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As shown in the figure, the power of inductive load p = 10kW, power factor cos & = 0, 6, connected to the power supply with voltage of 220V, power frequency f = 50Hz, if the power factor (2) if the power factor is increased from 0,95 to 1, how much should the capacitance of the shunt capacitor be increased?
1. Before compensation: capacity S1 = P / cos & 1 = 10 / 0.6 = 16.67 KVA
Reactive power = (kq1.5-13) * P
Line current I1 = S1 / u = 75.77 a
After compensation: capacity S2 = P / cos & 2 = 10 / 0.95 = 10.53 KVA
Reactive power Q2 = (S2 * s2-p * P) ^ 0.5 = 3.3 kvar
Line current I 2 = S2 / u = 47.86 a
Capacitance C = (q1-q2) * 1000 / u * u * 314 = 660.6 PF
2. Power increase to 1: capacity S3 = P / cos & 3 = 10 / 1 = 10 kVA
Reactive power Q3 = (S3 * s3-p * P) ^ 0.5 = 0 kvar
Line current I 2 = S3 / u = 45.45 a
Capacitance C = (q2-q3) * 1000 / u * u * 314 = 217.1 PF
A conductor has a resistance of 20 kiloohm and a voltage of 220 V applied to both ends of the conductor
I = u / r = 220 V / 20000 Ω = 0.011 a
You mean communication? If RMS current is calculated, 220 / 20000 is OK. If peak current is calculated, 220 * 1.414/20000 is OK
A small bulb marked with "220V 100W", its filament resistance is () ohm, and the current passing through it is ()
r=220*220/100=484
I=100/220=0.45
I=P/U=2.2,R=U/I=100
The 484 ohm current is about 0.45a
The voltage is 220 V, the power is 140 W, how to calculate the current
According to the formula I = P / u
I=140/220=0.64A
When the current passing through a 4 ohm resistor is 0.5A, the voltage at both ends of the resistor is (?) v,
The current through a 12 ohm resistor is () A. the resistance in the circuit is () ohm and the total voltage is () v
The resistance of 4 ohm and 12 ohm is connected in series in the circuit. When the current through the resistance of 4 ohm is 0.5A, the voltage at both ends is (2) V, and the current through the resistance of 12 ohm is (0.5) A. the resistance in the circuit is (16) ohm, and the total voltage is (8) v
A: the current through a 12 ohm resistor is (0.5) a. The resistance in the circuit is (16) ohm and the total voltage is (8) v.
A 200V 100W bulb is connected to the circuit of 220V. Calculate the filament resistance
The filament resistance is independent of the circuit to which it is connected,
R = u ^ 2 / P = 200 ^ 2 / 100 = 400 Ω
Strange, won't it burn? If it burns, the resistance is infinite
Rated current I = 100 / 200 = 0.5A
Filament resistance R = 200 / 0.5 = 400 Ω
When connected to 220 V circuit
Rated current I = 220 / 400 = 0.55a
The light bulb will burn out after a long time.
3. The output voltage of a substation is 220 V, and its apparent power is 220 kVA. For example, the output voltage is 220 V, the power factor is 0.8, and the rated power is 44 kW
3. The output voltage of a substation is 220 V, and its apparent power is 220 kVA. For example, to supply power to factories with voltage of 220 V, power factor of 0.8 and rated power of 44 kW, how many such factories can be supplied? If the user increases the power factor to 1, can the substation supply power to several similar projects? How to improve the power factor of the load? What is the significance of improving power factor?
220KVA = 220kw, 220kw * 0.8/44kw = 4, 220kw * 1 / 44kw = 5, that is to say, it can supply power for four such factories. If the user increases the power factor to 1, the substation can supply power for five similar projects
Seeking great spirit in calculation~
(- 4 7 / 9) - (- 3 1 / 6) - (+ 2 2 / 9) + (- 6 1 / 6) 5.6 + (- 0.9) + 4.4 + (- 8.1) + (- 1) (- 9) - [(- 10) - (- 2)] (- 1.5) - (1.4) - (3.6) + (- 4.3) - (5.2) - 3 / 4 * (- 1 and 1 / 3-0.4 + 12)
(- 4 7 / 9) - (- 3 1 / 6) - (+ 2 2 / 9) + (- 6 1 / 6) = (- 4 7 / 9 2 / 9) + (3 1 / 6 1 / 6) = - 7-3 = - 105.6 + (- 0.9) + 4.4 +
Don't understand. Is this a question? What grade are you in
A bulb marked with 220 V and 100 W is connected in series with a resistor in the 220 V circuit, and the electric power of the bulb becomes 81 W. calculate the electric power of the resistor in series
9W
The 1000 Watt heater is connected to the 220 V and 380 V circuit respectively, and the current passing through is
The current of 1000W heater is 4.545 (a) when it is connected to 220V circuit, while the current of 1000W heater is 2.631 (a) when it is connected to 380V circuit
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