# Calculation of current in driven series RLC circuit

• zenterix
zenterix
Homework Statement
I am a little confused about some calculations in some notes introducing the concept of an RLC series circuit.
Relevant Equations
Below I go through the reasoning in the notes.
The notes are in an attached pdf on pages 10-13.

The AC source voltage is ##V(t)=V_0\sin{(\omega t +\phi)}## and we would like to find the current ##I(t)=I_0\sin{(\omega t)}##.

$$V(t)-V_R(t)-V_C(t)=L\dot{I}=V_L(t)\tag{1}$$

$$V_0\sin{(\omega t+\phi)}-IR-\frac{Q}{C}=L\dot{I}\tag{2}$$

After some algebra and differentiation we obtain the following differential equation

$$\ddot{I}+\frac{R}{L}\dot{I}+\frac{1}{LC}I=\frac{V_0\omega}{L}\cos{(\omega t+\phi)}\tag{3}$$

This particular electromagnetism course doesn't assume the student has taken differential equations. It seems that for this reason they sometimes find solutions in a way that involves intuition rather than standard methods.

My question in this post is about how they are solving it in these notes.

First they consider the instantaneous voltages across the resistor, the inductor, and the capacitor.

Here are the phasor diagrams for the relationships between current and voltage in these three circuit elements.

For the record, phasor diagrams are new to me.

Using the phasor representation, Eq. (1) can be written as

$$\vec{V}_0=\vec{V}_{R0}+\vec{V}_{L0}+\vec{V}_{C0}\tag{4}$$

as shown in the next figure

Again we see that current phasor ##\vec{I}_0## leads the capacitive voltage phasor ##\vec{V}_{C0}## by ##\pi/2## but lags the inductive voltage phasor ##\vec{V}_{L0}## by ##\pi/2##.

The three voltage phasors rotate counterclockwise as time increases, with their relative positions fixed.
I understand this last paragraph on some level because before analyzing the current circuit, I analyzed three different circuits: an AC source with just a resistor, with just an inductor, and with just a capacitor.

I think I just need to be explicitly told why the phasors are the same in those three previous circuits as they are in the current circuit. Is it superposition because each circuit element is linear?

It is also not clear why we can justify going from (1) to (4).

After this, it is simple to obtain an expression for the amplitude of the current.

We simply look at the rhs plot above and compute ##V_0## using Pythagorean theorem.

Now, I found the notation used a bit confusing, let me try to clarify.

Let

$$I_{R0}=\frac{1}{X_R}V_{R0}\tag{4}$$
$$I_{C0}=\frac{1}{X_C}V_{C0}\tag{5}$$
$$I_{L0}=\frac{1}{X_L}V_{L0}\tag{6}$$

where ##X_R, X_C##, and ##X_L## are resistive, capacitive, and inductive reactances.

These three expressions simply relate current and voltage amplitudes for three separate circuits, each with an AC source and a single circuit element (a resistor, a capacitor, and an inductor, respectively).

$$X_R=R\tag{7}$$

$$X_C=\frac{1}{\omega C}\tag{8}$$

$$X_L=\omega L\tag{9}$$

So again, from the rhs plot above, we get

$$V_0=I_0\sqrt{X_R^2+(X_L-X_C)^2}\tag{10}$$

and so the amplitude of the current is

$$I_0=\frac{V_0}{\sqrt{X_R^2+(X_L-X_C)^2}}\tag{11}$$

From the same rhs plot above we can determine that the phase constant satisfies

$$\tan{(\phi)}=\frac{1}{R}\left ( \omega L-\frac{1}{\omega L}\right )\tag{12}$$

Therefore the phase constant is

$$\phi=\tan^{-1} \frac{1}{R}\left ( \omega L-\frac{1}{\omega L}\right )\tag{13}$$

In the next post I show how I went about solving the differential equation (3).

#### Attachments

• 12 - Driven RLC Circuits.pdf
2.7 MB · Views: 12
Last edited:
Here is how I solved the differential equation (3).

Complexifying (3) we have

$$\ddot{z}_p+a\dot{z_p}+bz_p=ce^{i(\omega t+\phi)}\tag{14}$$

where ##a=\frac{R}{L}##, ##b=\frac{1}{LC}##, and ##c=\frac{V_0\omega}{L}##.

We try a solution ##z_p(t)=Ae^{i(\omega t+\phi)}## and find

$$A=\frac{c}{p(i\omega)}\tag{15}$$

where ##p(s)=s^2+as+b## is the characteristic polynomial for the differential equation.

##p(i\omega)## introduces another phase constant. This is probably not what we want, but I am not sure what to do about it.

$$p(i\omega)=-\omega^2+a\omega i+b\tag{16}$$

Therefore

$$A=\frac{c}{-\omega^2+a\omega i+b}=\frac{1}{\sqrt{a^2\omega^2+(b-\omega^2)^2}}ce^{-i\gamma}\tag{17}$$

Our complex solution is then

$$z_p(t)=Ae^{i(\omega t+\phi)}=\frac{1}{\sqrt{a^2\omega^2+(b-\omega^2)^2}}ce^{i(\omega t+\phi-\gamma)}\tag{18}$$

The real solution is

$$I_p(t)=\frac{1}{\sqrt{a^2\omega^2+(b-\omega^2)^2}} c\cos{(\omega t+\phi-\gamma)}\tag{19}$$

$$=\frac{V_0}{\sqrt{R^2+\left (\omega L-\frac{1}{\omega C}\right )^2}} \cos{(\omega t+\phi-\gamma)}\tag{20}$$

$$=\frac{V_0}{\sqrt{X_R^2+\left (X_L-X_C\right )^2}} \cos{(\omega t+\phi-\gamma)}\tag{20}$$

Note that

$$\gamma=\tan^{-1}\left (\frac{a\omega}{b-\omega^2}\right )\tag{22}$$

$$=\tan^{-1}\left ( \frac{R\omega C}{1-\omega^2LC}\right )\tag{23}$$

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In the calculations in post #2, the phase constant is actually ##\phi-\gamma##.

The expression for ##\tan^{-1}{(\gamma)}## is similar to the one found for ##\tan^{-1}{(\phi)}## in the notes (shown in post #1), but the argument of the ##\tan^{-1}## is inverted.

The phase difference between the voltage and the current is, however, just ##-\gamma##, so taking this into account my calculations seem to make sense (except for the detail of the argument of ##\tan^{-1}## being inverted)

Last edited:
The issue of my expression ##\gamma=\tan^{-1}{\left ( \frac{R\omega C}{1-\omega^2LC}\right )}## is also a problem because of the following.

We found that ##I_p(t)=\frac{V_0}{Z}\cos{(\omega t+\phi-\gamma)}## where ##Z## is the impedance

$$Z=\sqrt{X_R^2+(X_C-X_L)^2}=\sqrt{R^2+\left (\frac{1}{\omega C}-\omega L\right )^2}$$

and thus the amplitude of the current is maximized at the resonant frequency

$$\omega_0=\frac{1}{\sqrt{LC}}$$

However, when we use this frequency in the expression for ##\gamma## we get ##\tan^{-1}{(\infty)}##.

But according to the notes I am reading, the phase is zero when we have the resonant frequency.

And just to be clear. The notes find a phase constant of

$$\tan^{-1}{\left (\omega L-\frac{1}{\omega C}\right )}=\tan^{-1}{\left (\frac{\omega^2LC-1}{R\omega C}\right )}$$

whereas the phase constant I found in post #2 was

$$\tan^{-1}{\left ( \frac{R\omega C}{1-\omega^2LC}\right )}$$

Last edited:
zenterix said:
We try a solution ##z_p(t)=Ae^{i(\omega t+\phi)}## and find

$$A=\frac{c}{p(i\omega)}\tag{15}$$

where ##p(s)=s^2+as+b## is the characteristic polynomial for the differential equation.

##p(i\omega)## introduces another phase constant. This is probably not what we want, but I am not sure what to do about it.
What you can do about it is not try a solution ##z_p(t)=Ae^{i(\omega t+\phi)}##. The constant phase ##e^{i\phi}## is part of A at this point. So try ##z_p(t)=Ae^{i \omega t}## and see what happens.

kuruman said:
What you can do about is not try a solution zp(t)=Aei(ωt+ϕ). The constant phase eiϕ is part of A at this point. So try zp(t)=Aeiωt and see what happens.
Already did that. As I expected, it doesn't make any significant difference to the results.

The only thing that changes is that we remove the ##\phi## from the cosine in the expression for the current.

##\phi## is after all just an arbitrary offset for the AC voltage.

For the record, here are my calculations

zenterix said:
And just to be clear. The notes find a phase constant of

$$\tan^{-1}{\left (\omega L-\frac{1}{\omega C}\right )}=\tan^{-1}{\left (\frac{\omega^2LC-1}{R\omega C}\right )}$$

whereas the phase constant I found in post #2 was

$$\tan^{-1}{\left ( \frac{R\omega C}{1-\omega^2LC}\right )}$$
Note that your ##\gamma## and their ##\phi## are related by ##\gamma+\phi=\frac{\pi}{2}.## The tangent of your angle is equal to the cotangent of theirs. Just swap the right sides of your triangle to bring it into agreement with theirs. Your ##b-\omega^2## horizontal side corresponds to their ##V_{L0}-V_{C)}## vertical side. Conventionally, the phase angle is measured relatively to the voltage across the resistor which is in phase with the current.

zenterix
Ok, so as @kuruman noted we need need to deal correctly with the angles.

Here is what is happening in more detail.

Recall that the AC source voltage is

$$V(t)=V_0\sin{(\omega t+\phi)}$$

and the steady-state current we computed was

$$I(t)=I_0\cos{(\omega t+\phi-\gamma)}$$

Let's write the voltage in terms of cosine.

$$V(t)=V_0\cos{\left (\omega t+\frac{\pi}{2}+\phi\right )}$$

We can see that in fact the phase difference between the voltage and the current is ##\beta=\frac{\pi}{2}+\gamma##.

This is the phase constant we are interested in.

$$\tan{(\beta)}=\tan{\left (\frac{\pi}{2}+\gamma\right )}=(...)=-\cot{(\gamma)}=-\frac{1}{\tan{(\gamma)}}$$

We already know what ##\tan{(\gamma)}## is: it was computed when we solved the initial inhomogeneous differential equation.

Therefore,

$$\tan{(\beta)}=-\frac{1}{\frac{R\omega C}{1-\omega^2LC}}$$

$$=\frac{\omega^2LC-1}{R\omega C}$$

This last expression matches what was in the notes.

Also, it makes sense now since this is the tangent of the phase difference between the voltage and the current.

What I was initially computing was the phase difference between the current and the input to the differential equation

$$\ddot{I}+\frac{R}{L}\dot{I}+\frac{1}{LC}I=\frac{V_0\omega}{L}\cos{(\omega t+\phi)}$$

But this input isn't the same as the AC voltage.

zenterix said:
But this input isn't the same as the AC voltage.
You need to adjust inhomogeneous solution to make is match. You can always add any homogeneous solution to the particular solution to match the boundary condition.

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