- #1

- 450

- 18

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

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)##?