Kneel and beg the electrician master?? Thanks in advance !!

1. The rated power of the power supply of the computer is 300 watts, which does not mean that the actual power is 300 watts when using the computer. The actual power of the computer is less than 300 watts, so if the 4 computers are turned on at the same time, they will not trip if the action value of the air switch current limit is not reached.

2. There are two possibilities for the air switch to trip: (1) overcurrent, that is, the current exceeds the action value of the air switch current limit; (2) Leakage, if your air switch is the one with leakage protection.

3. Use the water heater to trip, you have to figure out what kind of reason it is. If it is caused by electricity leakage, do not use it again, and repair the water heater as soon as possible; If it is because the power is too high, the power can be reduced.

4. How to reduce power: If your water heater has more than 2 gears, you can try it with a low power gear; If your water heater only has a 750 watt gear, you can buy a high-power diode and use it in series with the water heater, and your water heater will become about 375 watts. The power is reduced, and it may take longer for the water to heat up, in other words, the water heats up slowly.

5. The high-power diode should withstand voltage of 250V and current of more than 5A.

6. According to the questions you asked, it can be judged that you don't know much about electricity. Therefore, here is a special reminder to you, do not modify yourself, but find an electrician or engineer who understands electricity for consultation and help. For yourself, for your family, for your classmates, for your classmates' families, pay attention to safety!!

There are three possibilities:

1.Creepage. The air switch is leakage protection, and the electric water heater leaks.

Pad the pot with a wooden plank.

2.Overcurrent. The power of the electric water heater is too high at low temperature, and the overcurrent is protected.

Serial ballast. 3.There is also an air switch that you young people may not have experienced. This air switch is able to recognize the current curve of a resistive load such as a thermostat in an electric furnace, and once such a load is present in the circuit, the power is de-energized after ten seconds.

This kind of switch is widely used in the era of power shortage, but it is rare nowadays.

Serial ballast.

This is caused by the excessive inrush current of the electric water heater when it is started (the cold state resistance is small). Workaround:1Tandem one.

1. For a current-limiting load of 200 watts, it is enough to use a switch to short-circuit this load after it is energized. 2.Connect a diode (greater than 5A) in series, and then use a switch to short-circuit the diode after powering on.

Go for it. You're all college students, and you're sure to be able to solve this problem.

I wish you every success!

It should be more likely to be caused by excessive current when the water heater is started, in fact, if the 4 computers are started at the same time, they will definitely jump.

It is recommended that you find an inductive ballast, which is the one used in the old fluorescent lamps in the past, and connect it in series to the power line of the water heater.

Remember to wrap the thread well, and be sure to pay attention to safety.

Bypass the current limiting method: first turn on more electrical appliances, try to see what power it trips, such as 1000W, then only 1000W large electrical appliances can be used. You start by gradually driving to close to 1000W, then disconnect and quickly close the switch of the big appliance.

You can buy a single-pole double-throw switch for just a few bucks) or: you use a thyristor (this one is more specialized).

1.The power limit in the dormitory is generally limited to the current of the electric heater (pure resistance load). You can change the load to escape the limit.

2. Leakage protection, try to change the battery or don't let it leak.

The actual total power of a desktop computer is less than 750W.

4. There is a problem with the air switch.

Ha, we electrical students can't do anything about this problem. We pulled a cord from the toilet in the hallway, but it was very hidden. Otherwise, if you have the money, you can replace his air short.

Or you can just short the air shorter. But the consequence of this is that it is not safe.

Our solution in the dorm room was to wiring the lights in the hallway or the activity room on each floor. The electric water heater in our school is punishable by expulsion.

The text is garbled.

Power rationing in the dormitory is a current limit, so stealing electricity is the fundamental solution.

It's complicated to say.

There is an alternating current foundation that should be able to solve it by itself, that is, to take electricity in front of the gate, which is actually very simple to do.

The water heater is leaking, and the air switch is very sensitive, and it will jump if there is a slight leakage.

The total power of a computer is 300W, in fact, it is less than 300W, according to my estimate, the leakage is because the leakage switch current is too small, and it should be feasible to change one.

This problem is easy to solve, just buy a 400w 500w rice cooker and boil water.

<> from the **, this gas stove plug comes with a protective device, the white wireframe is the reset key (that is, the button you said), the pink wireframe is the test key, and the yellow wireframe is the indicator light. If the reset button cannot be pressed, it means that when you insert the gas stove plug, there is leakage or short circuit, and the protection device works to prevent the continued use of electricity, that is, the reset button cannot be pressed, and only when the reset button is pressed, can it be used normally.

The analysis is that there is a problem with the connection between the gas stove plug and the socket, you unplug the gas stove plug, check and clean the plug and socket panel, whether there is water, oil and dirt, etc., the lead wire, the zero line is short-circuited, and the grounding protection line will trigger automatic protection if there is leakage in the grounding protection.

First, unplug the ** upper plug.

Second, find the reset button of the main switch, press the reset button of the main switch, and then close the switch. (The reset button is shown in the figure below, and the button should be pressed before closing).

Third, you should be able to close the gate at this time, which means that there is a problem with the equipment carried by your **upper plug.

Hope it helps.

In this case, there must be a short circuit or leakage on the plug (wire) of the gas stove, and other sockets, including the gas stove socket, are no problem. Due to multiple short circuits tripping, the brake cannot be closed now, and it is best to ask an electrician to check the insulation condition of the line.

In this case, it should be that your home appliances or sockets have a short circuit or grounding.

The solution is to unplug all the electrical appliances plugged into the socket, which should be able to open the air.

Plug in the appliances one by one, and pay attention to which devices will trip, and you can determine which one is problematic.

Unplug the gas stove and close it again.

It is suspected that it is caused by a gas stove or a leakage of the line.

I personally feel that several important operational topics, 4What is an electrical appliance that can cut off the load current and the short-circuit current? 1. Circuit breaker 4

The operation ticket shall be based on the duty officer or (c.).Shift Leader) issued the operation 7If the motor is started by Y-δ, its starting current is 8 times that of full-voltage starting

In (a.).When overhauling electrical equipment in a high-temperature environment, it should be avoided as much as possible to work under live conditions.

1) In the experiment, the results can be obtained by directly measuring the current i2;

2) first find the power supply and resistance in the Thevenin equivalent circuit;

uab = i*r3 - us2； i = us1/(r1+r3)；

Then the open-circuit voltage Uoc = UAB, the equivalent resistance RO = R1 R3;

3) Take a power supply equal to UOC and a resistor equal to RO, and form the circuit in the following figure with R2;

4) The current i2 can be measured by calculation and experiment, and the results can be compared;

The Thevenin equivalent of the AB terminal is shown in the figure, let point C be the reference point, then the open circuit voltage: UOC = UAB = UAC - UBC = US*R2 (R1+R2) -US*R4 (R3+R4) ;

Equivalent resistance: RO = R1 R2 + R3 R4;

Okay, do the math yourself;

Since the internal resistance of the galvanometer is very low, the resistors are equivalent to parallel, and the power supply and the resistor form a voltage divider circuit, then one side of the resistance is regarded as the internal resistance of the power supply, and the other side of the resistance is regarded as the load. Calculating the current on each resistor yields a current of 0 on the galvanometer. Since the internal resistance of the galvanometer is very low, the resistors are equivalent to parallel, and the power supply and the resistor form a voltage divider circuit, then one side of the resistance is regarded as the internal resistance of the power supply, and the other side of the resistance is regarded as the load.

Calculating the current on each resistor yields a current of 0 on the galvanometer. Since the internal resistance of the galvanometer is very low, the resistors are equivalent to parallel, and the power supply and the resistor form a voltage divider circuit, then one side of the resistance is regarded as the internal resistance of the power supply, and the other side of the resistance is regarded as the load. Calculating the current on each resistor yields a current of 0 on the galvanometer.

Let's talk about the answer first, your bridge arm is balanced, and the output is 0

The core of the calculation method in all cases is to assume that some sources are 0 outputs, leaving only one output, and for the power supplies that have not been counted into the power supply for the time being, the impact should be treated as 0!

Why is the voltage source short-circuited and the current source open-circuited?

The external impact of the voltage is 0 = 0, and the voltage is equal to 0, which is a wire! As for the current source, a current of 0 must be an open circuit. Got it?

Another important thing is the shunt of current, you have to artificially set the direction of the current, the incoming is positive, the outgoing is negative, they should be positive and negative offset equal to 0, which is equivalent to the conservation of current.

In addition, the sum of voltages in any one of the closed-loop loops must also be equal to 0.

Your internal resistance is to short the power source, and the internal resistance is equivalent to the parallel connection of 4 resistors.

Under the above premise, take the back of the current you require as the load end and open the circuit, and count the power supply end after the open circuit as a power supply with internal resistance, and calculate the current.

Let's start the math and you're done!

On the basis of the above theory, I will change the voltage and imbalance treatment, you can understand it.