In electromagnetic induction, is Joule heat transformed from the work done by Ampere force? In other words, in terms of energy conservation, the work done by Ampere force [assuming no friction] is equal to the kinetic energy obtained by conductor rod and Joule heat generated in closed circuit? Can you explain the energy transformation relationship clearly,

(the following answer is pure resistance circuit) work is a measure of energy conversion. As long as the external force does work, it will be accompanied by energy conversion. Therefore, the amount of work done by the external force will be converted into energy. Therefore, it is correct to say that joule heat is not converted from the work done by Ampere force, but through the work done by Ampere force, other forms of energy are converted into internal energy
Does ampere force produce Joule heat? What force produces heat
Work is a measure of energy transformation. Only by doing work can energy be transformed. (not transfer, energy transfer, not work) there are two ways to get heat. One is that heat is transferred from one object to another, and the other is that other forms of energy are transformed into heat by doing work. This is the reason why heat is generated by doing function
The work done by Ampere force must be the conversion of electric energy into other forms of energy. It can be converted into mechanical energy and also into heat energy. Therefore, the work done by Ampere force can produce Joule heat. If the mechanical energy of Ampere force does not change in the process of doing work, all the work done by Ampere force will be converted into Joule heat
There are questions about the work done by Ampere force and Joule heat,
Your problem is to make clear "work and energy conversion". Work and energy are a pair of interdependent physical concepts, just like the "edge and edge" of a knife. Definition of work: when a force acts on an object and makes the object move a certain displacement in the direction of the force, it is said that the force does work on the object. W = F* SCOS@. Work must be accompanied by energy conversion
12V 15W bulb current 1 a actual power
The resistance of 12V 15W bulb is: r = u ^ 2 / P = 12 * 12 / 15 = 9.6 ohm
The current is 1 A and the actual power is p = I ^ 2R = 1 * 9.6 = 9.6w
12W
R=12*12/15=9.6
P = 1 * 1 * 9.6 = 9.6, so the actual power is 9.6 watts
Actual power: 12 × 1 = 12W
Rated power: 15W
R = u ^ 2 / P = 12 × 12 / 15 = 9.6 Ω
Actual power P = I ^ 2 * r = 1 * 1 * 9.6 = 9.6 (W)
What's the difference between Ohm's law and Ohm's law in some circuits? What about resistance and resistivity?
Ohm's law I = u / R of partial circuit usually calculates the current and voltage at both ends of conductor and the relationship between current and resistance. Or the voltage of a certain branch in circuit and the relationship between current Ohm's law I = E / (R + R) where e is electromotive force, R is external circuit resistance, R is internal resistance of power supply, and internal voltage uine = IR
Is the motor universal for 48V and 60V electric vehicles
It can't be used in general use. The 48 volt motor can't be used in the 60 volt electric vehicle. The voltage difference is more than 20%. It's dangerous to burn the motor. But the 60 volt motor can be used in the 48 volt electric vehicle
A "6V, 6W" light bulb is connected in series with a certain resistance. The power supply voltage is 9V, and the light is normally emitting. Calculate the bulb resistance, R resistance and total power of the circuit
So first of all, it's pure resistance
(1) Bulb resistance = u ^ 2 / P = 6 ^ 2 / 6 = 6 Ω (because the lamp works normally, power P = 6, voltage U = 6)
(2) The power supply voltage is 9 V, the bulb voltage is 6 V, so the resistance voltage is 9-6 = 3 V
The series circuit current is equal, so the current is: bulb voltage / bulb resistance = 6 / 6 = 1a
So r = resistance, voltage / current = 3 / 1 = 3 Ω
(3) There are many ways to calculate the total power of the circuit, directly according to P = UI = 9 * 1 = 9
You can also add up the resistance and the power of the bulb,
Two uniform wires X and y of the same material, X length is l, y is 2L, are connected in series in the circuit, and the potential change along the length direction is shown in the figure, then the ratio of cross-sectional area of X and Y wires is ()
A. 2：3B. 1：3C. 1：2D. 3：1
According to Ohm's law u = IR, the ratio of resistance is rxry = 32. According to the resistance law, r = ρ ls, then s = ρ LR. Then the ratio of cross-sectional area is sxsy = ρ LRX: ρ· 2lry = 1:3. So B is correct, a, C and D are wrong. So B is selected
Do you want to change the electric motor from 60V to 48V?
Does electric vehicle motor need to distinguish 36v48v60v?
The voltage of one battery is 12V, two batteries are 24V, three batteries are 36V, four batteries are 48V and five batteries are 60V. If you don't change the motor, only the battery will burn the motor
The specific principle is as follows:
Current (I) = voltage (U) / resistance (R),
If u increases and R remains unchanged, I will increase,
And the heat generated (W) = I * I * r * t
So the heat will be doubled, it's easy to burn the motor
The electromotive force of the power supply is 12 V, the internal resistance is 1 ohm, R1 = 1 ohm, R2 = 6 ohm, the coil resistance of the motor is 0.5 ohm, the current is 3 ohm, and the power consumed by the motor
Voltage at both ends of motor: Um = U-I (R1 + R) = 12-3 * (1 + 1) = 6V
Current through R2: I2 = um / r2 = 6 / 6 = 1A
Current through motor: Im = i-i2 = 3-1 = 2A
So the power of motor: PM = um * im = 6 * 2 = 12W
If two motors are connected in series, even if there is no resistance, the problem should be solved. The meaning of the title is not clear, or there are other numbers wrong. Inquiry: R2 and motor parallel current 3A

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