A series LR circuit dissipates 60 W.The circuit is powered by a 120 V rms 60 Hz supply.If the current is 2.0 A rms,what is the phase constant?
A) 14$^{\circ}$
B) 25$^{\circ}$
C) 60$^{\circ}$
D) 70$^{\circ}$
E) 76$^{\circ}$

Question

Quizplus · Accepted Answer

The power dissipated in a series LR circuit can be calculated using the formula:

P = I^2 * R = I^2 * (ωL)^2

where I is the rms current, ω is the angular frequency (2πf) and L is the inductance.

We are given that P = 60 W, I = 2.0 A rms, V = 120 V rms and f = 60 Hz. Therefore, we can calculate the inductive reactance:

X_L = ωL = 2πfL

L = X_L / ω = X_L / (2πf) = (V/I) * sin(φ)

where φ is the phase angle.

Using these equations and the given values, we get:

60 W = 2.0 A^2 * (2π*60 Hz*L)^2

L = 0.04 H

φ = sin^-1(P / VI) = sin^-1(60 / (120 * 2.0)) = 76.04°

Therefore, the answer is E: 76°.

A Series LR Circuit Dissipates 60 W ∘^{\circ}∘ B) 25 ∘^{\circ}∘ C) 60

A Series LR Circuit Dissipates 60 W $^{\circ}$ B) 25 $^{\circ}$ C) 60