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Light bulb in the home - alternating current (A/C) Light bulb in a car - direct current (D/C) Output of a battery charger - direct current Input of a battery charger - usually alternating current
To find the resistance necessary, one would need to know how much current the bulb draws. If one knows the current the bulb draws, then one would subtract the 14 volts from 120 volts then divide that by the current the bulb draws and one will find the resistance needed. Once this has been done, one would need to multiply the current drawn by the voltage drop to get the wattage rating necessary. Another important detail to note is that the power dissipated by the resistor will be much greater than the power consumed by the bulb itself. Finally if the bulb burns out the voltage across the contacts will be 120V. I would not recommend using this method to drop the voltage for the bulb.
Basically, Power = Current*Voltage Current = Power/Voltage Current = 15/120 Current = 0.125A or 125mA
No, a higher wattage INCANDESCENT light bulb uses more current than a lower wattage INCANDESCENT light bulb. Some CF and LED bulbs are rated by the amount of light that an incandescent bulb would produce, but they are also rated by the wattage that they use.
When lighting a light bulb, it is changed into light and heat/thermal energy.
V = irv = (0.5)(250)v = 125
the light bulb has changed our lives because with out it the world would be dull
The new bulb has higher resistance than the old one did, and it's designed to use less electrical energy than the old one used.
Light bulb in the home - alternating current (A/C) Light bulb in a car - direct current (D/C) Output of a battery charger - direct current Input of a battery charger - usually alternating current
Ohm's law applies: Current = Voltage / Resistance As such if you double the resistance of the light bulb you end up with half as much current.
An example of a parallel circuit would be the light bulbs in track lighting. Each bulb has the same voltage applied. The current through any one light bulb equals the voltage divided by the resistance of the bulb. The current also equals the wattage of the bulb divided by the voltage. So if all the bulbs had exactly the same resistance the current would be the same. However, your question says "always" so in general the answer is no. In the case of track lighting if you had a 60 watt bulb in parallel with a 120 watt bulb, the 60W bulb would draw 1/2 amp and the 120W bulb would draw 1 amp. The sum of the current flowing in a parallel circuit equals the sum of the current in each leg of the circuit.
The current of one bulb (two bulbs shorted) would be about1 three times the current of three bulbs.1 I say "about" because resistance is a function of temperature, and running three times the current through one bulb will make that one bulb much hotter, increasing its resistance. It might also burn out the bulb.
You know if current is flowing in a bulb circuit because, if there is enough power (voltage times current), the bulb will illuminate. If there is current, but not enough power to illuminate the bulb, you will need to measure the current with an ammeter to see if there is any current.
The brightness of a bulb would not change if you added a second bulb in parallel with the first.Unless, of course, the increased current exceeded the power supply's capacity causing a reduction in voltage.
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.
To find the resistance necessary, one would need to know how much current the bulb draws. If one knows the current the bulb draws, then one would subtract the 14 volts from 120 volts then divide that by the current the bulb draws and one will find the resistance needed. Once this has been done, one would need to multiply the current drawn by the voltage drop to get the wattage rating necessary. Another important detail to note is that the power dissipated by the resistor will be much greater than the power consumed by the bulb itself. Finally if the bulb burns out the voltage across the contacts will be 120V. I would not recommend using this method to drop the voltage for the bulb.
To increase the current you either need to reduce the resistance of the load or increase the voltage. Typically a higher wattage light bulb will decrease resistance or you could put multiple batteries in series.