Question

The instantaneous voltage across a circuit element is v(t) = 350 cos(t + 50°) volts, and the instantaneous current entering the positive terminal is i(t) = 125 cos(t + 30°) A. The instantaneous power p(t)=21875cos(20)-21875cos(2w0t+80) W

a.)Calculate the real power absorbed in kW (load convention)
b.)Calculate the reactive power absorbed in kvar (load convention)
c.) Give the power factor

Answer 1

a) The real power absorbed is approximately 8.75 kW (load convention). b) The reactive power absorbed is approximately 4.39 kVAR (load convention). c) The power factor is approximately 0.939.

Step-by-step explanation:

a) To calculate the real power absorbed, we need to find the average power. The average power is given by:

Pavg = (1/2) * Vrms * Irms * cos(θ)

Where Vrms is the root mean square voltage, Irms is the root mean square current, and θ is the phase angle between the voltage and current.

Using the given values, we have Vrms = 350/sqrt(2) volts, Irms = 125/sqrt(2) A, and θ = 50° - 30° = 20°. Substituting these values into the formula, we get:

Pavg = (1/2) * (350/sqrt(2)) * (125/sqrt(2)) * cos(20°)

Simplifying this expression gives us:

Pavg ≈ 8.75 kW

b) To calculate the reactive power absorbed, we can use the formula:

Q = Vrms * Irms * sin(θ)

Substituting the given values, we get:

Q = (350/sqrt(2)) * (125/sqrt(2)) * sin(20°)

Simplifying this expression gives us:

Q ≈ 4.39 kVAR

c) The power factor is given by:

pf = cos(θ)

Substituting the given value of θ, we get:

pf = cos(20°)

Simplifying this expression gives us:

pf ≈ 0.939