpha\:B_2\right)\operatorname{sin}\left(\omega_d\:t\right)\:\right] \end{align*}$$


$$\begin{align*} \frac{\text{d}V_{\left(0\right)}}{\text{d}t}&=\omega_d\:B_2-\alpha\:B_1 \end{align*}$$

You know that:

$$\begin{align*}\frac{V_{\left(0\right)}}{R}+C\frac{\text{d}V_{\left(0\right)}}{\text{d} t}+I_{L\left(0\right)}&=0\:\text{A}\\\\\therefore\\\\\frac{V_{\left(0\right)}}{R\:C}+\omega_d\:B_2-\alpha\:B_1+\frac{I_{L\left(0\right)}}{C}&=0\:\text{A}\\\\\omega_d\:B_2-\alpha\:B_1&=-\frac{V_{\left(0\right)}}{R\:C}-\frac{I_{L\left(0\right)}}{C}\\\\\omega_d\:B_2-\alpha\:V_{\left(0\right)}&=-\frac{V_{\left(0\right)}}{R\:C}-\frac{I_{L\left(0\right)}}{C}\\\\\therefore\end{align*}$$$$\begin{align*}B_1 &=V_{\left(0\right)}&B_2&=\frac{V_{\left(0\right)}\left(\alpha-\frac{1}{R\:C}\right)-\frac{I_{L\left(0\right)}}{C}}{\omega_d}=-\frac{\alpha\:V_{\left(0\right)}+\frac{I_{L\left(0\right)}}{C}}{\omega_d}\\\\&&&\text{note above: }\tfrac{1}{R\:C}=2\alpha\\\\&&\text{If }V_{\left(0\right)}\ne 0\:\text{V then }\\\\&&\eta&=-\frac{\alpha+\frac{I_{L\left(0\right)}}{V_{\left(0\right)}\:C}}{\omega_d}\\\\&&B_2&=\eta\: V_{\left(0\right)}\end{align*}$$

And that solves the time-dependent voltage equation with the use of initial conditions.

$$\begin{align*}V_{\left(t\right)}=\left\{\begin{array}{l}V_{\left(0\right)}\ne 0\:\text{V},&&V_{\left(0\right)}\:e^{-\alpha\:t}\big[\operatorname{cos}\left(\omega_d\:t\right)+\eta\:\operatorname{sin}\left(\omega_d\:t\right)\big]\\&&V_{\left(0\right)}\:e^{-\alpha\:t}\:\sqrt{1+\eta^2}\operatorname{sin}\left(\omega_d\:t+\operatorname{tan}^{-1}\left(\frac{1}{\eta}\right)\right)\\\\V_{\left(0\right)}= 0\:\text{V},&&-\frac{I_{L\left(0\right)}}{\omega_d\,C}\:e^{-\alpha\:t}\:\operatorname{sin}\left(\omega_d\:t\right)\end{array}\right.\end{align*}$$

At this point, the remaining question is the time-dependent current equation for the inductor.

$$\begin{align*} I_{L\left(t\right)}&=\frac{1}{L}\int V_{\left(t\right)}\:\text{d} t\\\\ &=\frac{V_{\left(0\right)}}{L}\left[\int e^{-\alpha\:t}\operatorname{cos}\left(\omega_d\:t\right)\:\text{d} t+\eta \int e^{-\alpha\:t}\operatorname{sin}\left(\omega_d\:t\right)\:\text{d} t\right]\\\\ &=\frac{V_{\left(0\right)}}{L}\cdot\frac{-e^{-\alpha\:t}}{\alpha^2+\omega_d^2}\bigg[\left(\alpha\operatorname{cos}\left(\omega_d\:t\right)-\omega_d\operatorname{sin}\left(\omega_d\:t\right)\right)+\eta\cdot\left(\alpha\operatorname{sin}\left(\omega_d\:t\right)+\omega_d\operatorname{cos}\left(\omega_d\:t\right)\right) \bigg]+D_0\\\\ &=\frac{V_{\left(0\right)}}{L}\cdot\frac{-e^{-\alpha\:t}}{\alpha^2+\omega_d^2}\bigg[\left(\alpha+\eta\:\omega_d\right)\operatorname{cos}\left(\omega_d\:t\right)+\left(\eta\:\alpha-\omega_d\right)\operatorname{sin}\left(\omega_d\:t\right)\bigg]+D_0 \end{align*}$$


$$\begin{align*} I_{L\left(0\right)}&=-\frac{V_{\left(0\right)}}{L}\frac{\alpha+\eta\:\omega_d}{\alpha^2+\omega_d^2}+D_0\\\\\therefore\\\\D_0&=I_{L\left(0\right)}+\frac{V_{\left(0\right)}}{L}\frac{\alpha+\eta\:\omega_d}{\alpha^2+\omega_d^2}\end{align*}$$

Resulting in,

$$\begin{align*}I_{L\left(t\right)}=\left\{\begin{array}{l}V_{\left(0\right)}\ne 0\:\text{V},&&I_{L\left(0\right)}+\frac{V_{\left(0\right)}}{L}\left[\frac{\alpha+\eta\:\omega_d-e^{-\alpha\:t}\big[\left(\alpha+\eta\:\omega_d\right)\operatorname{cos}\left(\omega_d\:t\right)+\left(\eta\:\alpha-\omega_d\right)\operatorname{sin}\left(\omega_d\:t\right)\big]}{\alpha^2+\omega_d^2}\right]\\&&I_{L\left(0\right)}+\frac{V_{\left(0\right)}}{L}\left[\frac{\alpha+\eta\:\omega_d}{\alpha^2+w_d^2}-e^{-\alpha\:t}\sqrt{1+\eta^2}\frac{\operatorname{sin}\left(\omega_d\:t+\operatorname{tan}^{-1}\left[\frac{\alpha+\eta\:\omega_d}{\eta\:\alpha-\omega_d}\right]\right)}{\sqrt{\alpha^2+w_d^2}}\right]\\\\V_{\left(0\right)}= 0\:\text{V},&&I_{L\left(0\right)}+\frac{B_2}{L}\left[\frac{\omega_d-e^{-\alpha\:t}\big[\omega_d\operatorname{cos}\left(\omega_d\:t\right)+\alpha\operatorname{sin}\left(\omega_d\:t\right)\big]}{\alpha^2+\omega_d^2}\right]\\&&I_{L\left(0\right)}+\frac{B_2}{L}\left[\frac{\omega_d}{\alpha^2+w_d^2}-e^{-\alpha\:t}\frac{\operatorname{sin}\left(\omega_d\:t+\operatorname{tan}^{-1}\left[\frac{\omega_d}{\alpha}\right]\right)}{\sqrt{\alpha^2+w_d^2}}\right]\end{array}\right.\end{align*}$$

  • \$\begingroup\$ Should the final equations you have for B2 be divided by Wd as you seem to have done in the final equation for V(t)? Also, where did A1 and A2 come from and what are they? I've just skimmed becaue it's my turn to be in a hurry but it looks like your answer for v(t) matches mine. \$\endgroup\$
    – Andy aka
    May 9, 2018 at 19:49
  • 1
    \$\begingroup\$ @Andyaka \$A_1\$ and \$A_2\$ arrive from finding all possible solutions. This is standard mathematics. Start at this Khan Academy 2nd video and follow through to the 3rd one. It's not long. Second order linear equations \$\endgroup\$
    – jonk
    May 9, 2018 at 19:53
  • \$\begingroup\$ Wow that hidden dependency in finding B2 though. +1, great answer. Taking back my answer @Andyaka \$\endgroup\$
    – Mitu Raj
    May 9, 2018 at 19:54
  • 1
    \$\begingroup\$ @MITURAJ I think that is the sensible thing to do unlike some people. \$\endgroup\$
    – Andy aka
    May 9, 2018 at 19:55
  • 1
    \$\begingroup\$ @MITURAJ I just went back to the original equation again and tossed in the new derivative equation when \$t=0\$. Like a walk in the park, so to speak. \$\endgroup\$
    – jonk
    May 9, 2018 at 20:01

Digging up old notes, I found these solutions for initial conditions in the case of the underdamped parallel RLC, natural response:

$$B_1=V_C^{t=0}$$ $$\omega_d B_2-\alpha B_1=-\frac{I_L^{t=0}}{C}-\frac{V_C^{t=0}}{RC}$$ $$=> B_2=\frac{(\alpha RC-1)V_C^{t=0}-R I_L^{t=0}}{\omega_d RC}$$

where \$V_C^{t=0}\$ is the initial condition for C, and \$I_L^{t=0}\$ is the initial condition for L. With these, \$y(t)\$ is the expression you gave. For the currents:

$$I_R(t)=\frac{y(t)}{R}$$ $$I_C(t)=C\frac{\text{d}y(t)}{\text{d}t}$$ $$I_L(t)=-I_R(t)-I_C(t)$$

Testing this with some random values R=12.34, L=0.618, C=2.718, with initial condition of 1.618 for L and 3.14 for C, gives these results in LTspice:


and in wxMaxima:


If you don't want the derivative in the expression for the capacitor current, you can use this:

$$I_C(t)=-C\exp(-\alpha t)\left[(B_1\omega_d+B_2\alpha)\sin(\omega_d t)+(B_1\alpha-B_2\omega_d)\cos(\omega_d t)\right]$$

then use the difference.

Here are the results of the simulation and the mathematical derivation given your input: R=3k, L=256u, C=1u, IC(L)=2, IC(C)=10:



If deducing the current through L is too much given the difference above, then use this:

$$I_L(t)=\frac{(B_1\omega_d-B_2\alpha)\sin(\omega_d t)-(B_2\omega_d+B_1\alpha)\cos(\omega_d t)}{(\omega_d^2+\alpha^2)\exp(\alpha t)L}$$

which is simply the integral of \$y(t)\$ divided by L. Complicated.

  • \$\begingroup\$ Comments are not for extended discussion; this conversation has been moved to chat. \$\endgroup\$
    – Dave Tweed
    May 8, 2018 at 19:13
  • 4
    \$\begingroup\$ @DaveTweed I'm sorry if this turned out this way, but I know from a previous, recent experience (can provide links) that moving this to chat would have had zero effect. \$\endgroup\$ May 8, 2018 at 19:20
  • \$\begingroup\$ In your relation of wd, alpha, B2,B1, it is not Ic(0) ....It is actually IL(0). Also Vc(0) = VL(0) = V(0) simply ... \$\endgroup\$
    – Mitu Raj
    May 10, 2018 at 5:05
  • \$\begingroup\$ @MITURAJ You're right, thanks for spotting the error. The formulas I have use different notations: \$\omega_n\$, IC(C), IC(L), and I translated on the spot. Corrected. Still, the solution is correct, as seen in the pictures (which used the same formulas). \$\endgroup\$ May 10, 2018 at 7:18

You folks know most of what I will write and at the risk of being pedantic ...:

The target is some knowledge about "\$B_{1}\$ and \$B_{2}\$" in:

$$v(t)=e^{-\alpha t}[B_{1}cos(\omega_{d}t)+B_{2}sin(\omega_{d}t)]$$

and about "\$B_{3}\$ and \$B_{4}\$" in: $$i_{L}(t)=e^{-\alpha t}[B_{3}cos(\omega_{d}t)+B_{4}sin(\omega_{d}t)]$$

Starting with the inductor current, the Laplace domain target is: $$I_{L}(s)=\frac{B_{3}(s+\alpha)+B_{4}\omega_{d}}{(s+\alpha)^{2}+\omega_{d}^{2}}\tag{1}$$ KCL


Compare to standard form:

$$\frac{d^{2}i_{L}}{dt^{2}}+2\zeta\omega_{N}\frac{di_{L}}{dt}+\omega_{N}^{2}i_{L}=0 \tag{2}$$

The \$2^{nd}\$ Order parameters are: $$\zeta=\frac{1}{2R}\sqrt{\frac{L}{C}},\omega_{N}^{2}=\frac{1}{LC}$$

\$2^{nd}\$ Order underdamped poles are: $$p_{1},p_{1}^{*}=-\alpha\pm j\omega_{d}; \alpha = \zeta\omega_{N}; \omega_{d}=\omega_{N}\sqrt{1-\zeta^{2}}$$

Using the Laplace property for derivatives Equation 2 is transformed into the Laplace domain. Notice that the initial conditions are now revealed such that \$I_{0}\$ is the initial inductor current, \$V_{0}\$ is the initial capacitor voltage, and \$\frac{V_{0}}{L}\$ is the initial rate of change of the inductor current.


Solving for the inductor current and completing the square in the denominator reveals where \$B_{3},B_{4}\$ can be identified by arranging the numerator. $$I_{L}(s)=\frac{I_{0}s+2\zeta\omega_{N}I_{0}+V_{0}/L}{(s+\alpha)^{2}+\omega_{d}^{2}}=\frac{I_{0}(s+\alpha)+\frac{\alpha I_{0}+V_{0}/L}{\omega_{d}}\omega_{d}}{(s+\alpha)^{2}+\omega_{d}^{2}}$$

Comparing to equation 1 reveals the values: $$B_{3}=I_{0};B_{4}=\frac{\alpha I_{0}+V_{0}/L}{\omega_{d}}$$

The values:

$$B_{1}=V_{0};B_{2}=\frac{\alpha V_{0}+I_{0}/C}{\omega_{d}}$$ are found by observing symmetry between the inductor current solution and the capacitor voltage solution and then making the substitutions without having to do all the math again.

Hope this helps.


This is a partial answer to the solution of the terminal voltage across the RLC circuit.

The formula is: -

\$v(t) = e^{-\alpha t}\cdot[B1\cdot \cos(\omega_d t) + B2\cdot\sin(\omega_d t)]\$


  • \$\alpha\$ is 1/2CR (see below)
  • B1 is the initial voltage (at t = 0) i.e. \$V(0)\$


  • B2 is \$\dfrac{\dfrac{I_L(0)}{C}-\dfrac{V(0)}{CR}+\alpha\cdot V(0)}{\omega_d}\$

\$\dfrac{I_L(0)}{C}\$ is the dominant driver of the rate of change of voltage at t = 0 and this is modified by the load resistor adjusting the dominant dv/dt by \$\dfrac{V(0)}{CR}\$.

This gives a spreadsheet graph result that is as close to simulation as possible. For instance with I(0) at 2 amps, V(0) at 10 volts, L = 256 uH, C = 1 uF and R = 3000 ohm, the spreadsheet tells me the next voltage maxima is 33.387722 volts and micro-cap tells me it is 33.388 volts: -

enter image description here

enter image description here

50% solved so can I solve for inductor current or will someone step in?


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