Question

In a parallel circuit, ET = 277 V, R = 56 kΩ, and XL = 68 kΩ. What is the phase angle (angle theta)?

Answer 1

Phase angle is 50.527°

Explanation:

Given that in a parallel circuit
`E_(T)=277V`, R=56kΩ and
`X_(L) =`68kΩ.

And we are asked to find phase angle.

In any circuit, phase angle
`\theta=tan^(-1)(\frac{X_L-X_C} {R})`

Since we don't have any capacitor,
`X_(C)=0`

Hence phase angle,
`\theta=tan^(-1)((68000)/(56000))`

`=tan^(-1)(1.2143)`=50.527°≈50.53°