How to obtain amplitude of current in parallel RLC circuit?

[zenterix]

Homework Statement

Consider the circuit below which is an RLC circuit with the circuit elements in parallel.

Relevant Equations

I would like to find the amplitude of the current ##I##.

If we apply KVL to the three separate loops involving the AC voltage we obtain expressions for ##I_R, I_L##, and ##I_C##.

$$-V(t)+I_R(t)R=0$$

$$\implies I_R(t)=\frac{V_0}{R}\sin{(\omega t)}$$

$$-V(t)=-L\dot{I}_L(t)$$

$$\implies I_L(t)=\frac{V_0}{\omega L}\sin{(\omega t-\pi/2)}$$

$$-V(t)+\frac{q(t)}{C}=0$$

$$\implies I_C(t)=V_0C\omega\sin{(\omega t+\pi/2)}$$

By KCL we have

$$I=I_R+I_L+I_C=\frac{V_0}{R}\sin{(\omega t)}+\frac{V_0}{\omega L}\sin{(\omega t-\pi/2)}+V_0\omega C\sin{(\omega t+\pi/2)}$$

How do we find the amplitude of ##I##?

In the notes I am following, they use phasor diagrams. I would like to know how to obtain the amplitude using analytical methods.

 

[DaveE]

Phasors (complex impedance) is one good way. You'll just do algebra instead of drawing pictures, I guess.

https://www.khanacademy.org/science...-analysis-topic/ee-ac-analysis/v/ee-impedance

 

[zenterix]

Finding the amplitude of ##I## is actually simple.

$$I(t)=V_0\left ( \frac{1}{R}\sin{(\omega t)}+\cos{(\omega t)}\left (\omega C-\frac{1}{\omega L}\right )\right )$$

where I have used the fact that

$$\sin{(\omega t-\pi/2)}=-\cos{(\omega t)}$$

$$\sin{(\omega t+\pi/2)}=\cos{(\omega t)}$$

Thus, we have an expression of the form ##a\sin{\omega t}+b\cos{\omega t}##.

If we set

$$a=A\sin{\phi}$$

$$b=A\cos{\phi}$$

then

$$A=\sqrt{a^2+b^2}$$

Thus, we can write

$$I(t)=a\sin{\omega t}+b\cos{\omega t}=A\sin{\phi}\sin{\omega t}+A\cos{\phi}\cos{\omega t}$$

$$=A\cos{(\omega t+\phi)}$$

where

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

which is the amplitude of the current.

 

[zenterix]

For the record, perhaps one might ask why I am not embracing phasor diagrams currently.

I guess it is because I don't have a solid grounding in using them.

For the example in the OP we have

It seems that the vectors are

$$\vec{I}_R=\frac{V_0}{R}\langle \cos{\omega t},\sin{\omega t}\rangle$$

$$\vec{I}_L=\frac{V_0}{\omega L}\langle \cos{(\omega t-\pi/2)},\sin{(\omega t-\pi/2)}\rangle$$

$$\vec{I}_C=V_0\omega C\langle \cos{(\omega t+\pi/2)},\sin{(\omega t+\pi/2)}\rangle$$

Apparently, we can add these vectors to get the vector for the current ##I##.

The diagram above seems to be the situation at times ##t=2\pi k## for integer ##k##.

The vectors are thus

$$\vec{I}_R=\langle V_0/R,0\rangle$$

$$\vec{I}_L=\langle 0,-V_0/\omega L\rangle$$

$$\vec{I}_C=\langle 0,V_0\omega C\rangle$$

and thus

$$\vec{I}_C+\vec{I}_L=\langle 0,V_0(C\omega+1/\omega L\rangle$$

which is the little pink vector in the picture above.

The black vector is the sum of all three phasors and is

$$\vec{I}=\langle V_0/R, V_0(C\omega+1/\omega L$$

The angle ##\phi## is the phase of the current and

$$\tan{\phi}=\frac{c\omega-1/\omega L}{1/R}$$

These are the same results we found algebraically.