A coil has a resistance of 75 and an inductance of 1.4 Henry. When the applied voltage is 240 Volt AC , at what frequency is the current 0.3 Ampere? What is the power factor at this frequency?

Answer 1

To solve this problem, we can use the formulas for impedance and power factor in an RL circuit:

Impedance (Z) = √(R^2 + (ωL)^2)

Power factor (PF) = cos(θ) = R/Z

where R is the resistance, L is the inductance, ω is the angular frequency in radians, and θ is the phase angle.

First, let's find the angular frequency using the given information:

I = V/Z, where I is the current, V is the voltage, and Z is the impedance.

Impedance (Z) = √(75^2 + (ω*1.4)^2)

0.3A = 240V / √(75^2 + (ω*1.4)^2)

Solving for ω:

(0.3A)^2 = (240V)^2 / (75^2 + (ω*1.4)^2)

0.09 = 240^2 / (75^2 + (ω*1.4)^2)

(75^2 + (ω*1.4)^2) = 240^2 / 0.09

75^2 + (ω*1.4)^2 = 240^2 / 0.09

(ω*1.4)^2 = (240^2 / 0.09) - 75^2

ω^2 = [(240^2 / 0.09) - 75^2] / (1.4^2)

ω = √([(240^2 / 0.09) - 75^2] / (1.4^2))

Now that we have found the angular frequency ω, we can calculate the power factor using the formula:

PF = cos(θ) = R/Z

PF = 75 / √(75^2 + (ω*1.4)^2)

Substituting the value of ω we found earlier, we can calculate the power factor.