What is the equivalent of 1214000 kwh?

What is the equivalent of 1214000 kwh?

Kilowatt hour is usually called "degree", is the unit of electric work
For daily use, 1 kwh is 1 degree
1214000 kwh = 1214000 kWh
How many million kwh is 75711.2 kwh?
KWh = 75711.2 kwh = 7557112 kWh
One kilowatt hour is equal to one thousand watt hours
1 kwh = 1 kWh
75711.2 degree = 75711.2 kwh
Is one degree equal to one kilowatt hour
In physics, degree is defined as kilowatt hour
1 degree = 1 kWh
That is, a power of 1 kilowatt electrical appliances for 1 hour of continuous power consumption is 1 degree
The calculation formula is: power consumption (KWH) = power (kw) × service time (H)
The total resistance of two resistors in series is 32 ohm, and the total resistance of two resistors in parallel is 6 ohm
Let two resistors be R1 and R2 respectively, and the total resistance when two resistors are in series is (R1 + R2): R1 + R2 = 32 ① The total resistance of two resistors in series is (1 / R1 + 1 / r2): 1 / R1 + 1 / r2 = 1 / 6 ② Fang
R1+R2＝32
1/R1 + 1/R2 = 1/6
Solution: R1 = 24 Ω, R2 = 8 Ω
How many square wires do you need for 100A current``
And how big is the load switch
In a parallel circuit, it is known that resistance 1 is equal to 12 ohm, resistance 2 is equal to resistance 3, and the main current is 2 amperes. The current flowing through resistance 1 is 1 amperes. Calculate resistance 2 and resistance 3
R2 = R3, so the equivalent resistance of R2 and R3 in parallel is R2 / 2;
If the main current is 2a and the current flowing through R1 is 1a, the equivalent resistance flowing through R2 and R3 in parallel is 1a;
Therefore, the equivalent resistance of R2 and R3 in parallel is equal to R1, that is, R2 / 2 = R1;
So R2 = R3 = 2r1 = 24 Ω
It's simple. You don't have a self-study attitude. What's the point of giving you the answer? To tell you the truth, it's too simple. It's meaningless for you and me to help you. I don't want to waste my time. I'm willing to help you with the difficult points! After 90, wish you study hard!
Can you add which resistors are in parallel, or are they all in parallel. This problem is not difficult, the key point is that the voltage at both ends of the parallel circuit is equal. The voltage at both ends of resistor 1 is 12 V, which depends on whether there is series connection.
If there is no series connection and all are parallel connection, the result is as follows.
The voltage at both ends of resistor 1 is 12 volts. The total resistance in the circuit is 12 volts divided by 2 amps, making a total of 6 ohm. Then 1 / 6 = 1 / 12 + 1 / R + 1 / R, where R is the resistance of resistance 1 and resistance 2. The solution shows that r = 24 ohm.
That is the sum of resistances 2 and 3. ... unfold
Can you add which resistors are in parallel, or are they all in parallel. This problem is not difficult, the key point is that the voltage at both ends of the parallel circuit is equal. The voltage at both ends of resistor 1 is 12 V, which depends on whether there is series connection.
If there is no series connection and all are parallel connection, the result is as follows.
The voltage at both ends of resistor 1 is 12 volts. The total resistance in the circuit is 12 volts divided by 2 amps, making a total of 6 ohm. Then 1 / 6 = 1 / 12 + 1 / R + 1 / R, where R is the resistance of resistance 1 and resistance 2. The solution shows that r = 24 ohm.
That is the sum of resistances 2 and 3.
If I'm wrong, you can leave me a message. Can continue to answer, I hope you can be satisfied, thank you
Parallel, so the current flowing through the other two branches is 1a. If the resistance of the two branches is equal, the current will be divided equally, one is 0.5A. The resistance is inversely proportional to the current. There is R1 / r2 = 0.5/1, so R2 = 2r1 = 2 * 12 = 24 ohm
Known a induction cooker 380V voltage, power 15kw, how to calculate the current?
Motor power: P, voltage: u, power factor: cos φ, efficiency: η
Calculation of three-phase motor current:
I＝P/(1.732×U×cosφ×η)
But I don't know how much COS and η I give you,
P＝1.732×U×I×cosφ×0.8
I=P/(1.732×U×cosφ×0.8)
=15000(1.732×380×0.85×0.8)
=33.51
cosφ=0.85
η=0.8
When the two resistors are 9 ohm and 6 ohm in parallel, the circuit resistance is 0
The parallel resistance calculation formula is 1 / r = 1 / R1 + 1 / R2. 1 / r = 1 / 9 + 1 / 6 = 15 / 54. The reciprocal of the total resistance is equal to the reciprocal sum of the branch resistance, so it is 54 / 15 = 18 / 5 = 3.6 Ω. Do you understand!
If you are satisfied with my answer, I hope you can adopt it
How many square cables do you need for a three-phase motor with a total current of 100A
I'm not very clear about your question. (the sum of three-phase current of three-phase asynchronous motor should be zero, that is, the instantaneous value is o). So I can only give a self explanatory answer to your question: if the rated current of your motor is 100A, then the cross-sectional area of each phase line should be 25 square mm. The details also depend on the ambient temperature and heat dissipation, If the rated current of your motor is more than 30a, you can choose 10mm2 insulated wire
A light bulb and a resistance box are connected into the circuit. As shown in the figure, it is known that the supply voltage is 6 volts and the resistance of light bulb L is 6 ohm
How to adjust the resistance of the rheostat box if the voltage at both ends of the bulb L is to make the current in the circuit reach 0.5 A?
There should be a process
First question: u bulb = E / (6 + R variable resistance); (is to find the formula expression)
The second question: known I = 05A; R total = E / I = 6 / 0.5 = 12 Ω;
R rheostat = R total - R bulb = 12 - 6 = 6 Ω. Adjust rheostat box to 6 Ω
Series 6 ohm resistance, that is to say, the resistance value of the transformer box should be 6 ohm
What about the picture