The current drawn by the lamp can be calculated using the formula I = P/V, where P is the power (40 W) and V is the voltage (220 V). Substituting the values, I = 40 W / 220 V = 0.182 A. Therefore, the lamp will draw approximately 0.182 amperes of current from the main line.
The formulas you are looking for is I = E/R.
To calculate the current, you can use the formula: current (in amps) = power (in watts) / voltage (in volts). So, for a 50W halogen lamp operating at 12V, the current draw would be approximately 4.17 amps (50W / 12V = 4.17A).
Using the formula Power = Voltage x Current, we can calculate the current: Current = Power / Voltage. Plugging in the values, we get 1500W / 120V = 12.5A. So, a 1500W resistance heater would draw 12.5A of current at 120V.
This will be marked on the label on the iPad and its original charger.
Since power is volts time amps, the current in a 60W lamp connected to 120V is 0.5A. Since a lamp is a resistive load, there is no need to consider power factor and phase angle, so that simplifies the explanation. ======================== Assuming this is an incandescent or halogen lamp (using a filament to make the light) there is a trick here: the resistance of a lamp filament varies with temperature and does not follow Ohm's law. The resistance will be much lower, thus the current will be much higher when the filament is cold, when the lamp is first connected. As the filament heats up, the resistance increases until it gets to a steady operating point of 0.5A. For a halogen lamp, the operating temperature is about 2800-3400K, so the R at room temperature is about 16 times lower than when hot... so when connected, the current is about 8A but drops rapidly. The current could be even higher if the lamp is in a cold environment. Non-halogen lamps operate at a lower temperature and would have a lower initial current--about 5A. And this all assumes the lamp is rated for 120V. If it is a 12V/60W lamp, the filament will probably break and create an arc, which may draw a very large current.
The current drawn by the lamp can be calculated using the formula I = P/V, where P is the power (40 W) and V is the voltage (220 V). Substituting the values, I = 40 W / 220 V = 0.182 A. Therefore, the lamp will draw approximately 0.182 amperes of current from the main line.
The formulas you are looking for is I = E/R.
40W Bulb will spiol due to over current passing through its coilAnswerSince the 40-W lamp has a higher resistance than the 100-W lamp, the greater voltage drop will appear the 40 W lamp. As a result the 40 W lamp will be subjected to a voltage beyond its 100-V rating, and the 100-W lamp will be subjected to a voltage below its 100-V rating. Therefore, the 40-W lamp will burn much more brightly than the 100-W lamp.Incidentally, the symbols for the 'watt' and 'volt' are upper, not lower, case: W and V.
Either a short to ground, or too much current draw in that circuit.Either a short to ground, or too much current draw in that circuit.
5500Watts/220V=25 Amps
An ammeter is connected in series with the load to measure the current flowing through the load. By placing the ammeter in series, it becomes a part of the circuit path so that all the current flowing through the load also passes through the ammeter, allowing for an accurate measurement of the current.
How much what? Power? 102mW (0.102W). How much light? A filament bulb produces 1 to 3 percent of the power input as light if the proper voltage is applied. If the lamp is rated for a higher voltage, it won't get hot enough and the light out will be a lot less.
To calculate the current, you can use the formula: current (in amps) = power (in watts) / voltage (in volts). So, for a 50W halogen lamp operating at 12V, the current draw would be approximately 4.17 amps (50W / 12V = 4.17A).
Without the battery hooked up, the current draw is too much for the alternator to keep fire on the plugs.
Yes you can. The battery supplies only as much current (amps) as the lamp draws when connected to 6 volts. The "12 amp" battery won't supply any more current when the lamp is shining than the "6 amp" battery did, but it'll last twice as long between charges.
It would be pretty much undefined, since the filament of the halogen bulb would fail immediately then there would be an open circuit with no current draw. <<>> The formula for current is Amps = Watts/Volts. The lamp itself would draw 4.16 amps. Since the voltage of the lamp is 12 volts there is a internal transformer involved in the fixture itself. It doesn't matter what the input (primary) voltage to the transformer is, so long as it meets the manufacturer's specification as to the proper voltage to operate the fixture.