Semi-Parallel RLC Circuit

Hi everyone, I find myself stumped on how to approach this problem since there is a resistor in series with the capacitor for what is otherwise a parallel RLC circuit.

First off, I am attempting to find the step response equations for i_L(t) and V_c(t).

As far as the diagram is concerned, the 6V square wave is just there to act like a switched DC voltage source.

I've been farting around with Kirchoff's laws and DEqs but can't figure out how to set up anything that would find a solution.

That said, I've found the initial conditions for before switch closure, just after and final conditions in case those will be needed later.

Before switch closure: $$V_L(0-)=0V, i_L(0-)=2A$$ $$V_c(0-)=0V, i_c(0-)=0A$$ $$V_{R_{eq}}(0-)=0V, i_{R_{eq}}(0-)=0A$$

Just after closure: $$V_L(0+)=2.8708V, i_L(0+)=2A$$ $$V_c(0+)=0V, i_c(0+)=0.01435A$$ $$V_{R_{eq}}(0+)=2.8708V, i_{R_{eq}}(0+)=0.50718A$$

Final conditions: $$V_L(\infty)=0V, i_L(\infty)=3A$$ $$V_c(\infty)=0V, i_c(\infty)=0A$$ $$V_{R_{eq}}(\infty)=0V, i_{R_{eq}}(\infty)=0A$$

Let $$V_a$$ represent the node voltage of the parallel elements. Setting up the node voltage equation we have


Note that, $$i_{C,R_2}=\frac{V_a-V_c}{R_2}$$ and $$i_{R_{eq}}=\frac{V_a}{R_{eq}}$$



Note that $$L\frac{d}{dt}i_L=V_a$$ so I could differentiate both sides and multiply by L, to to eliminate all the currents. However, I still have $$V_C$$ which is what is stopping me from going forward.

I am starting to think I need more equations...

Any help on this would be very much appreciated.


2 Answers 2


one more for you:

\$i_{cR_2} = C \frac{dV_c}{dt} = \frac{V_a-V_c}{R_2} = f(V_a,V_c) ----(3) \$

Now you have three independent equations and three unknowns (\$i_L,V_c,V_a\$) to solve everything.

  • \$\begingroup\$ Thanks for the reply. So from $$C\frac{dV_c}{dt} = \frac{V_a-V_c}{R_2}$$ I get that $$R_2C\frac{dV_c}{dt}+V_c=V_a$$ However, when I substitute this into the equation and then take the derivative of both sides and multiply through by L, to eliminate $$i_L$$ don't seem to get the same answer as my PSPICE simulation. But I may have mixed something up along the way. Does this seem like the right way to move forward? \$\endgroup\$
    – Big Gulps
    May 2, 2018 at 4:28
  • 1
    \$\begingroup\$ Your way of solving is wrong cz this is not typical linear equation. They are differential equations and hence boundary conditions will matter too. Also you cannot differentiate on both sides of an equation always and say it holds true, unless the equation is true for all real values of dependent variables. Better solve it by converting all these to laplace equations. \$\endgroup\$
    – Mitu Raj
    May 2, 2018 at 5:00

enter image description here

It so happens that my original answer was correct while my PSPICE simulation was wrong. Resimulating the original circuit above gives the same results as my analytical solution. So my solution process was indeed valid.

In the hopes this helps others like myself, I've written out my solution and the process I used to obtain it below.

STEP 0: Identify the Order of the Differential Equation/s

Since this circuit contains R, L, and C, it is a 2nd order ODE.

Note: You may end up with a system of ODEs rather than a single ODE.

STEP 1: Find Initial Conditions

Before switch closure. Solve by treating inductor as zero resistance and capacitor as an infinite resistance.

$$V_L(0^-)=0V, i_L(0^-)=2A$$ $$V_c(0^-)=0V, i_c(0^-)=0A$$

Just after closure. Solve by treating capacitor as zero resistance and inductor as an infinite resistance. Note that the current through an inductor cannot change instantaneously, nor can the voltage across a capacitor. So these values are the same as the left handed limits.

$$V_L(0^+)=2.8708V, i_L(0^+)=2A$$ $$V_c(0^+)=0V, i_c(0^+)=0.01435A$$

Then we have that the initial conditions for i_L are

$$\begin{equation}\begin{cases} i_L(0^+)=2\\ i^{'}_L(0^+)=\frac{V_L(0^+)}{L}=28.708 \end{cases}\end{equation}$$

STEP 3: Simplify the Circuit after Switch Closure

Use source transformations and parallel/series reductions.

enter image description here

STEP 4: Use KVL and KCL to obtain System of Linearly Independent Equations

$$\begin{equation}\begin{cases} i_L+i_{R_{eq}}+i_{C,R_2}=3 & \text{(KCL)}\\ -V_L+V_C+R_2CV^{'}_C=0 & \text{(KVL)} \end{cases}\end{equation}$$

Note any helpful identities:

$$\begin{equation}\begin{cases} V_L=Li^{'}_L \\ i_{R_{eq}}=\frac{V_L}{R_{eq}}=\frac{L}{R_{eq}}i^{'}_L \\ i_{C,R_2}=CV^{'}_C \end{cases}\end{equation}$$

Modify system of equations to as few of independent variables as possible. In this case there will be two independent variables.

$$\begin{equation}\begin{cases} i_L+\frac{L}{R_{eq}}i^{'}_L+CV^{'}_C=3 & \text{(Eq1)}\\ -Li^{'}_L+V_C+R_2CV^{'}_C=0 & \text{(Eq2)} \end{cases}\end{equation}$$

STEP 5: Elimination of Variables

It's arbitrary which variable to eliminate, but since we have the initial conditions for i_L, we will eliminate V_C.

Perform the linear operation (Eq2)-R_2(Eq1), then we have


Rearranging to obtain V_C, we have

$$V_C=-3R_2+L(1+\frac{R_2}{R_{eq}})i^{'}_L+R_2i_L \text{.....(Eq3)}$$

We now take the derivative of (Eq3)

$$V^{'}_C=L(1+\frac{R_2}{R_{eq}})i^{''}_L+R_2i_L^{'} \text{.....(Eq4)}$$

and substitute (Eq4) into (Eq1) to obtain

$$\begin{equation}\begin{cases} i_L(0^+)=2\\ i^{'}_L(0^+)=28.708\\ LC(1+\frac{R_2}{R_{eq}})i^{''}_L+(\frac{L}{R_{eq}}+CR_2)i^{'}_L+i_L=3 \end{cases}\end{equation}$$

You then solve this nonhomogeneous equation for i_L and substitute this solution back into (Eq3) to obtain V_C. I would write out the solution but it is a real mess of latex to do.


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