RLC circuit analysis

  • #1
Titan97
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Main Question or Discussion Point

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)##?
 

Answers and Replies

  • #2
f95toli
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Are you familiar with phasor notation?
 
  • #3
cnh1995
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Why are they considering a phase difference of ϕϕ\phi?
That phase difference is between voltage and current.
 
  • #4
Titan97
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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:
  • #5
SammyS
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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.
There's ##\ i\ ## and then there's ##\ i\ ## .

You probably should write ##\ i(t)\ ## for the current
 
  • #6
Titan97
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