A series LR circuit dissipates an average power of 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°
B) 25°
C) 60°
D) 70°
E) 76°

Question

Quizplus · Accepted Answer

The average power dissipated in a series LR circuit can be expressed as:

P = I^2R = I^2X sin^2(θ)

where I is the rms current, R is the resistance, X is the inductive reactance, and θ is the phase angle between the current and the voltage.

Since the circuit is powered by a 120 V rms supply and the current is 2.0 A rms, we can calculate the total impedance of the circuit as:

Z = V/I = 120/2.0 = 60 Ω

We can also express the impedance in terms of resistance and reactance as:

Z = sqrt(R^2 + X^2)

Since the circuit has only resistance and inductance, we can write:

60 = sqrt(R^2 + X^2)

Squaring both sides:

3600 = R^2 + X^2

We are given that the average power dissipated in the circuit is 60 W. We can express the average power in terms of the rms voltage and current as:

P = VrmsIrmscos(θ)

Substituting the given values:

60 = 120*2.0*cos(θ)

Simplifying:

cos(θ) = 60/(120*2.0) = 0.25

Taking the inverse cosine:

θ = acos(0.25) = 76°

Therefore, the phase constant is 76°.

A Series LR Circuit Dissipates an Average Power of 60