Final answer:
The power factor of a load with impedance Z = 60 + j40Ω and an applied voltage of 320 cos(377t + 10)° V is 0.832. The apparent power is 710.61 VA, the real power is 591.23 W, and the reactive power is 356.22 var.
Step-by-step explanation:
To find the power factor (PF), apparent power (S), real power (P), and reactive power (Q) of a load with an impedance of Z = 60 + j40Ω when the applied voltage is 320 cos(377t +10)° V, we proceed as follows:
The impedance Z = 60 + j40Ω can be understood to have a resistance (R) of 60Ω and a reactance (X) of 40Ω. To calculate the power factor, you use the formula PF = R/Z, where Z is the magnitude of the impedance.
Magnitude of Z (|Z|) = √(R2 + X2) = √(602 + 402) = √(3600 + 1600) = √5200 = 72.11Ω
PF = cos(φ) = R/|Z| = 60/72.11 = 0.832, where φ is the phase angle.
Next, we calculate the apparent power (S) which is S = Vrms * Irms. Since we're given the peak voltage Vp of 320 V, Vrms = Vp/√2 = 320/√2 = 226.27 V. To compute Irms, we use Irms = Vrms/|Z| = 226.27/72.11 = 3.14 A. So, S = 226.27 V * 3.14 A = 710.61 VA (volt-amps).
Real power (P) is calculated using P = Vrms * Irms * PF = 226.27 V * 3.14 A * 0.832 = 591.23 W (watts).
Reactive power (Q) is calculated using Q = Vrms * Irms * sin(φ) = 226.27 V * 3.14 A * √(1 - PF2) = 226.27 V * 3.14 A * √(1 - 0.8322) = 356.22 var (volt-amps reactive).