Well, that basically depends on the kind of load. If you have a passive load (without power electronics) e.g. a basic heater, the current in the device would be reduced to the half and its power to a quarter as given by Ohm´s law. If you have an active load where its power is kept constant by power electronics, it will consume double current to compensate for the reduction of voltage. This could eventually damage your device as well.   

Well, that basically depends on the kind of load. If you have a passive load (without power electronics) e.g. a basic heater, the current in the device would be reduced to the half and its power to a quarter as given by Ohm´s law. If you have an active load where its power is kept constant by power electronics, it will consume double current to compensate for the reduction of voltage. This could eventually damage your device as well.   

What happens will depend upon the appliance. A toaster, a resistive heater, an incandescent bulb will function at roughly one-fourth of its capacity as the power input is proportional to the square of the voltage. An incandescent bulb will function as if its output has been dimmed. A CFL, an electric motor (a fan), a compressor (air-conditioner) will simply not run. 

Either the appliance performance will be low (low light intensity in case of bulb, less heating in case of heater, low speed operation in case of an electric motor) or it will not run at all if the appliance current requirements are not met. overheating due to excessive current drawn which is way above the appliance current rating will damage the appliance. This is possible only if  by some means the appliance resistance is reduced. 

I agree with all of the above - it depends on what load you are running.

You will find some switch mode supplies, e.g. Laptop power supply, CRT displays, will function just fine, as these are designed to run on a supply from 90(or 100)-230V.

Some CFL's and Fluorescent lamps, when started on 220V, continue to glow when turned to 120V (try it) - during brown outs, we have seen the voltage drop as low as 85V before they would drop out.

it is depend which type of instrument you r going to connect.

if it is fan or motor then may winding get burn due to low back emf development.

and tube light not get on and may get damage its filament by heating more.

Dear Mohammad,

I agree with the colleagues above. They excited important issues on the effect of reducing the ac voltage to 50 percent of its nominal value. Really this question has many answers as indicated by the colleagues depending on the type of the load.

I would only here point out that every load has its nominal operating conditions in which this load can operate safely and efficiently. Normally for motors the operating voltage range is from 90 percent to 110 percent of the nominal operating voltage of 220 voltage the mains. The operating frequency can be changed +- 5 percent of the nominal frequency. Also, heaters and lamps have specific voltage and current ratings where they perform satisfactory.

All load can be modeled by a more or less with a resistance and  others can be modeled by an impedance composed of a reactive load and resistive load. The motors has a mechanical load that can be represented by an output power required to be delivered by the power supply . One can analyse the effect of reducing the  supply voltage by using these simple models.For resistive and impedance loads the current will be reduced to its half nominal value. This reduction has no consequence for the safety of the appliance. It detracts only its operation such that most time it is unacceptable. In case of motors, The situation is different for motors it depends on the nature of the mechanical load. Since the power is IV cosphi= the Pm where Pm the motor power, then reducing the voltage to half it rated value may increase the current to a value => double the rated current. this current if greater than rated the maximum rated current, the motor may be burned if there is no suitable overheat protection.

wish you the best

One must consider the amount of current that will pass or go into your appliance. With a lower voltage, a greater amount of current (and heat) will pass through it. Your appliance will "over heat".

@Eddie,

only if the automatic regulation will ask for a special power.

Hanno, the appliance will "pull" the amount of power specified of it. P - EI. With E divided by 2, I will be multiplied by 2, to maintain the specified power. It is I that flows and causes friction and of course, heat. With heat doubled than that the 240 v appliance can carry, the appliance will overheat, melt, or explode, or burn.

Or the appliance will not function having a voltage (push) that is insufficient. Ed

You are right Eddie, that´s exactly what I wanted to hint to. Take a simple lamp, apply half the voltage, the lamp will stay dark or show only a low light. No explosion etc.

Be careful. If the load is a constant power load (as an induction motor), the result will be to double the current absorved by the device. this can damage the device.

Sorry, the answers are repeated! 

Thank you very much for your helpful and clear answers.

I appreciate Mr.  E. Arroyo for comprehensive answer.

Yes, in the same but reverse way if you connect 120V appliances to 240V supply then insulation designed for 120V supply get damage under 240V supply.

it will be broken

only concern is if the current at the input increases fuse gets dameged