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In the study of microscopic particles, electron volt (EV) is often used as the unit of energy. 1eV is equal to the kinetic energy of an electron accelerated by 1V voltage. Please deduce the conversion relationship between electron volt and Joule
According to the electric field force work formula w = Qu: 1eV = 1.6 × 10-19 × 1J = 1.6 × 10-19j
The relationship between EV and Joule: 1eV = 1E * 1V = 1.60 * 10 ^ - 19C * 1V = 1.6 * 10 ^ - 19j
Why not negative?
Isn't the electron negative?
Here e refers to the amount of charge, a quantity
Joule is a scalar and there is no negative
In fact, negative charge is relative to positive charge, because they can be neutralized
In the study of microscopic particles, electron volt (EV) is often used as the unit of energy. 1eV is equal to the kinetic energy of an electron accelerated by 1V voltage. Please deduce the conversion relationship between electron volt and Joule
According to the electric field force work formula w = Qu: 1eV = 1.6 × 10-19 × 1J = 1.6 × 10-19j
When we study microscopic particles, we usually use electron volt (EV) as the unit of energy. 1eV is equal to the kinetic energy of an electron accelerated by 1V voltage, then
Why isn't a negative electron with a negative sign?
How can kinetic energy be negative? This is a unit, because electrons run from negative potential to positive potential. There is a negative sign on the voltage. Negative leads to positive
Connect a bulb with rated voltage of 220 V and rated power of 40 W to a 110 V circuit. What is its actual power?
R = U2 / P amount = 1210w, P = u actual square / r = 12100 / 1210 = 10W
Since P = u ^ 2 / R, r = u ^ 2 / P = 220 ^ 2 / 40 = 1210 Ω.
Actual power p '= u' ^ 2 / r = 110 ^ 2 / 1210 = 10W
The total output power is 0.5KW through the resistance of transmission line
How to calculate load voltage and line power loss?
Line current I = P / u = 5 * 1000 / 220 = 22.73 a
Load voltage U1 = u - I * R1 = 220 - 22.73 * 0.4 = 210.91 V
Line power loss: P1 = I * I * r = 206.66 w
Line current = P / V=
5000w÷220v≈22.73A
Line voltage drop = I × R=
22.73×0.4Ω≈9.1v
The load voltage is equal to the supply voltage minus the line voltage drop=
220v-9.1≈210.9v
Line loss = line voltage × line current=
9.1v×22.73A≈206w
The total power of Xiaoming household appliances is 3340 watts. The specifications of the air switch wires of the household energy meter are 220 V 40 A, 220 V 20 A and 220 V 20 a respectively
How to solve the problem by calculating the danger in use? (2) if Xiaoming's air conditioning and refrigeration works normally for 200 hours in summer, how much power consumption
1) P = UI; I = 3340 / 220 = 15.18a, meters and switches are enough. Choose 4 square copper wire
2) W = Pt; w = P (power consumption of air conditioning) × 200 = reference: junior high school physics book
The resistance of an electric furnace is 48.4 ohm, and it works in a circuit with a voltage of 220. What is the heat generated when it is electrified for 1 minute?
fast
Q = u square * t / r = 220 square * 60 / 48.4 = 60000j
Two identical "220 V 100 W" incandescent lamps are connected in parallel and connected in 220 V circuit. What is the total power of the two lamps
Two identical "220 V 100 W" incandescent lamps are connected in parallel in the 220 V circuit, both lamps can work normally, so the power of the two lamps is 100 W, and the total power of the circuit is 200 W
Parallel connection is 200W, series connection is 50W
Resistance of lamp r = u ^ 2 / P = (220 V) ^ 2 / 100 W = 484 Ω
After two lamps are connected in parallel, the total resistance R = 1 / 2R = 242 Ω
Total power P total = u ^ 2 / R total = (220 V) ^ 2 / 242 Ω = 200 W
How much cable is needed for 80 kW power?
It's 380V
380V, power 80kW, current about 160A, DIN VDE 0298-4 standard, 50 square line rated current is 168a, can be used
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