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

Question

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

1 Answer

Impact of this question

3868 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-06-22T22:27:46

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

$Z = \sqrt{R^{2} + X_{C}^{2}}$ $Z = sqrt(R^2 + X_C^2)$

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

$tan ϕ = \frac{X_{C}}{R}$ $tanphi = (X_C)/R$

thus the phase shift $ϕ$ $phi$ is:

$ϕ = arctan (\frac{X_{C}}{R})$ $phi = arctan((X_C)/R)$

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

Science

Math

Humanities

... and beyond

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

1 Answer

Related questions

Impact of this question