There are 66 incandescent lamps with 220V 110W, and the line voltage should be 380V

Put the 66 incandescent lamps in 22 groups, each group has three incandescent lamps, take a wire from each of the three incandescent lamps, screw the three wires together, and then connect the other wire of the three incandescent lamps to the three phase wires respectively, and the 33 groups of incandescent lamps are connected in this way.

Because the incandescent lamps of the two are 66, so in the process of accessing the chain home, you should first measure its total voltage in advance.

380 volts is not necessarily a bit too high, if you want to step it down, you must first have your own transformer.

If you plug this bulb into a voltage of 238, you need a converter.

You have to reveal the system voltage to be able to operate, otherwise there is no way to operate.

Hello, don't connect like that. with parallel. You can choose according to your actual needs.

The voltage of 25 is recorded directly from the client.

225, this is specific to you to look at the incandescent lamp for those professional explanations, or look for information on this.

There are 220 feed-110 watts of incandescent lamps 66 people how to access uh to achieve a voltage of 380 on it smart connection,

Good night master after 110 watts of incandescent lamps 66 should be connected to the voltage of 380 volts, you can connect in series.

If you want to pick this up, you can find a special person to pick it up for you, and then you can save trouble and use it with confidence.

They cannot be used in series on a 220V power frequency AC power supply.

Because their rated voltage is the same, but the rated power is different, which shows that their filament resistance is different: the incandescent lamp filament resistance with large rated power is small, and the incandescent lamp filament resistance with small rated power is large, and their voltage distribution is proportional to the resistance value after series connection, therefore, the voltage added to the 100W incandescent lamp is lower than the rated value and cannot work normally; A voltage higher than the rated value applied to a 40W incandescent lamp will damage the lifespan.

Technical principle. An incandescent lamp is a source of heat radiation, and the energy conversion efficiency is very low, only 2% to 4% of the electrical energy is converted into light that the eye can perceive. However, incandescent lamps are still widely used because of their good color rendering, continuous spectrum and easy use.

A lit incandescent lamp has a filament temperature of up to 3000 °C. It is because of the light radiation produced by the red-hot filament that the electric light shines brightly. Because some tungsten atoms will evaporate into gas at high temperatures and deposit on the glass surface of the bulb, making the bulb black, so incandescent lamps are made into a "pot-bellied" appearance, which is to make the deposited tungsten atoms disperse on a relatively large surface.

It is known that the resistance of the two lamps is equal, Ohm's law.

The actual voltage of the two lamps is 220 2 = 110V

W=Ur It is obtained that when U is halved, the power W is reduced by 1 4, and at U=110V, the power of each lamp is the original 1 4, that is, 25W, because the resistance of the two bulbs is the same, according to the series circuit.

The voltage of each bulb is 110V, and according to P=U2 R, the actual power of each bulb is only the rated power.

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In series, the voltage is half of the rated voltage, and the resistance is unchanged, then the power = U 2 R = 1 4 * rated power = 1 4 * 100 = 25 watts.

p=u2r, the total power is the original 1 4, and the two bulbs each account for 1 2, i.e. 100w 8=.

250W is much brighter, because the same voltage is small power, that is, the resistance value is large, after connecting them in series, we know that the current in the series circuit is equal, because the voltage drop of some resistors is also large, so the one with small power will be brighter.

w=(u*u)/r。It can be calculated that the internal resistance of the two lamps is 484R and 1936R. i is equal in series. So the higher the voltage, the brighter it is. According to Ohm's theorem. The smaller the resistance in series, the lower the voltage, and the higher the voltage with an internal resistance of 1936r.

There are two incandescent lamps of "220V 100W" and "220V 25W", which are connected to the 220V AC power supply after being connected in series, and their brightness is (a).

A 25W bulb is the brightest.

For series circuits, the voltage distribution is proportional to the resistance value. The resistance of a 25W lamp is 4 times that of a 100W lamp. So the voltage after series connection is 100W 4 times. It also bears 4 times as much power, so the 25wr bulb is the brightest.

100W Bulb Resistance: 220 (100 220)=440 25W Bulb Resistance: 220 (25 220)=1936 2 bulbs in series in series rear loop current:

220 (440+1936)=100W bulb in series:

Power of 25W bulbs in series:

It should be a 25W bulb that is brighter than a 100W bulb.

The series power is different and does not light up, so choose D

Answer]: 66 bulbs can be divided into three groups, each group of 22, parallel after the front annihilation is connected between the phase line and the neutral wire respectively, to obtain the phase voltage of 220V.

When the load symmetry is delayed, the line current i=p u=22*100 220a=10a.

The rated voltage of the bulb is 220V, the power is 100W, if it is connected to the "three-phase 380V power supply", how to connect.

Hello dear, the rated voltage of the lamp Kai bubble is 220V, the power is 100W, if it is connected to the "three-phase 380V power supply", how to connect to 380V must be connected in series with two lamps in series. One of the 220V live wires and the other zero wire can be connected. 1. A bulb with a rated voltage of 220V and a power of 100W, when connected to the three-phase 380V system, because there is no zero line N, there are three methods, one is to use three bulbs and wire according to the "star", the other is to make a zero line by yourself, and the other is to lead the zero line of the system over, which is the best way not to refer to; 2. A bulb with a rated voltage of 220V and a power of 100W is connected to the three-phase 380 220V system, and it is easier to connect it by monitoring the slag to have a neutral line, and only need to connect a live wire (A, B, C phases can be) and connect a neutral line; It is also possible to wire with three bulbs and press the "star".

The rated voltage of the incandescent lamp is 2V, the rated power is 4W, and the direct current of 2V is connected.

Hello Kiss is happy to answer this question for you. The rated voltage of the incandescent lamp is 2V, the rated power is 4W, and the direct current of 2V is connected (1) The noise voltage and the rated voltage of the incandescent lamp are equal to the rated voltage when the incandescent lamp is emitting normally, and the resistance of the filament is calculated according to P=UI=U2R; (2) When two incandescent lamps of this specification are connected in series to the power supply of 220V, the voltages they share are equal, the current through the bulb is read out according to the image, the total actual electric power of the two lamps is calculated according to P=UI, and the total electric energy consumed by the two lamps is calculated according to W=PT; (3) When the incandescent lamp is connected to the 12V student power supply, the circuit is the path, and the electrical energy is judged according to w=uit, and the electrical energy is converted into other forms of energy when the electrical appliance is working

rl=ul2pl=（220v）240w=1210ω；(2) When two incandescent lamps of this specification are connected in series to the power supply of 220V, the voltage they get is equal, that is, UL = U2 = 220V2 = 110V, it can be seen from the image that the current in the circuit is I=, then the total actual electric power of the two lamps is P=UI=220V, which can be obtained from P=WT, and the total electric energy consumed by the two lamps for 2H is w=pt=degree;

Because the resistance of a 100W bulb is: R1 110 Square 100 121 Eurom The resistance of a 60W bulb is: R2 110 Square 60 1210 6 Eurom 100W and 60W bulbs are connected in series, and the voltage of a 100W bulb is divided into:

u1＝220/（121＋1210/6）x121＝u2＝220－u1＝。

Therefore they cannot be used in tandem.

The resistance of two 100W bulbs is the same, and the voltage of each is 110V when used in series, so it can be used normally.

R1 - > 100W resistor.

R2 – >40W resistor.

r1 = U2 P = 220V2 100W = 484 euros (48400 was 220 square meters in that year).

r2=...= 1210 euros.

First of all, there is no doubt that if you want to burn it in series, you need to burn it with a large resistance, because p=i2*r, so if you want to burn 40w, it will be burned first;

Then judge whether it will be or not

i=380v/(r1+r2)=

p1=i2*r=;

p2=...= >40w rating, so it hangs.

It's really not a problem to study your own.

The internal resistance of 100W bulb is 404 ohms.

The internal resistance of the 40W bulb is 1210 ohms.

The total resistance of the series is 1614 ohms, and the current is ampere.

A 100 watt bulb gives a voltage of about 100 volts.

A 40 watt bulb gives a voltage of about 280 volts.

So the 40w bulb burned out.

Ignore the temperature coefficient of resistance.

First, calculate the respective resistance with the rated voltage and power: r = the square of the voltage divided by the rated power, 100W, 40W are 484, and the two resistors are added together to 1694 after series connection;

After connecting to 380V, use Ohm's law to find the current, voltage and power

The voltages of 100W and 40W bulbs are; The power is 24W and 61W, respectively.

Since the terminal voltage of the 40W bulb is, it will burn out.

The specific calculation method is that the resistance is inversely proportional to the power, so the resistance ratio of the two bulbs is 2:5, so 380 is divided into seven points, 100w accounts for 2 points, and 40w accounts for 5 points, respectively, and the voltage is the same as that, so 40w will burn out, and the specific calculation formula is r=u 2 p

r=U2 p to find the incandescent lamp resistance r1,r2. After series connection, the current i=380 R1+R2. Actual power p = 380 2 r. If it is calculated that it exceeds the rated power, it will be burned.

Rated current i1 = up p1 = i2 = up p2 = resistance r1 = u 2 p1 = 484 ohms r2 = u 2 p2 = 1100 ohms after series current i = us (r1 + r2) =

Actual power p1 = i 2 * r1 = p2 = i 2 * r2 = from the rated power and actual power or the rated current and the actual current can be seen that the 40W bulb will burn out (i1> i> i2, p1p2).