RLC circuit analysis

• B
Gold Member
In an RLC series circuit let applied EMF be given ##V=V_0\sin\omega t##, $$Z=Z_C+Z_R+Z_L=R+i\left(\frac{1}{\omega C}-\omega L\right)$$
$$|Z|=\sqrt{R^2+\left(\frac{1}{\omega C}-\omega L\right)^2}$$

Then $$i(t)=\frac{V(t)}{Z}=\frac{V_0e^{i\omega t}}{R+i\left(\frac{1}{\omega C}-\omega L\right)}$$

Its given in my book that
$$i(t)=\frac{V_0(\sin\omega t+\phi)}{\sqrt{R^2+\left(\frac{1}{\omega C}-\omega L\right)^2}}$$

Why are they considering a phase difference of ##\phi##?

Also, why are they taking modulus of ##Z## and only the imaginary part of applied voltage?

What is the difference between the first ##i(t)## and the second ##i(t)##?

f95toli
Gold Member
Are you familiar with phasor notation?

cnh1995
Homework Helper
Gold Member
Why are they considering a phase difference of ϕϕ\phi?
That phase difference is between voltage and current.

Gold Member
I get it. I can write $$R+i\left(\omega L-\frac{1}{\omega C}\right)=\sqrt{R^2+\left(\omega L-\frac{1}{\omega C}\right)^2}e^{i\phi}$$
Hence,
$$i(t)=\frac{V(t)}{\sqrt{R^2+\left(\omega L-\frac{1}{\omega C}\right)^2}e^{i\phi}}$$
$$i(t)=\frac{V(t)e^{-i\phi}}{\sqrt{R^2+\left(\omega L-\frac{1}{\omega C}\right)^2}}$$

Now it's in phasor notation.

Last edited:
SammyS
Staff Emeritus
I get it. I can write $$R+i\left(\omega L-\frac{1}{\omega C}\right)=\sqrt{R^2+\left(\omega L-\frac{1}{\omega C}\right)^2}e^{i\phi}$$ Hence,$$i(t)=\frac{V(t)}{\sqrt{R^2+\left(\omega L-\frac{1}{\omega C}\right)^2}e^{i\phi}}$$$$i(t)=\frac{V(t)e^{-i\phi}}{\sqrt{R^2+\left(\omega L-\frac{1}{\omega C}\right)^2}}$$ Now it's in phasor notation.