# When the resistance is greater than the capacitive reactance in a series RC circuit, what is the value of the phase angle?

Jun 22, 2015

The resistance ($R$), reactance (${X}_{C}$), and overall impedance ($Z$) are related by:

$Z = \sqrt{{R}^{2} + {X}_{C}^{2}}$

Thus, there is a right triangle such that $Z$ is the hypotenuse, and $R$ and ${X}_{C}$ are two legs. Using the angle between $Z$ and $R$, we have:

$\tan \phi = \frac{{X}_{C}}{R}$

thus the phase shift $\phi$ is:

$\phi = \arctan \left(\frac{{X}_{C}}{R}\right)$

Let's say that $R > {X}_{C}$. That means the argument of $\arctan$ is less than $1$, so $- \frac{\pi}{4} < \phi < \frac{\pi}{4}$.