The answer to this is a bit more complicated than you might think due to the strong positive temperature coefficient of resistance of the filament material.
When operating at its stable temperature (and voltage) this is easy to compute using Ohm's Law and the Power Formula, as follows:
At intermediate temperatures the resistance will be somewhere in between these values.
It makes no difference if the bulb is powered by 24VDC or 24VAC. as AC is specified in RMS (root mean square) which is equivalent to DC. It's worth pointing out that, because the resistance of a tungsten lamp varies with variations in voltage, it is considered to be a 'non-linear' or 'non-ohmic' device, meaning that it does not obey Ohm's Law (unless the temperature is constant, but as every change in voltage changes the temperature and the temperature takes time to change following the voltage change making the actual resistance as a function of voltage very difficult to determine).
no , it will burn out
A 100 watt 220 volt light bulb (or anything consuming 100 watts on 220 volts) draws 100/220, or .45 Amps. It will also have about 220²/100, or 484 ohms resistance. A 60 watt 220 volt light bulb (or anything consuming 60 watts on 220 volts) draws 60/220, or .27 Amps. It will also have about 220²/60, or 807 ohms resistance.
The current flowing through a bulb is equal to the (voltage across the bulb) divided by the (bulb resistance), and can be expressed in Amperes. The rate at which the bulb dissipates energy is equal to (voltage across the bulb) times (current through the bulb), and can be expressed in watts.
If you add one extra bulb and the voltage remains constant, then you have doubled the current drained from the regulator. 12 Volt and One 12 Watt light bulb drains 1 Ampere Current. 12 Volt and Two 12 Watt light bulbs drains 2 Ampere Current. However: If having a 24 volt power source and you add two 12 Volt 12 Watt in serial, then you still only drain 1 Ampere Current. NOTE: Wattage and Voltage of bulbs may be different even if the sockets are the same. Lower voltage on the bulb will increase the current drain, if voltage is a lot lower it might cause the circuit delivering voltage to burn out or blow a fuse. It can also quickly burn the bulb, sometimes in a fraction of a second. It will however do little damage to add a bulb with higher voltage than the circuit is designed for. You will then only observe that you do not get the light you might hope for. Total Current/Ampere= Combined Wattage divided by Voltage Total Wattage = Combined Current or Ampere multiplied by Voltage. In simpler words: If you double the bulbs, twice the current is drained from the battery
V = I x R so current I = 1/2 amp. I bet the bulb is rated at 60 W because Watts = Current x Voltage. Where V = voltage (volt) I = current (ampere) R = resistance (ohm) Your question isn't easy to answer. A lamp has two 'resistances': a 'cold' resistance, and a 'hot' resistance. Before it is energised, it is cold, so its resistance is low; when it is energised, it becomes very hot, and its resistance increases significantly. So, the question is whether your '240 ohms' is the cold resistance or the hot resistance. If it is the cold resistance, then a current of 0.5 A will flow through it for a fraction of a second, then its resistance will increase significantly, and the current will fall to a very much smaller value.
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.
yes the bulb will actually last longer
no , it will burn out
No, they do not draw the same current. The current drawn by an electrical device is determined by the power (Watts) and voltage (Volts) using the formula: Current (amps) = Power (Watts) / Voltage (Volts). So, the 12 volt 50 watt bulb will draw higher current compared to the 230 volt 50 watt bulb.
A 100 watt 220 volt light bulb (or anything consuming 100 watts on 220 volts) draws 100/220, or .45 Amps. It will also have about 220²/100, or 484 ohms resistance. A 60 watt 220 volt light bulb (or anything consuming 60 watts on 220 volts) draws 60/220, or .27 Amps. It will also have about 220²/60, or 807 ohms resistance.
12 volts is enough for a 12-volt 100-watt light bulb. It would not be enough for a 120-volt or 240-volt bulb.
the 220 volt bulp in 220 volt ac current
On this calculation I am assuming that the light bulb is using a 120 volt source. Watts = Amps x Volts. Amps = Watts/Volts, 40/120 = .33 amps. R = Volts/Amps, 120/.33 = 363.6 ohms resistance in the 40 watt light bulb.
Yes, you can replace a 110 volt 35 watt halogen bulb with a 120 volt 35 watt bulb, as the wattage is the same. The slight difference in voltage (110V vs. 120V) generally won't affect performance, as most bulbs can operate within a range of voltages. However, ensure the fixture is rated for the wattage to avoid overheating. Always check the specifications of your fixture for compatibility.
That depends on what voltage it's designed to operate from. Power = (voltage)2 / R R = Voltage2 / power If it's a 117-volt bulb, R = (117)2 / 28 = 489 ohms. If it's a 240-volt bulb, R = (240)2 / 28 = 2,057 ohms.
The current through a 220 volt 150 watt bulb is I = W/E = .68 amps. The resistance of that bulb is R = E/I = 324 ohms. The wattage used by the 220 volt bulb when 110 volts is applied W = E(sqd)/R = 37 watts. Half the voltage with the same resistance will quarter the wattage output.
No, the highest wattage bulb will have the lowest resistance.