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Solving problems about electric work, electric power or Joule's law
Calculation of electric work, electric power or Joule's law
1、 Fill in the blanks:
1. If 3 Coulomb of electricity passes through a consumer in 5 seconds and the current does 18 joules of work, the current intensity passing through the consumer is a; the voltage at both ends of the consumer is v
2. Electric heater uses the effect of current to work. The main component of electric heater is that it is made of resistance wire with resistivity and melting point
3. The electric energy of 1 kilowatt hour can supply the normal working hours of "220 V, 100 W" electric soldering iron
4. A "220 V, 40 W" incandescent lamp works at rated voltage every minute
2、 Multiple choice questions:
1. In the home circuit, the physical quantities usually measured by the electric energy meter are:
A. Electric power B. electric work C. voltage D. current intensity
2. Among the following units, the unit of electrical work is:
A. Kilowatt B. kilowatt h C. Joule D. Watt
When two lights marked "220 V, 40 W" and "36 V, 40 W" light up normally, compare their brightness, then: A, "220 V, 40 W" light is brighter; B, "36 V, 40 W" light is brighter; C, two lights are equally bright; C, two lights are equally bright; D. It is impossible to judge which light is brighter
2. If the power of the electric wire is 20 watts, it will work normally
If the light is bright, then: - - - - - - - - - - - - - (a), "220 V, 40 W" light is brighter; B, "36 V, 40 W" light is brighter; C, two electric lights are equally bright; D, it is impossible to judge which electric light is brighter. 2. When a section of resistance wire works normally, the electric power is 20 watts. If this section of resistance wire is evenly elongated, the, If it is connected to the original power supply, its actual power will be: () a, equal to 20W; B, greater than 20W; C, less than 20W; D, can not be compared. 3. The electrical appliance of a is marked with "220V, 60W", and the electrical appliance of B is marked with "220V, 15W". They all work under the rated voltage, so the correct judgment is as follows: - () a, electrical appliance of a does more work; B. The work done by the appliance a must be fast; C. It takes more time for the appliance a to complete the same work; D. It takes less time for the appliance a to complete the same work. 4. The appliances a and B are marked with "220 V, 15 W" and "110 V, 100 W" respectively. They all work under the rated voltage, 5. A lamp is marked with "220 V, 100 W". If it is connected to both ends of a 110 V power supply, it will be connected to a 110 V power supply, The following statements are correct: A, its rated power becomes half of the original; B, the current intensity through it becomes half of the original; C, its actual power becomes 1 / 4 of the original; D, its actual power becomes 1 / 2 of the original, Bulb L and a certain value resistor are connected in parallel at both ends of a power supply u, and the voltage U remains unchanged. When the electric key is turned off to close, the brightness of lamp l will be: - () a, bright; B, dark; C, constant brightness; D, unable to judge. 7. After bulb L and a sliding rheostat are connected in series, close the electric key to make sliding rheostat slide P from end a to end B In the process of end sliding, the brightness of lamp l will be () a, from bright to dark; B, from dark to bright; C, brightness unchanged; D, unable to judge. 8. Two electric heating wires, R1: R2 = 1:2, through their current intensity ratio I1: I2 = 2:3, in the same time, the heat release ratio Q1: Q2: - - - (a), 1:3; B, 3:1; B, 3:1; C. The ratio of the heat generated by the two electric heaters in series is 25:9, 4:9, 25:9, In order to make the power of the electric heater half of its original value, the following methods are correct: A. keep the voltage unchanged, If half of the original resistance is cut off, the voltage becomes the same as that of the original resistance, 13. If the voltage at both ends of a circuit is increased by three times and the resistance in the circuit is reduced by one half, the current will change in the same time
14. The four units listed below, which are not electric power units, are: A, Watt; B, kilowatt hour; C, Joule / second; D, volt ampere 15. There is a "220 V, 5 a" electric energy meter, which can be connected with "22 ov, 40 W" electric lamps at most according to the regulations: - - - - - - - - - - - - - - - - (a), 25; B, 27; C, 28; D, 30; C, 28; D, 30; a, 25; B, 27; C, 28; D, 30; 16. The main basis for judging the brightness of an electric lamp is: - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - (a), electric power; B, electric power; C, current intensity; D, voltage. 17. The resistance of bulb L1 is twice that of bulb L2, Then the electric power ratio of bulb L1 and L2 is: - - -) a, 1:2; B, 2:1; C, 1:4; D, 4:1.18. As shown in the figure, the power supply voltage of the circuit remains unchanged. After closing the key K, if the slide P of the sliding rheostat is moved to the right, the voltage of the rheostat u, The change of the current intensity I and the electric power P of the constant resistance R in the circuit is as follows: - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - (a, u increase, I increase, P increase; B, u decrease, I decrease, P decrease; C, u increase, I decrease, P decrease; D, u remain unchanged, I decrease, 19. Three electric lamps with specifications of "220 V, 40 W", "36 V, 40 W" and "110 V, 40 W" respectively work under the rated voltage, and their luminous conditions are as follows: - () a, "220 V, 40 W" electric lamp is the brightest; B, "36 V, 40 W" electric lamp is the brightest; C, "110 V, 40 W" electric lamp is the brightest; D, three electric lamps are the same, When a "18W" bulb is connected to a certain circuit, the measured current intensity through the bulb is 0.4 a, then its actual power is: - - -) a, less than 18 watts; B, more than 18 watts; C, equal to 18 watts; D, equal to 11.52 watts. 21. In the two circuits as shown in the figure, the power supply voltage remains unchanged, when the slide P of the rheostat slides to the right, the, The brightness of L1 and L2 will be as follows: - - - - - (a), all will be bright; B, all will be dark; C, L1 will be bright, L2 will be dark; D, L2 will be dark, L1 will not be changed. 22. After two wires with the same resistance are connected and connected to the power supply with constant voltage, the heat generated in series is 1 / 4 times of that in parallel in the same time; B. 23. After a "220 V, 100 W" electric lamp is connected in parallel with a resistor, it is connected to the 220 V line. If the voltage remains unchanged, the brightness of the electric lamp will be: - - - (a), bright; B, dark; C, unchanged; D, undetermined. 24. After the electric furnace is powered on, the furnace wire is hot and red, But the copper wire connected with the electric furnace wire is not so hot, because: - - - - - - - - - - - - - (a) the current through the electric furnace wire is large, but the current through the copper wire is small; b) the specific heat of the electric furnace wire is larger than that of the copper wire; c) there is the thermal effect of the current in the electric furnace wire, but there is no thermal effect of the current in the copper wire; D. The resistance of the electric furnace wire is much greater than that of the copper wire. 25. After the filament of the incandescent lamp is broken, if it is put on again, it will still emit light. At this time, the electric power consumption is compared with that of the original: - - - - - - (a) increases; b) decreases; and;
C. D. the increase or decrease of electric power consumption depends on the position of filament lap
If the area of a right triangle is 6, then the analytic expression of the function between the lengths of its two right angles y and X is?
y=12/x
Prove (a1 + A2 +... + an) * (1 / A1 + 1 / A2 +... 1 / an) > = n ^ 2 by mathematical induction
Where a1, A2,... An are positive integers
Please don't occupy the space, leave the opportunity to the friends who have the ability and patience
Prove (a1 + A2 +... + an) * (1 / A1 + 1 / A2 +... 1 / an) > = n ^ 2 by mathematical induction
prove:
When n = 1, A1 * (1 / A1) = 1 > = 1 ^ 2 holds
Suppose that k = n is true
That is: (a1 + A2 +... + AK) * (1 / A1 + 1 / A2 +... 1 / AK) > = k ^ 2
Then when n = K + 1,
(a1+a2+...+ak+a)*(1/a1+1/a2+...1/ak+1/a)
=(a1+a2+...+ak)*(1/a1+1/a2+...1/ak)+a*(1/a1+1/a2+...1/ak)+1/a*(a1+a2+...+ak) +1
>=K ^ 2 + A * (1 / A1 + 1 / A2 +... 1 / AK) + 1 / A * (a1 + A2 +... + AK) + 1 {the conclusion when n = k}
>=K ^ 2 + 2 * radical [a * (1 / A1 + 1 / A2 +... 1 / AK) * 1 / A * (a1 + A2 +... + AK)] + 1 {arithmetic mean not less than geometric mean}
=K ^ 2 + 2 * radical [(1 / A1 + 1 / A2 +... 1 / AK) * (a1 + A2 +... + AK)] + 1 {the conclusion when n = k}
>=k^2+2*k+1
=(k+1)^2
Therefore, when n = K + 1, the proposition holds
The proposition is proved
The following bracket is greater than 2 when n = 4. Then the product of the following bracket and the preceding bracket is greater than n (n + n). Greater than holds.
The difference between electric power, power and electric power
Refuse to copy! As long as you don't want to distinguish any units and so on, "the key is the difference between power and electric power."
Electric work refers to the work done by the current, which is the process of converting electric energy into other forms of energy when the current flows through the consumer. The amount of energy consumed is related to the time. That is, w = uit. Electric power refers to the work done by the current in unit time, which reflects the speed of the current's work, that is, P = w / T, For example, electric power and mechanical power, the formula of mechanical power is still p = w / T, the difference is that w = FS
Electric work: the work done by the electric field force is called electric work. Power: the work done per unit time can be electric work or the work done by other forces. Electric power only refers to the electric work per unit time
There is a right triangle with an angle of 45. Find the analytic expression of the function between its area y and the length x of the hypotenuse
y=(1/2)*x^2
Proving A1 ^ 2 + A2 ^ 2+
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