R1=v square/50
R2 = v square/25 = v square/25 multiplied by 2/2
= 2v square/50
R2 = 2 multiplied by R1
R2 is twice more than R1
there were three reasons, first, an effective incandescent material, next he was able to achieve a higher vacuum and third a higher resistance lamp that made power distribution from a central source an economical proposition
because there is a correlation between resistance and voltage and current. The equation resistance = voltage divided by current shows that the higher the voltage, the bigger the resistance,, and the bigger the resistance the hotter the filament lamp will get because of the electrons bumping into each other which means there is a loss of energy and that energy is being transferred to the filament making the actual filament bulb hot since there is more thermal energy wasted at the end.
Lamp have required high rating to glow & have a specific luminus area.exa-table lamp,florucent lamp,sodium vapur lamp,BUT but bulb have no required rating exa- halogin bulb
Because it takes more current and power and produces more light power. Running a 100 w lamp costs 4 times more than a 25 w lamp.
If you have a lamp, you can assume that the resistance of the lamp when it is under power will follow the ohms law. BUT, one thing you must remember is, when a lamp is under load, it is glowing HOT. When metal is HOT, the molculoes of the meals are in much more active state. When this happens, the resistance will increase. Conversely, when the lamp is NOT on ON state, the filaments are cold. Moleculoes in the filaments are not as active. Thus, the resistance is lower. There is almost 10 to 1 difference in resistance from hot to cold. Taking out a multimeter and measuring the resistance of the lamp will not help you determine the resistance of the lamp when it is actually under load (with voltage applied) Really, the only thing you can do is to measure the voltage, measure the current, then arrive at the resistance mathmatically.
As per the formula for power (Power (Watt) = Voltage (V) x Current (i) & Resistance (R) = V / i), 25w lamp bulb would have higher resistance compared to that of 5w lamp bulb.
Yes. A 60W bulb has a higher resistance than the 40W buld. The extra resistance requires more current to light up the bulb. The fillament then glows brighter.
You should not use a higher wattage bulb in a lamp that says 40 watts. The higher wattage could cause the lamp to catch fire due to the excess heat and could cause you serious electrical problems.
Depends on the lamp and bulb if you are referring to temperature. Directly related to energy wasted as heat and not converted to lumens. The element or filament has a higher resistance and displaces heat. Inefficient unless heat is your goal as with a heat lamp. If you are referring to degrees of diffusion; that depends on the bulb or the shades diameter, the shades reflectivity and distance of the light source to the end of the shade.
Because the filament of a 25-W lamp has a higher resistance than that of a 60-W lamp and, therefore, will experience a greater voltage drop -the lamp with the voltage drop closer to its rated voltage (in this case, the 25-W lamp) will be the brighter.
Yes, the resistance of the filament of a light bulb is what generates enough heat to make the filament glow and produce light.
Since the bulb has a constant resistance, increasing the voltage will increase the current which will increase the brightness until the bulb blows in a flash of light.Ohm's Law governs this process where Volts = Current x Resistance.Another Answer Incandescent lamps do not have a constant resistance, as their resistance increases significantly with temperature. In fact, the 'hot resistance' of a lamp filament is around 10-18 times higher than its 'cold resistance'. The temperature increase, of course, is due to the current passing through the filament. This means incandescent lamps do notobey Ohm's Law and are, thus described as being 'non-ohmic' or 'non-linear'. However, to answer your question directly, the rated power of a lamp only occurs at the lamp's ratedvoltage. Even a small reduction in voltage will result in a significant loss of power and, thus, brightness. So, in this sense, a lamp's brightness is determined by the voltage applied across its filament.
there were three reasons, first, an effective incandescent material, next he was able to achieve a higher vacuum and third a higher resistance lamp that made power distribution from a central source an economical proposition
If the filament really was made from a material that has a negative temperature coefficient (as temperature increases, resistance decreases) then the decreasing resistance would cause more and more current to be taken as the lamp heated up and the temperature would get higher and higher in a runaway manner until either the power supply's breaker would trip or (more likely) the light bulb's filament would simply burn open. In fact the filament has to be made from a material that has a positive temperature coefficient. (As temperature increases, resistance increases.) Then, as the bulb's temperature rises, its filament's increasing resistance causes less current to be taken than when it was cold. Quite quickly a stable "steady-state" temperature and "running" resistance is reached so that the bulb simply continues to give out a steady amount of light according to the current it is taking from the electricity supply.
because there is a correlation between resistance and voltage and current. The equation resistance = voltage divided by current shows that the higher the voltage, the bigger the resistance,, and the bigger the resistance the hotter the filament lamp will get because of the electrons bumping into each other which means there is a loss of energy and that energy is being transferred to the filament making the actual filament bulb hot since there is more thermal energy wasted at the end.
A desk lamp which takes a regular size bulb. You could easily change it to a higher watt.
It depends on the voltage rating of each lamp, and the value of the supply voltage. It's important to understand that a lamp will only operate at its rated power (therefore at its full brightness) when subject to its rated voltage.So, let's assume each lamp is rated at, say, 24 V.If connected in parallel across a 24-V supply, then they will both operate of full brightness.If connected in series across the same 24-V supply, then each lamp will be subject to half its rated voltage, and will be very dim.On the other hand, if connected in series across a 48-V supply, then they will each be subject to 24 V, and will both operate at full brightness.