On the 220 V line, 20 40 W fluorescent lamps with power factor of 0.5 and 100 40 W incandescent lamps are connected in parallel. The total current, total active power, reactive power, apparent power and power factor of the line are calculated

On the 220 V line, 20 40 W fluorescent lamps with power factor of 0.5 and 100 40 W incandescent lamps are connected in parallel. The total current, total active power, reactive power, apparent power and power factor of the line are calculated

U = 220 V, P = 20 * 40 W = 800 W, cos φ = 0.5, I = P / u * cos φ = 800 / 220 * 0.5 = 800 / 110 = 7.27 a, 2. U = 220 V, P = 100 * 40 W = 4000 W, cos φ = 1 (incandescent lamp), I = P / u = 4000 / 220 = 18.18 a, total power P = 1 + 2 = 800 + 4000 = 4800 W, total current I = 1 + 2 = 7.27 + 18

A 40 W fluorescent lamp is connected in series with a ballast and connected to a 220 V AC power supply. The current is 0.41 a. the power factor of the fluorescent lamp is calculated

40W fluorescent lamp and ballast are connected in series, the loss of ballast itself is less than 8W, and the total power is about 47.5w based on 7.5W
Power factor of fluorescent lamp:
P＝U×I×cosΦ
cosΦ＝P/(U×I)＝47.5/(220×0.41)≈0.53

In the three-phase symmetrical circuit, the measured value of power supply voltage is 380V, the current is 10a, and the power factor is cos φ = 0.8. The apparent power, active power and reactive power are calculated, and the vector diagram is drawn

The apparent power s = 1.732 * u * I, in VA
The active power is calculated according to P = 1.732 * u * I * cos φ
Reactive power q = 1.732 * u × isin φ
According to sin φ square + cos φ square = 1, the reactive power is calculated
What about vector graphs,