Physics of senior two! In the chapter of electromagnetic induction, we need to find Joule heat and use the conservation of energy. Why not calculate the work done by Ampere force?

Physics of senior two! In the chapter of electromagnetic induction, we need to find Joule heat and use the conservation of energy. Why not calculate the work done by Ampere force?

Because the ampere force appears in the form of resistance in the electromagnetic induction phenomenon, the ampere force received by the induced current always does negative work in the electromagnetic induction phenomenon. While the ampere force does negative work, it is the process of converting mechanical energy into electrical energy. From the perspective of energy conversion, the electromagnetic induction phenomenon is the process of converting mechanical energy into electrical energy
How many joules is one degree of electricity?
One kilowatt hour equals 1000 * 3600 = 3600000j
How many joules does one degree equal
I want to know how many joules 20 degrees will equal
Degree is the unit of electric energy in life, and Joule is the international unit of electric energy. 1 degree = 1 kwh. According to w = Pt = 1 kW * 1 hour = 1000 watts * 36000 seconds = 3600000 Joule, 20 degree = 20 kwh = 20 * 3600000 Joule = 72000000 Joule
How many joules is one degree of electricity
1 degree = 1 kWh (kW / h) 1W = 1 joule / S (J / s) 1 degree = 1000W * 3600s / h = 3600000 Joule
How much is the resistivity of common metals such as copper, iron and aluminum
The resistivity of common metal conductors at 20 ℃ (1) Ag 1.65 × 10-8 (2) Cu 1.75 × 10-8 (3) al 2.83 × 10-8 (4) w 5.48 × 10-8 (5) Fe 9.78 × 10-8 (6) Pt 2.22 × 10-7 (7) Mn Cu 4.4 × 10-7 (8) Hg 9.6 × 10-7 (9)
How to calculate battery capacity according to load power
(for example, I have a 200W load and need to be powered by the battery for 4 hours. How many ah - 48V batteries do I need to choose)
200W load, 48V battery current is 200 / 48 = 4.2a, you need four hours, so the capacity should be greater than 4.2a * 4H = 16.8ah
If a 12V 6W bulb is connected to a 36V power supply, how many Ω resistor needs to be connected in series for its normal lighting. The power consumption of the resistor is w
1. It is known that the maximum resistance of sliding rheostat is 1210 Ω, and the rated voltage of bulb is 220 V and 40 W
(1) Minimum voltage at both ends of bulb
2. A light bulb is marked with 9V 3W. It is connected in series with sliding rheostat R (20 Ω 2am) and ammeter (0-0.6a)
(1) The position of picture P remains unchanged. When S2 is closed, what is the indication of ammeter? What is the power consumed by sliding rheostat?
36-12 = 24 V, I = P / u = 6 / 12 = 0.5 A, a resistor R = u / I = 24 / 0.5 = 48 Ω needs to be connected in series
P=I^2R=0.5X0.5X48=12W
1. Resistance of bulb r = u ^ 2 / P = 220x220 / 40 = 1210 Ω
The resistance of sliding rheostat is also 1210 Ω
Obviously, when the resistance of the sliding rheostat is adjusted to the maximum, the resistance of the sliding rheostat and the bulb are both 1210 Ω, each sharing half of the voltage of 220 v
Therefore, the minimum voltage at both ends of the bulb is 110V. When the resistance of the sliding rheostat gradually decreases from 1210 Ω, the voltage at both ends of the bulb gradually increases from 110V
2. It's not easy to give you an answer without a picture
Known P = 6W u = 12V
∵ P = 6W u = 12V
∴R=UxU/P=12Vx12V/6w
∴R=24Ω
∵ connected in series with 36V power supply
∴12V/36V=24Ω/R
∴R=48Ω
One
∵ the bulb is connected in series with the sliding rheostat
And ∵ r max = 1210 Ω
When u light is minimum, u slide is maximum
∵ P amount = 40W u amount... Expansion
Known P = 6W u = 12V
∵ P = 6W u = 12V
∴R=UxU/P=12Vx12V/6w
∴R=24Ω
∵ connected in series with 36V power supply
∴12V/36V=24Ω/R
∴R=48Ω
One
∵ the bulb is connected in series with the sliding rheostat
And ∵ r max = 1210 Ω
When u light is minimum, u slide is maximum
∵ p value = 40W U value = 220V
∴R=1210Ω [ P=UxU/R ]
U lamp / u forehead max = R lamp / R forehead = 1 / 1
∵ u total = 220 V
The minimum value of u lamp is 110V
The second question is not easy to answer, dizzy typing to me for 20 minutes, I am also a junior high school student Oh, put it away
What is the resistivity of all kinds of metals
Material temperature T / ℃ resistivity temperature coefficient AR / ℃ - 1
AG 20 1.586 0.0038 (20 ℃)
Copper 20 1.678 0.00393 (20 ℃)
Au 20 2.40 0.00324 (20 ℃)
Al 20 2.6548 0.00429 (20 ℃)
Calcium 20 3.91 0.00416 (0 ℃)
Beryllium 20 4.0 0.025 (20 ℃)
Mg 20 4.450.0165 (20 ℃)
Iridium 20 5.3 0.003925 (0 ℃ ~ 100 ℃)
Tungsten 27 5.65
Zinc 20 5.196 0.00419 (0 ℃ ~ 100 ℃)
Cobalt 20 6.64 0.00604 (0 ℃ ~ 100 ℃)
Nickel 20 6.840.0069 (0 ℃ ~ 100 ℃)
Cadmium 200 6.83 0.0042 (0 ℃ ~ 100 ℃)
Indium 20 8.37
Fe 20 9.71 0.00651 (20 ℃)
Platinum 20 10.6 0.00374 (0 ℃ ~ 60 ℃)
Tin (0.0011 ~ 100 ℃)
Rubidium 20 12.5
Chromium 20 12.9 0.003 (0 ℃ ~ 100 ℃)
Gallium 20 17.4
18 0
Cesium 20
Lead 20.684 (0.0037620 ℃ ~ 40 ℃)
Antimony 0 39.0
20.0 titanium
Mercury 50 98.4
Mn 23 ~ 100 185.0
The resistivity of common metal conductors at 20 ℃ (1) Ag 1.65 × 10-8 (2) Cu 1.75 × 10-8 (3) al 2.83 × 10-8 (4) w 5.48 × 10-8 (5) Fe 9.78
How to calculate the power of 12v7ah battery
The ideal power is 12 * 7 = 84w
Generally, the battery has one hour rate after labeling, such as 12v7ah (20HR) or (12V 7.2A / 20HR)
Here, HR = hour rate, 20rh = 20 hour rate, that is, the battery needs constant current discharge for 20 hours to discharge 7Ah (power 84w)
7Ah 20 hour rate discharge current 0.36A discharge power 7.2ah
7Ah 10 hour rate discharge current 0.684a discharge power 6.84ah
The discharge current of 7Ah is 1.152a, and the discharged power is 5.76ah
7Ah 1 hour rate discharge current 4.68a discharge power 4.68ah
In addition, temperature is also one of the factors that affect the discharge power
The two bulbs of a "12V 8W" and B "6V 2W" are connected in series in a 12 V circuit, and the power is reduced
Two bulbs of a "12V 8W" and B "6V 12W" are connected in series in a 12 volt circuit. If one bulb lights normally and the other is darker than the normal light, which bulb works normally? What is the power supply voltage? What is the actual power of the other bulb? (regardless of the change of filament resistance)
This title is in conflict with the title! Is it 6v2w or 12W? I made it according to 2W. If it is connected in series, the resistance will be calculated first. If it is connected in series, the partial voltage is 6V. P = u * U / R. therefore, the power of a is 2W and the voltage is 6V. If it is connected in series, the resistance will be calculated first. If it is 18, the partial voltage will be 6V
For a lamp, the power supply voltage is 12V, and the voltage at both ends of the other bulb is 6V. The actual voltage is half of the rated voltage, so the actual power is 1 / 4 of the rated power, and the actual power is 2W
U1 = 12V, P1 = 8W, so I1 = P1 / U1 = 2 / 3a, U2 = 6V, P2 = 12W, then I2 = P2 / U2 = 2A
Because in the series circuit I = I1 = I2, the rated current is I1