I = E ÷ R = 120V ÷ (60Ω + 40Ω + 20Ω) = you figure it out now
A 120V power supply connected to a 30 Ohm resistor will produce 120/30 or 4 amps of current.
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.
Because they are "in-phase". In order to get 240v, you need two 120v Alternating Current lines that are 180° out of phase, that is, opposite phases. Only when one line is +120v and the other -120v will you see 240v between the wires.
Because the white wire on a 120 volt circuit is the neutral wire that is connected to the silver screw on outlets and switches. It is connected to the neutral bar in the service panel.
R = E / I = 120/2 = 60 ohms.
The current in the light bulb connected to the 400 V source would be higher compared to when connected to the 60 V source, assuming the resistance of the light bulb remains constant. This is because current is directly proportional to voltage according to Ohm's Law (I = V/R), so a higher voltage will result in a higher current flow in the circuit.
A high watt lamp draws more current than a low watt lamp because power (watts) is the product of current and voltage. If the voltage remains constant, a higher power rating means a higher current is required to supply that power. This is in accordance with Ohm's Law, which states that current is directly proportional to power and inversely proportional to resistance.
without knowing load, cannot say
A reverse connected 120V ballast is a type of ballast that can operate a fluorescent lamp with reversed polarity. This means that it allows the lamp to be connected to the power source in reverse compared to a standard ballast, providing flexibility in certain lighting applications.
All transformers are designed to work on AC. They do not work on DC.If you connect an inductor to DC, the current will increase until the capacity of the source or the conductance (1/resistance) capacity of the inductor and conductors is reached. Often, this condition will overheat and destroy the inductor, or destroy the source. A transformer is not an exception, as it is a form of inductor.
A 100 watt bulb typically draws around 833 milliamps (mA) of current. This calculation is done using Ohm's Law formula: Current (I) = Power (P) / Voltage (V). With a standard 120V power source, a 100W bulb would draw approximately 833mA.
I = E ÷ R = 120V ÷ (60Ω + 40Ω + 20Ω) = you figure it out now
If a 227V power source is connected to a 120V ballast, the ballast may get damaged due to the excessive voltage. It is important to always ensure that the voltage input matches the rated voltage of the ballast to prevent such issues.
A lamp doesn't care whether it operates from a 12-V d.c. supply or from a 120V a.c. supply -all that matters is the supply must be 12 V!If it's connected to a 12-V d.c. supply, then it will draw a current of 8.33 A; if it's connected to a 12-V a.c. supply, it will still draw a current of 8.33 A.
The formula you are looking for is V = IR where V = Voltage I = Current R = Resistance With some formula manipulation and numbers plugged in you get I = 120V / 9.6Ω I = 12.5A The kettle would have 12.5 volts of current running through it.
Assuming that you mean connected to 120 Volts (V) supply, start with Power(P)=Current(I) x Voltage(V), and if P=V*I, then I=P/V. I=12W/120V=0.1A or 100mA and that is your current.