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I know that the current in an inductor would lag by a 90 degrees' phase angle w.r.t. the voltage and the opposite would occur with capacitors; while in resistors, both of them would be in the same phase. But what if we have an inductor or a capacitor with some internal resistance? Would the current in the inductor lag/lead or remain in phase with the voltage? Why?

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Using a simple model, for an inductor with resistance the phase would be determined as for an ideal inductor in series with a resistor and for a capacitor with resistance you would be analysing an ideal capacitor in parallel with a resistor.

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Well, when we have an RL-circuit with the voltage being at the real axis we get:

$$\arg\left(\underline{\text{I}}_{\space\text{in}}\right)=\arg\left(\frac{\underline{\text{V}}_{\space\text{in}}}{\underline{\text{Z}}_{\space\text{in}}}\right)=\arg\left(\underline{\text{V}}_{\space\text{in}}\right)-\arg\left(\underline{\text{Z}}_{\space\text{in}}\right)=$$ $$0-\arg\left(\text{R}+\text{j}\omega\text{L}\right)=-\arctan\left(\frac{\omega\text{L}}{\text{R}}\right)\tag1$$

And for the RC-circuit, we get:

$$\arg\left(\underline{\text{I}}_{\space\text{in}}\right)=\arg\left(\frac{\underline{\text{V}}_{\space\text{in}}}{\underline{\text{Z}}_{\space\text{in}}}\right)=\arg\left(\underline{\text{V}}_{\space\text{in}}\right)-\arg\left(\underline{\text{Z}}_{\space\text{in}}\right)=$$ $$0-\arg\left(\text{R}+\frac{1}{\text{j}\omega\text{C}}\right)=-\arg\left(\text{R}-\frac{\text{j}}{\omega\text{C}}\right)=-\left(\frac{3\pi}{2}+\arctan\left(\frac{\text{R}}{\omega\text{C}}\right)\right)\tag2$$

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