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?
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.