To solve for P (active power), Q (reactive power), S (apparent power), and power factor in a single-phase circuit with a given load, we need to use the following formulas:
P = V * I * cos(θ)
Q = V * I * sin(θ)
S = V * I
Where:
V is the voltage magnitude (in this case, 120 volts)
I is the current magnitude
θ is the phase angle between voltage and current
To calculate the current magnitude (I), we can use the impedance triangle:
Z = √(R^2 + X^2)
Where:
R is the resistance (5 ohms)
X is the reactance (2 ohms)
Plugging in the values:
Z = √(5^2 + 2^2) = √29 ≈ 5.39 ohms
Now we can calculate the current magnitude (I):
I = V / Z = 120 V / 5.39 Ω ≈ 22.27 A
Next, we need to find the phase angle (θ) between the voltage and current. Since the load is resistive and capacitive, the power factor is leading (due to the capacitive reactance).
θ = arctan(X/R) = arctan(2/5) ≈ 21.8 degrees (leading)
Now, we can calculate the active power (P), reactive power (Q), and apparent power (S):
P = V * I * cos(θ) = 120 V * 22.27 A * cos(21.8°) ≈ 2,408.8 W or 2.41 kW (rounded to two decimal places)
Q = V * I * sin(θ) = 120 V * 22.27 A * sin(21.8°) ≈ 890.9 VAr or 0.89 kVAr (rounded to two decimal places)
S = V * I = 120 V * 22.27 A ≈ 2,672.4 VA or 2.67 kVA (rounded to two decimal places)
Finally, we can calculate the power factor (PF):
PF = P / S = 2.41 kW / 2.67 kVA ≈ 0.903 (rounded to three decimal places)
The power factor of the circuit is approximately 0.903, indicating a leading power factor due to the capacitive load.