What Power The Lamp Of 200w 230v Will Drow If It Is Connected To 115v Supply

A 200 W, 230 V lamp is connected across a 115 V supply, will the lamp draw power?

The lamp is basically resistive load.Power for resistive load or power (W) will be = VxV / R, so R = VxV / WFor 230V application R = 230x230/200For 115V application R = 115x115 / W1 (W1 = Power consumed for 115V application)So, R = 230x230/200 = 115x115 / W1or W1 = ( 115x115x200) / (230 x 230) = 50 Watt.So a 200 W, 230V lamp will consume 50 Watt if connected across 115V

A 200 W, 100v lamp is to be operated on a 250v supply. What will be the additional resistance required to be connected in this series?

The answer is indeed 75 ohmsPower (P) = Voltage (V)x Current(I)substituting for P and V,200 = 100 x II = 2 ampsThe resistance of the lamp = V/I = 100/2 = 50 ohmsnow the circuit will contain 250 V power supply in series with above lamp and another resistor R and lamp requires 2 amps of current to flow in the circuit. So, we just need to find total resistance for that,R + 50(resistance of lamp) = V/I = 250/2 = 125 ohmsR + 50 = 125R = 75 ohms

Two 100W, 200V lamps are connected in series across a 200V supply. The total power consumed by each lamp will be watts? How?

Power depends on three factors :VoltageCurrentResistanceOf these three, the first two are liable to change according to a given circuit, but the resistance of a device is fairly constant.Let's take a 200 volts - 100  watts bulb and calculate the resistance of its filament :We know,[math]P = \dfrac {V^2}{R}[/math]or[math]R = \dfrac {V^2}{P}[/math]When V=200 volts and P = 100 watts,R = (200*200)/100 = 400 ohms.The reason why we calculated out the resistances is because R does not change with the applied voltage. It may be considered as a constant at a given temperature and pressure.Now, let's connect the two bulbs in series :Since now the two bulbs are in series, the total resistance offered by them is :R' = 400 Ω + 400 Ω = 800 Ω Let them be powered by a 200 volts DC source.Now the current that flows through either of them is :[math]I = \dfrac {V}{R'} [/math][math]= \dfrac {200}{800} = 0.25 A[/math]Since the bulbs are in series, the same amount of current flows through each of them. Now note that the potential across the each bulb in this case is not 200 volts. Instead, it is the voltage across the combination of bulbs.Since the two bulbs are identical, the voltage across each bulb is :[math]V=IR[/math][math]V=0.25*400[/math]V=100 volts In this case, the power at which each bulb is working is given by :[math]P[/math][math]=[/math][math]V*[/math][math]I[/math]If we substitute V ( = I*R) from Ohm's law, we get[math]P=I^2∗R[/math][math]P = (0.25*0.25)*400[/math]P = 25 wattsThus, on connecting the bulbs in series, the power they consume decreases significantly.This happened because the bulb which was originally rated at 100W at 200V has now been connected to a voltage supply of 100V. In case of a parallel circuit : The voltage across each bulb is 200 volts, and each of them work at their full power, 100 watts.In fact, the power rating is just a relative term.On appliances, you see labels like these :Well, that means that only when this device is connected to a 230 volts supply, will it work at 2000 watts. If the source voltage is different, the power consumed will change.On connecting to different voltage sources, the current through the appliance changes, but it's resistance remains fairly constant. And because of that, it's power changes.

What will happen if we connect a 120V appliance to a 115V socket?

The appliance which you are about to connect is of a higher voltage rating than you are actually applying to it. Be rest assured that it will work. But the output of the appliance will go down drastically (this will happen only when it has been designed to do so!). For eg. If the appliance is a fan and it should technically rotate at 1000 rpm (just a rough number) when you are applying 120V (according to its design), it will rotate at around 450 rpm (again a rough figure) when you are applying 115V.Basically you have to beware of exposing your equipment to any voltage higher than its Rating. However undervoltages have also been known to reduce the lifespan of Electrical equipments.

Electric bulb marked with 100 watt and 230 volt, if the supply voltage drops to 115 volt what is the heat energy production by the bulb in 20 minutes?

Ruvian, below, is of course correct.But what I’m guessing ‘they’ are expecting as an answer is along the lines of the following:-If the bulb dissipates 100W at 230V & the supply is halved to 115V, the power dissipated will quarter to 25W. [P=V^2/R]20 minutes is 1200 seconds & 25w x 1200 seconds is 30,000Ws or 30kJ of energy.But it’s not true.Firstly, assuming you ARE implying an incandescent ‘bulb’ with probably a tungsten filament:-(1) Of the power dissipated by the bulb, some is dissipated in the form of visible light, but a significant proportion is dissipated in the form of heat.(2) A hot tungsten filament is a non-ohmic, or non-linear resistance. its resistance depends on its temperature, with a positive temperature coefficient of resistance. So, the bulb’s resistance will be lower when running on 115V (& it is at a lower temperature & glowing less brightly) than when it is running on 230V. It will draw relatively more current, thus dissipate more power, than expected above.So, you cannot directly, linearly relate the power used by the bulb simply as a function of applied voltage, since its resistance does not remain constant when the voltage varies.I guess you could calculate it roughly by reference to tables etc. It would be better to actually MEASURE it. But you would somehow need to find the PROPORTION of the total power/energy consumption which is dissipated as HEAT, not light.As I’ve said before in these things, it sounds like your typical ‘dumb’ question set by some teacher who should not be teaching.If, instead of a 100 watt, 230 volt bulb, the question were posed using a 529 ohm 100 watt RESISTOR, you could give the answer as above.BUT a light bulb is NOT a resistor, so it comes back to what Ruvian said - not enough information given to calculate.