Effect on phase angle due to resistance When an ac voltage source is applied to a resistor, current and voltage are in phase in the circuit.
When an ac voltage is applied to a pure inductor, voltage leads the current in the circuit by 90 degree.
But
When we apply ac voltage to a circuit where both resistor and inductor are in series, voltage leads the current not by 90 degree but by another angle.
My question is that when resistor is alone in circuit it doesnt effect the phase angle, then how can it causes change in phase angle of the overall current when it is with an inductor?
 A: Well, we have the following circuit:

If the input voltage is given by the following time depended function:
$$\text{V}_\text{in}\left(t\right)=\hat{\text{v}}\cos\left(\omega t+\varphi\right)\tag1$$
If we write that voltage in a complex way, we get:
$$\underline{\text{V}}_{\space\text{in}}=\hat{\text{v}}\exp\left(\varphi\text{j}\right)\tag2$$
Where $\text{j}^2=-1$.
The input impedance is given by:
$$\underline{\text{Z}}_{\space\text{in}}=\text{R}+\text{j}\omega\text{L}\tag3$$
So, the complex input current is given by:
$$\underline{\text{I}}_{\space\text{in}}=\frac{\underline{\text{V}}_{\space\text{in}}}{\underline{\text{Z}}_{\space\text{in}}}=\frac{\hat{\text{v}}\exp\left(\varphi\text{j}\right)}{\text{R}+\text{j}\omega\text{L}}\tag4$$
So, the time dependent input current is given by:
$$\text{I}_\text{in}\left(t\right)=\left|\underline{\text{I}}_{\space\text{in}}\right|\cos\left(\omega t+\arg\left(\underline{\text{I}}_{\space\text{in}}\right)\right)\tag5$$
Where:
$$\left|\underline{\text{I}}_{\space\text{in}}\right|=\frac{\hat{\text{v}}}{\sqrt{\text{R}^2+\left(\omega\text{L}\right)^2}}\tag6$$
$$\arg\left(\underline{\text{I}}_{\space\text{in}}\right)=\varphi-\arctan\left(\frac{\omega\text{L}}{\text{R}}\right)\tag7$$
