Electric energy meter and electric power At present, I have used 3 * 10 (40) a, and now the equipment needs to be upgraded to use speed regulating motor 30kW + temperature control heating, with a total of 30kW. Can I barely use it for a period of time? If not, what may be the consequences?

Electric energy meter and electric power At present, I have used 3 * 10 (40) a, and now the equipment needs to be upgraded to use speed regulating motor 30kW + temperature control heating, with a total of 30kW. Can I barely use it for a period of time? If not, what may be the consequences?

1. U = 220 V, P = UI, I = 136.37 a > 120 a, and 120 a is only the maximum current that the meter can bear in a short time. So it can't. The result is to burn the meter

A certain resistance R is connected in series with a sliding rheostat whose maximum resistance is 2R, and the power supply voltage remains unchanged. When the resistance of the sliding rheostat connected to the circuit changes from zero to the maximum, the constant resistance R ()
A. The current is reduced to 2 / 3B. The voltage is reduced to 2 / 3C. The power consumed in the same time becomes 1 / 3D. The power consumed in the same time is only 1 / 9 of the original

The fixed resistance R and the sliding rheostat with the maximum resistance of 2R are connected to the circuit in series, and the power supply voltage remains unchanged. When the resistance of the sliding rheostat is zero, the current in the current I = ur, and the voltage at both ends of the fixed resistance u; when the resistance of the rheostat reaches the maximum, the resistance in the circuit is 3R, and the current I ′ = u3r, so option a is incorrect; the fixed resistance R The voltage at both ends u ′ = RI ′ = R × u3r = U3, so option B is incorrect. Because the current in the constant resistance R becomes the original 13, and the voltage at both ends of the resistance becomes the original 13, according to the formula w = uit, the power consumed in the same time is only the original 19, so option C is incorrect. Therefore, option D is selected

From 8:00 to 9:00 on March 26, 2011, Yueyang and many cities in the world participated in the "Earth Hour" activity. The activity proposed to turn off the unnecessary lights and cut off the power supply of TV, computer and other electrical appliances. Xiao Ming learned in the publicity of the activity: as long as the power plug is not pulled out, the power of each computer is 5.8W, and the power of each TV is 8.2w. (1) There are 100 computers and 100 TV sets in Xiaoming's community. If the original standby mode is changed to cut off the power for one hour, how many joules of electric energy can be saved. (2) the saved electric energy is completely absorbed by water, how many kilograms of 20 ℃ water can be raised to 60 ℃? [cwater = 4.2 × 103J / (kg ·℃)] (3) when the electric water heater marked with "220V & nbsp; 1000W" works normally, it takes 100min to raise the same temperature of water with the same mass as that in (2), what is the efficiency of the electric water heater? If the actual voltage of the electric heater is only 200V during the peak period, what is its actual power? (the resistance of the electric heater is constant, and only one decimal place is reserved)

(1) The power consumption of 100 computers and 100 TV sets in one hour w = Pt = (5.8W + 8.2w) × 100 × 3600s = 5.04 × 106j. A: a total of 5.04 × 106j can be saved. (2) because the saved power is completely absorbed by water, according to the deformation of q = cm (t-t0), the mass of water M = q absorption C (t − t0) = 5.04 × 10

A problem of electric energy meter and electric work calculation
After learning the section "electric energy meter and electric power", Xiao Ming went home to observe the electric energy meter at home and found that there was 3000r (kW. H) on it. He only turned on the bulb, and then counted the time it took for the turntable to turn three times. If the voltage at both ends of the bulb was 220V at this time, it would be better
(1) How much energy does this bulb consume in 36 seconds
(2) What is the current through this bulb

(1)3r/3000r（kw.h)=0.001（kw.h）
A: within 36 seconds, the power consumption of this bulb is 0.001 (kW. H)
(2) Known: u = 220 V, t = 36 s, w = 0.001 kW. H = 3.6 × 10 & # 179; J
Request: I
I=W/Ut=3.6×10³J/(220V×36s)≈0.45A
A: the current through the bulb is 0.45a