When a motor works normally, the coil voltage is 220V, the coil resistance is 2 ohm, the current is 10a1s, the electric energy consumed is j, and the heat generated in 1min is J

When a motor works normally, the coil voltage is 220V, the coil resistance is 2 ohm, the current is 10a1s, the electric energy consumed is j, and the heat generated in 1min is J

W=I*I*R*T=10*10*2*1=200(J)
Q=I*I*R*T=10*10*2*60=12000(J)
200 joules per second, 12000 joules per minute
The resistance of a motor coil is 1 ohm. When the voltage applied at both ends of the coil is 2V, the current is 0.8A, and the motor works normally. Calculate the power consumption in 1min
p=0.8*0.8*1*1/60=0.01wh
When the resistance of a motor coil is 1 ohm and the voltage applied at both ends of the coil is 2V, the current is 0.8A, and the motor works normally,
When the resistance of a motor coil is 1 ohm and the voltage applied at both ends of the coil is 2V, the current is 0.8A, and the motor works normally, calculate the mechanical electric energy converted in 1min?
Electric energy = u * a * t = 2 * 0.8 * 1 / 60 = 0.027w
Heat energy = I ^ 2 * r * t = 0.8 ^ 2 * 1 * 1 / 60 = 0.011w
Mechanical energy = electric energy thermal energy = 0.016w
W (Mechanical) = w (Electrical) - Q
=Uit-i ^ 2 RT / / ^ 2 for square
=2V×0.8A*60s-(0.8A)^2×1Ω×60s
=96J-38.4J
=Question: you're right!! I should give it to you!! % >
What is the resistivity of copper winding at 75? Or how to calculate it?
The resistivity of pure copper is 0.0175 ohm. Mm2gm and the temperature coefficient of resistance is 0.00393 / ℃
ρ=ρo(1+at)
Where ρ - resistivity at t ℃
ρ O -- resistivity at zero centigrade
T -- temperature
Then the resistivity of copper winding at 75 degrees is 0.0172 (1 + 0.00393 * 70) = 0.0219 ohm. Mm2 / m
When the ordinary light bulb of "pz220-40" is connected to 220 V and 110 V respectively, what is the actual power consumption of the light bulb in the circuit?
In the circuit of "pz220-40" ordinary bulb connected to 220 V and 110 V respectively, what is the actual power consumed by the bulb? (assuming that the resistance of the filament does not change with the change of temperature)
When the ordinary bulbs of "pz220-40" are respectively connected to 220 V, the actual electric power consumed by the bulbs in the circuit is 40 W, so r = uxu / w = 220 x220 / 40
In the circuit of "pz220-40" ordinary bulb connected to 110V, the actual power consumed by the bulb is w = uxu / r = 110x110 / (220x220 / 40) = 10W
PZ220-40,
When receiving the voltage of 220 V, it is 40 W rated power
When receiving 110V voltage, it is 40 / 4 = 10W, because the power of the lamp is proportional to the square of the voltage.
R = U2 / P P real = U2 / R
You can use the proportional method. The first 220 V works normally, so the power is 40 W
Because it's the same light, R must be
Because P = u ^ 2 / R
So r = u ^ 2 / P
So 220 ^ 2 / 40 = 110 ^ 2 / P real
There is the word "220V 10A 600r / kWh" on a household electric energy meter. The electric energy meter turns six times in one minute to calculate the power
600r / kWh = 600 cycles / degree, 6 cycles per minute, 360 cycles per hour, 360 / 600 = 0.6 degrees, that is to say, the power consumption per hour is 0.6 degrees
The electric power of an electric appliance is 100 watts, and its physical meaning is
Electrical appliances work 100 joules per second
Electrical appliances do 100 joules of work per second.
Electrical appliances work 100 joules per second
How to calculate the resistivity of conductor
R is the resistance, l is the length of the conductor and S is the cross-sectional area of the conductor
From the formula r = ρ * L / s, ρ = RS / L (ρ is the resistivity of conductor)
When a lamp with a rated voltage of 220 V is connected to a 110 V circuit, the actual power consumption is 25 W, then the rated power of the bulb is 25 W______ w．
The actual voltage of the bulb is u = 110V, the actual power is p = 25W, according to P = u2r, the resistance of the bulb is r = u2p = (110V) 225W = 484 Ω, ∵ the actual power of the bulb is equal to the rated power under the rated voltage, ∵ the rated power of the bulb is p = u, 2R = (220V) 2484 Ω = 100W
What do you think of the 3 * 220 V / 380 V 3 * 10 (40) a 800 imp / kWh, 50 Hz meter?
3 * 220 / 380 means: applicable to 3-phase, working voltage of each phase is 220-380 v
3 * 10 (40) a means: the working current of each phase is 10a, and the maximum is 40A
800imp / kWh means that the rotary table rotates for 800 turns for 1 kwh
50 Hz means that the alternating frequency is 50 Hz, and the 220 or 380 V used in China is 50 Hz