The rated power of an electric appliance is 200W. When it is connected to a circuit with a voltage of 220V, a resistance of 60ohm should be connected in series A 60 ohm resistor should be connected in series to work normally, and the rated voltage of the appliance should be calculated Detailed explanation Can we be more specific and list the following equations

The rated power of an electric appliance is 200W. When it is connected to a circuit with a voltage of 220V, a resistance of 60ohm should be connected in series A 60 ohm resistor should be connected in series to work normally, and the rated voltage of the appliance should be calculated Detailed explanation Can we be more specific and list the following equations

120V
Just solve a quadratic equation of two variables
U*U-220U+12000=0
Suppose the rated voltage is u, the voltage on the resistor is 220-u, and the current through the resistor is (220-u) / 60, which is also the current through the electrical appliance. So 200 = u * (220-u) / 60
The solution is u = 100V or u = 120V
3 * 16 + 1 * 10 square copper cable is 50 meters long / electric appliance is 220 V, how many kilowatts can it load?
At present, I have 142 computers, 12 1.5p air conditioners, Internet bar servers, one 5p air conditioner, one 5p spare ceiling air conditioner, about 11W for each 250m2 energy-saving lamp, 6 smoking machines [4 * 40W] and one large suction exhaust fan [150W] with a total power within 60kW [according to my calculation], now there is only one 3 * 16 + 1 * 10 copper cable, and the grounding should be solved by myself, which is more than 40m long, Will this cable work?
If 3 * 16 + 1 * 10 square copper cable is laid in the air, the current carrying capacity is 74a, then it is estimated that 74 / 4 = 18.5kw according to a kilowatt 4 current of single-phase equipment. If each phase is 18.5, then three-phase is 18.5 * 3 = 55.5kw
The resistance wires with the same material, thickness and different length are made into AB, CD and EF wires, which are respectively placed on the smooth metal guide rail with negligible resistance and perpendicular to the guide rail. As shown in the figure, the direction of uniform magnetic field is perpendicular to the plane of the guide rail and inward. The external force makes the wire move horizontally to the right with uniform speed, and the power of each external force is the same. The length of the three wires between the guide rails is known If the relationship is lab < LCD < lef, then ()
A. Because the effective length of the magnetic induction lines cut by the three wires is the same, the induced electromotive force generated by them is the same. D. ignoring the change of the internal energy of the conductor, the heat generated by the three wires per second is the same
A. According to the law of conservation of energy, the heat generated per second is equal to the power of the external force. P = E2R = (BLV) 2R the external force makes the wire move horizontally to the right at a constant speed, and the power of each external force is the same, EF is longer. According to the resistance law, R is larger, and the uniform speed is the largest, so a is wrong, B is correct
A 121 ohm resistance wire connects her to a 220 V circuit. What is the power consumption per minute
It has a lot of power
p=U^2/R
=220^2/121
=48400/121
=400W=0.4KW
W=pt
=0.4*1/60
=0.00666 degrees
She uses such a little electricity. Of course she's a woman
ha-ha
P=U*U/R=48400/121=400(w)
W=P*t=400*60=24000(j)
Connect her to a 220 V circuit? How do you know the resistance wire is female? C
P=U^2/R=48400/121=400(w)
W=P*t=400*60=24000(j)
The plagiarism of clodydede
I didn't learn``
A copper wire with cross section of 0.10mm2 is 1.2m long. When a voltage of 0.6V is applied at both ends of the copper wire, the current passing through the copper wire is 0.1A, then the resistivity of the copper wire is______ .
According to Ohm's law, r = UI = 0.60.1 Ω = 6 Ω. According to resistance law, r = ρ ls, then resistivity ρ = RSL = 6 × 0.1 × 10 − 61.2 = 5 × 10 − 7 Ω· M
What is the effect of wire thickness on current or voltage or resistance?
If the voltage is constant, what is the effect on the current resistance?
Is there any formula?
The longer the wire, the greater the resistance and the smaller the current. The thicker the wire, the smaller the resistance and the larger the current
The resistivity ρ is related to the material. After the material is fixed, the resistivity will not change. The cross-sectional area of conductor s is calculated as follows: I = u / r = u / (ρ L / s) = US / ρ L. the voltage is constant, and the current is related to the cross-sectional area and length of conductor. The longer the conductor L is, the greater the resistance R is, the smaller the current I is, the thicker the conductor is (the larger the cross-sectional area s is), and the smaller the resistance R is, the greater the current I is.
The resistance of a resistance wire is 97 ohm, connected to the voltage of 220 V, how much current flows through it
I=U/R=220/97=2.27A
220 divided by 97
Pure resistance circuit..
I=U/R
Is it that simple?
There is a copper wire with a cross-sectional area of 0.5mm square and a length of 6m. When a voltage of 12V is applied at its two ends, the current passing through it is exactly 2a
There is a copper wire with a cross-sectional area of 0.5mm square and a length of 6m. When a voltage of 12V is applied at its two ends, the current passing through it is exactly 2A. The resistivity of this copper wire is () ohm,
When the voltage of 12V is applied at its two ends, the current passing through it is exactly 2a, indicating that the resistance R = u / I = 12 / 2 = 6 ohm
And the resistance R = P * L / s
So the resistivity of this copper wire P = R * s / L = 6 * [0.5 * 10 ^ (- 6)] / 6 = 5 * 10 ^ (- 7) ohm * M
5X10^-7
There is a copper wire with a cross-sectional area of 0.5mm square and a length of 6m. When a voltage of 12V is applied at its two ends, the current passing through it is exactly 2A. The resistivity of this copper wire is (0.5) Ω. Mm2gm.
6 ohm
Is the greater the resistance, the greater the resistivity
(1) Definition or interpretation
Resistivity is a physical quantity used to express the resistance characteristics of various materials. The resistance of a conductor made of a certain material with a length of 1 meter and a cross-sectional area of 1 square millimeter is called the resistivity of this material
(2) Unit
In the international system of units, the unit of resistivity is Ohm · m, and the commonly used unit is Ohm · mm 2 / m
(3) Explanation
① The resistivity ρ is not only related to the material of the conductor, but also related to the temperature of the conductor. In the range of little change in temperature, the resistivity of almost all metals changes linearly with temperature, that is, ρ = ρ o (1 + at). Where t is the temperature of centigrade, ρ o is the resistivity at O ℃, and a is the temperature coefficient of resistivity
② As the resistivity changes with temperature, the physical state of the resistance of some electrical appliances, such as 220 V, must be explained
The resistance of 100W lamp filament is 484 ohm when it is electrified and only about 40 ohm when it is not electrified
③ Resistivity and resistance are two different concepts. Resistivity is the property reflecting the blocking effect of material on current, while resistance is the property reflecting the blocking effect of material on current
The following table shows the resistivity of several metal conductors at 20 ℃
Material resistivity (Ω m)
(1) Silver 1.6 × 10-8 (5) platinum 1.0 × 10-7 (9) constantan 5.0 × 10-7
(2) Cu 1.7 × 10-8 (6) Fe 1.0 × 10-7 (10) Ni Cr alloy 1.0 × 10-6
(3) Al 2.9 × 10-8 (7) Hg 9.6 × 10-7 (11) Fe Cr Al alloy 1.4 × 10-6
(4) W 5.3 × 10-8 (8) Mn Cu 4.4 × 10-7 (12) al Ni Fe alloy 1.6 × 10-6
(13) Graphite (8 ~ 13) × 10-6
It can be seen that the resistivity of metal is small, the resistivity of alloy is large, and it is not gold
Germanium, silicon, selenium, oxide
The resistivity of copper and boron is smaller than that of insulator but larger than that of metal. We call these materials semiconductors
(semiconductors).
Conclusion: under normal conditions (except for special occasions), the best conductive materials are silver, copper and aluminum. These three materials are the most commonly used and are often used as wires. Among them, copper is the most widely used. Almost all the wires are made of copper (precision instruments, except for special occasions). Aluminum wire has been eliminated because of its unstable chemical properties and easy oxidation. Silver has the best conductivity, but it is rarely used because of its high cost, It is only used in high demand occasions, such as precision instruments, high-frequency oscillators, aerospace, etc.. By the way, gold is also used in some instrument contacts. That's because the chemical property of gold is stable, so it is used, not because of its low resistivity
Resistance R = ρ * L / s, where ρ is the resistivity, l is the length of the conductor and S is the cross-sectional area of the conductor.
If the resistivity and the area of the conductor are constant at the same temperature, the length of the conductor may be very long
The resistance of a conductor is related to its material, cross-sectional area and length. This is the calculation of the specified quantity. In fact, the resistance of the conductor is also related to the temperature.
The material of conductor is resistivity in calculation. Since resistance is related to many factors, there is no saying that the greater the resistance, the greater the resistivity.
The resistivity is related to the material, but it has nothing to do with the material
The resistivity is only related to the material and temperature used to make the resistance. The material temperature does not change and the resistivity does not change
As shown in figure a, the resistance of the heating wire is R1 = 484 Ω, R2 = 242 Ω, and the power supply voltage is u = 220 v
1. How to set the switches when the heating wires are required to be connected in parallel? 2. Close the switch S2 only, and find out the equivalent resistance of the circuit and the voltage ratio of R1 and R2. 3. Find out the ratio of the maximum heating power and the minimum heating power of the circuit. 4. The simultaneous closing of S1 and S2 in the circuit will form a short circuit. Please draw the improved circuit diagram in the dotted box
Please provide the drawing