In order to connect a "220V 1000W" electric furnace to a 110V power supply, the power consumed is still 1000W, which can be used 2. In order to connect a "220V 1000W" electric furnace to a 100V power supply, the power consumption is still 1000W, the method that can be adopted is () A. Cut 1 / 2 of the electric furnace wire into the circuit B. The electric furnace wire is folded in half and connected to the circuit C. The electric furnace wire is divided into three equal parts, and then connected to the circuit in parallel D. Cut off 3 / 4 of the furnace wire and connect it to the circuit

B
Connected to the 110 V power supply, the consumed power is still 1000 watts, and the calculated resistance is 12.1 ohm
The resistance of a "220V 1000W" electric furnace is 48.4 ohm. If you want to get 12.1 ohm, you can only fold the electric furnace wire and connect it to the circuit. After folding, each wire is 24.2 ohm, and then you can get 12.1 ohm in parallel
If a straight conductor is 2 m long, carrying a current of 5A and forming a 30 degree angle with a uniform magnetic field of B = 1.5T, the ampere force on the conductor is
A、 7.5N B、15N C、6.5N D、13N
Correct answer a
From F = iblsin30 ^ 0 = 7.5n
One appliance, 250 watts, how many kilowatts per hour
0.25 kwh per hour
The TV is 1000W / h, so it's one kilowatt hour for electric appliances
0.25 kWh
Theoretically, the electric appliance uses 0.25 degree electricity per hour.
1 kilowatt hour = 1 kilowatt hour
250 watts of electrical appliances, use 4 hours, that is 1 degree of electricity
So 250 Watt appliances use 0.25 kilowatt hour
The power of an electric furnace under the rated voltage of 220 V is 1000 W, and its resistance is 0______ When the actual voltage is only 80% of the rated voltage, if the resistance of the furnace remains unchanged, the actual power of the furnace is______ .
According to p value = U value 2R, r = U value 2p value = (220V) 21000w = 48.4 Ω, when U value = 80% U value, the resistance is unchanged, then the actual power: P value = (80% U value) 2R = 0.64 × U value 2R = 0.64p value = 0.64 × 1000W = 640w
In the uniform magnetic field of B = 2T, put an electrified straight wire with a length of 0.8m perpendicular to the direction of the magnetic field, and the current in the wire is 5A. Find out: (1) is the direction of the magnetic force on the wire parallel to or perpendicular to the direction of the current? (2) What is the magnitude of the magnetic force?
(1) If the length of 0.8m electrified straight wire is vertically placed in a uniform magnetic field and the input current is 5a, the direction of Ampere force can be determined by the left-handed rule to be perpendicular to the current direction. From the formula, the magnitude of Ampere force is f = bil = 2 × 5 × 0.8N = 8N. A: (1) vertical & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; (2) The magnitude of Ampere force is 8N
What does "12V, 100Ah / 20HR" on the battery mean
12V is the voltage of the battery
The battery capacity is 100Ah times the standard time
20HR is the discharge rate of the battery, which can be understood as the maximum output power. 100Ah / 20HR = 5a, which means the maximum allowable output current is 5A
An electric furnace marked with "220 V, 600 W" can be calculated as follows: (1) the current of the electric furnace in normal operation and the power consumption for 5 hours;
(2) If it is connected to a 110V circuit, what is its current and actual power?
Don't just give one answer. Please
W=Pt=0.6*5=3kW*h
The resistance of the filament does not change with temperature. The resistance is constant. It is known from P = u ^ 2 / R. P is proportional to u ^ 2
P‘/P=（110/220）^2=1/4 P'=1/4P=1/4*600=150W
As shown in the figure, there is a uniform magnetic field, and the magnetic induction intensity is b = 1t. There is a section of wire with length of L = 0.1M placed perpendicular to the direction of the magnetic field. When I = 0.5A horizontal right current is applied in the wire, (1) judge whether the direction of Ampere force on the wire is vertical to the wire upward or downward; (2) calculate the magnitude of Ampere force F on the wire
(1) According to the left-hand rule, the direction of Ampere force on the wire is perpendicular to the wire upward. (2) ampere force on the wire: F = bil = 1 × 0.5 × 0.1N = 0.05N A: (1) the direction of Ampere force on the wire is perpendicular to the wire upward; (2) the magnitude of Ampere force on the wire is 0.05N
Popcorn machine, power 220 V, 50 Hz, power 1.3 kW, how to use battery power supply?
I have an electric popcorn machine, "economical 8-Ampere popcorn machine, power 220 V, 50 Hz, power 1.3 kW". But now I want to sell in the street, there is no stable AC power supply, so I want to use battery and inverter for power supply. Excuse me, is there such a battery and inverter, how to operate it?
This machine can't use battery, and this machine is mainly heated by electricity. If you want to do mobile business, you can choose gas heating
A certain electric furnace is marked with the word "220V, 1000W". If it is connected to a 220V circuit, how much electric energy does the electric furnace consume when it works normally for one hour
t=1h
Rated power P = 1000W = 1KW
Because the electric furnace is marked with the word "220V, 1000W", connect it to the 220V circuit, and the electric furnace works normally
Therefore, the power consumption for one hour is: w = Pt = 1kW × 1H = 1kW. H = 1 degree
A: slightly
1 kwh,
1kW / h is one degree
1 degree
Electric energy is electric work, that is, electric current does work
The small lamp lights normally, so the power is 1000W
Then w = Pt = 1000 * (3600) = 3.6 * 10 ^ 6 J = 1 kwh
That's one degree of electricity
1000w=1kw
w=Pt=1Kw×1h=1Kw·h
1000 W is 1000 watts
1 degree = 1000W / h
That is, the electric energy consumed by one kilowatt working for one hour
Therefore, the furnace works normally for 1 hour and consumes one degree of electric energy
1 kWh
One degree
One kilowatt hour is 1kW / h. That is to say, the electricity consumption of 1kW electrical appliances in an hour is 1 degree.

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