Watts and amps measure different aspects of electricity and are not interchangeable.
If current is 3 amperes and resistance is 1.5 ohms, then voltage is 4.5 volts. (Ohm's law: voltage equals amperes times resistance) Power is volts times amperes, or 13.5 watts. A watt is a joule per second, so suppling 13.5 watts for 4.5 seconds produces 60.75 joules.
A multimeter measures current in amperes and potential difference in volts. Wattmeters are used to measure watts and the reading is a combination of current being drawn and the voltage applied. watt = volts x amps
If this is a homework assignment, please consider trying to answer it yourself first, otherwise the value of the reinforcement of the lesson offered by the assignment will be lost on you.If a 100-watt bulb draws 0.87 amperes of current, 17 of them will draw about 14.8 amperes, (0.87 times 17), if they were wired in parallel.However, wiring them in series would not give you 0.051 amperes, (0.87 divided by 17), as one might expect, because the resistance-temperature coefficient of bulbs is quite dramatic, so more current would actually be drawn because the bulbs would be much cooler. How much more would require testing. You could do this by supplying 6.8 volts to one bulb and seeing what you get, or just hook 17 of them up in series to 115V.
A 100 ohm resistor carrying a current of 0.3 amperes would, by Ohm's Law, have a potential difference of 30 volts. A current of 0.3 amperes through a voltage of 30 volts would, by the Power Law, dissipate a power of 9 watts. You need a 10 watt resistor, alhough it is better to use a 20 watt resistor. E = IR 30 = (0.3)(100) P = IE 9 = (30)(0.3)
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.
Using Ohm's law, W(Watts)=E(voltage) x I(current), the answer is 10 Amperes.
No, current is always measured in amperes (A). watt hours is the unit for power.
To determine the current drawn by a watt lamp when connected to a voltage ( V ), you can use the formula ( I = \frac{P}{V} ), where ( I ) is the current in amperes, ( P ) is the power in watts, and ( V ) is the voltage in volts. For example, a 60-watt lamp connected to a 120-volt supply would draw ( I = \frac{60}{120} = 0.5 ) amperes.
"Amps" is a steady thing. There's no such thing as "Amps per hour".The current through a 24-watt load is[ 24/the voltage across the load ] Amperes.
If you refer to the units, power (any power, not just electrical power) is energy divided by time. The SI unit is the watt, equal to 1 joule/second.
You can convert amperes by using the formula: Amperes = Watts / Volts. To find out how much current 'X' watt of electrical equipment at 'Y' volt consumes per hour, you would need to divide the wattage by the voltage to get the amperes, and then also consider the duration of the consumption in hours.
To determine how long a 12-volt battery will last powering a 20-watt light, first calculate the current draw in amperes using the formula: current (A) = power (W) / voltage (V). For a 20-watt light at 12 volts, the current is approximately 1.67 amps. If you have a typical 12-volt battery with a capacity of 50 amp-hours, you can estimate the runtime by dividing the capacity by the current draw: 50 Ah / 1.67 A ≈ 30 hours. However, actual runtime may vary based on battery age, discharge rate, and efficiency.
A 6000 watt toaster oven, if one could exist, would pull 50 amperes from a 120 volt supply.The question is unrealistic, because the maximum branch current for a normal circut would be about 16 amperes, using the 80% rule, and that would produce about 2000 watts.
A 60 watt bulb at 12 volts will pull 5 amps of current.
Yes, and it's proportional. V * I = P Where V is the voltage in Volts, I is the current in Amperes, and P is the power in Watts. So we get: I = P / V For example, with a 240 Volt supply, a 12 Watt lamp would draw: 12/240 0.05 Amperes, or 50 milliamps
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.
The amperage flowing through the 60-watt bulb connected to a 120-volt circuit can be calculated using the formula P (power) = V (voltage) x I (current). In this case, the current can be found by rearranging the formula to I = P / V, which gives a current of 0.5 amperes flowing in the bulb.