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How many joules is one degree of electricity
1kw*h=3.6*10^6J
As shown in the figure, the resistance in the circuit R = 10 Ω, the coil resistance of motor M r = 1 Ω, and the voltage at both ends of the loading circuit u = 100V
As shown in the figure, the resistance in the circuit is r = 10 Ω, the coil resistance of motor M is r = 1 Ω, the voltage at both ends of the loading circuit is u = 100V, and the reading of the ammeter is 30A. What is the current intensity through the motor coil? What is the useful work of the current on motor M after being electrified for one minute?
Figure: m and resistance R in parallel, total voltage u, ammeter to measure the main circuit current
Why can't the current through M divide 100 by 1 ohm directly? (total voltage divided by resistance)
First calculate the current I = 100 / 10 = 10A through the resistance R, according to the parallel shunt, the current I = 30-10 = 20A through the motor M can be obtained; the active work of the motor refers to non thermal work, so first calculate the voltage U0 = 100-20 * 1 = 80V used to do the active work, so the useful power P = U0 * I = 80 * 20 = 1600W, one minute
The resistance of a motor is 4 ohm. When the motor runs under the rated voltage of 220 V, the power consumed by heating is 400 W. if the motor works for 5 min, the resistance will do work_____ J [the answer is 660000, worry!]
Heating power P = I2R = I2 * 4 = 400W
Current I = 10A
W=UIt=220*10*5*60=660000J
What is the resistivity range of semiconductors?
About: (10 ^ - 3 ~ 10 ^ 9) Ω. Cm
(others put forward: (10 ^ - 3 ~ 10 ^ 8) Ω. Cm)
How to calculate the power of storage battery according to its capacity
Battery capacity is usually expressed by "ampere / hour". For example, 20A / h means that the discharge of 1 ampere can last for 20 hours, and the discharge of 2 amps can last for 10 hours. Some are marked with the maximum discharge current, such as 10a, 12V battery * 10A = 120W. As long as the power is not greater than this, the electric appliance can drive
Connect a bulb marked with 12V 15W into a circuit. If the current it passes is one ampere, its actual power is greater than or equal to 15W
I = P / u = 15 / 12 = 1.25A
R=U/I=12/1.25=9.6Ω
P = 9.6w
Its actual power is 9.6 watts, less than 15 watts
In determining the resistivity of metals, what are the principles for selecting power supply ammeter, voltmeter, sliding rheostat, current limiting circuit or voltage dividing circuit?
In most cases, the choice of partial voltage can not be wrong, in some cases can not be used to limit the current. 1: the resistance voltage to be measured starts from 0. 2: the current exceeds the rated current of each component. 3. The resistance of sliding rheostat is too small, the change of sliding in the current limiting connection method is very small, and the change of the whole circuit is not big
How to calculate the full charge of the battery and how many degrees of electricity it can discharge
For example, a battery is 24 V with a capacity of 45 ah. How many kilowatts can it store when it is fully charged?
24V * 45A * 1H = 1080wh = 1.08kwh (degree)
A constant value resistor and a small bulb L1 marked with 6V and 6W are connected in series into a circuit with a certain supply voltage. It is known that the electric power consumed by the small bulb L1 is just 6W,
Remove the small bulb L1 and connect the small bulb L2 with the word "6V 9W" into the circuit, then the electric power of the small bulb L2 () a must be less than 9W, B must be greater than 9W, C must be less than 9W, D is lack of conditions and cannot be judged
Wrong. A must be 9 watts
A correct
D because the resistance of L1 is greater than that of L2, on the same power supply, the resistance of constant value resistor remains unchanged, and the power of variable part is the maximum when the resistance is equal to that of constant value resistor, while the relationship between their organization size remains unchanged, so there is no condition to judge whether you have not looked at the problem, L1 is removed. Yes, remove the small bulb L1 and replace it with small bulb L2. Regard L1 and L2 as variable resistors and connect them with a constant resistor. The power of the variable part has a maximum value, which is the maximum value when it is equal to the constant resistor
D because the resistance of L1 is greater than that of L2, on the same power supply, the resistance of constant value resistor is unchanged, and the power of variable part is the maximum when the resistance is equal to that of constant value resistor, while the relationship between the size of their organization is unchanged, so there is no condition to judge. Ask: didn't you look at the problem? L1 is removed.
During the experiment of "measuring metal resistivity", an experimental group correctly operated to obtain the diameter of metal wire and the readings of ammeter and voltmeter as shown in the figure, then their reading values are in the order of______ mm、______ A、______ V
The reading of spiral micrometer is d = 0.5mm + 49.9 × 0.01mm = 0.999mm; the reading of ammeter is I = 0.42a; the reading of voltmeter is u = 2.28v, so the answer is 0.999, 0.42, 2.28
RELATED INFORMATIONS
- 1. The resistance of a motor coil is 1 ohm, When the voltage applied at both ends of the coil is 2V, the current is 0.8A, and the motor works normally. Calculate the electric energy consumed by the motor in normal operation for 1min The second question is how much mechanical energy is converted by the motor working for one minute?
- 2. When a motor works normally, the coil voltage is 220V, the coil resistance is 2 ohm, the current is 10a1s, the electric energy consumed is j, and the heat generated in 1min is J
- 3. When the coil resistance of a motor is 1 ohm, the voltage applied at both ends of the coil is 2V, and the current is 0.8A, the electric energy consumed by the motor in normal operation for one minute is____ J, What is the converted mechanical energy J
- 4. In the motor model, the coil resistance is 1 ohm, the voltage at both ends is u = 2 V, and the current in normal operation is 0.8 ampere 1. Electric energy consumed in 1min when small motor works normally 2. The internal energy produced in 1min when the small motor works normally? 3. The mechanical energy produced in 1min when the small motor works normally? 4. Heating power of small motor in normal operation? 5. Total power of small motor in normal operation? 6. Output power of small motor in normal operation?
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- 9. What can an ammeter measure besides the current?
- 10. In the circuit shown in the figure, both ammeter and voltmeter are ideal ammeter. If resistance R1 is open circuit, the following statements are correct () A. The current indicates that the number becomes smaller; B. the voltage indicates that the number becomes larger; C. the power consumed by R3 becomes larger; D. the power consumed by the circuit in the power supply becomes larger
- 11. 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?
- 12. 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,
- 13. Find the formula of electric heating
- 14. Now I need 10 questions about physical electric power in grade two, Better not be a textbook~
- 15. Is electric work equal to electric energy
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- 17. The relationship between current, resistance, power and voltage
- 18. As the circuit shown in the figure, R1 = R2 = 12 ohm, the power supply voltage is 18 V, how much is the total resistance after R1 and R2 are connected in parallel? What is the indication of the ammeter? ┌——│R1 │————A————— │ │ │ │ │ │ │——│R2 │—— │ │ │ │ │ Quick switch
- 19. Put R1 and R2 in parallel in the circuit. The resistance of R1 is 5 ohm and that of R2 is 10 ohm. What is the total resistance
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