The formula you are looking for is
V = IR
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.
The resistance of the filament in a light bulb is(voltage at which the bulb is designed to operate)2/(the rated power/watts of the bulb)
No. By definition a resistor provides opposition to the flow of "energy" in a circuit. Electric stoves and ovens operate by using the heat generated by the resistance to electric flow provided by resisting elements in the circuit.
ANSUL Systems operate automatically when fire linkages are connected and the tension lever is ON. ANSUL can also be deployed manually by pulling the handle.
1. The nature of the information flow designed into the system.2. The kind of components included into the system3. The relationship of control to the decision process.
I was also searching for the same but according to me the lamps with higher wattage should glow brighter. We know that, P=VI In series connection current is same through all the lamps but voltage across each lamp depends on the wattage of the bulb. So the bulb with higher wattage will draw more voltage and glow brighter.AnswerUnfortunately, the previous answer is incorrect, although it seems* to be the logical answer! The higher-wattage lamp has a lower resistance than the lower-wattage lamp. So, when they are connected in series, the larger voltage drop (IR)will appear across the lower-wattage lamp. As power is proportional to the square of the voltage, it is the lower-wattage lamp that will be the brighter.[*Many people have the mistaken belief that a higher-wattage lamp has a higher resistance than a lower-wattage lamp. That's the wrong way around!]
yes because of bulb resistance :)
ANY METER needs some kind of current flow to operate. Internal in the meter there are batteries that provide current that when passed trough a resistor will develop voltage as a function of the current. the meter will read this current and display the resistor size to cause this current to flow.
No it does not. A volt meter only reads the current that is passing through it.AnswerAll instruments draw some (albeit tiny) current from the circuit under test in order to operate. So, if this is what you mean by 'taking power from circuit', then the answer is yes, it does.Instruments also change the normal resistance of the circuit being tested -for example, ammeters increase the resistance of the circuit into which they are connected, while voltmeters decrease the circuit resistance across which they are connected. So adding a voltmeter (or an ammeter) to a circuit affects the operation of that circuit to some degree. To minimise this interference, it is important that an ammeter's internal resistance is very much lower than the circuit's resistance, and a voltmeter's resistance is very much higher than the circuit's resistance.
(R) Resistance = (E) Voltage / (I) Current in Amperes R = 12/3 R = 4 ohms
An incandescent light bulb is essentially a wire through which current flows. The wire gets heated up and glows giving off light. According to Ohm's Law Volts = Current x Resistance. In this case resistance is the resistance of the filament. Since in a residence the voltage remains more or less constant, the way more energy would be use is if more current passed through the filament. If the resistance of the filament decreased then the current would increase. Not sure how an aging bulb would have a decreased resistance. If this were somehow possible such that the composition of the filament changes and reduced resistance then the energy required to operate the bulb would increase.
When electric current travels through a conductor, there is always resistance. This resistance causes some of the energy of the current to express as heat. Additionally, the movement of the current causes a magnetic field to form around the current in a clockwise direction. This principle is what allows coil heaters and induction motors to operate.
Excessive resistance (drag), will cause motor to draw more amps (current flow) to operate.
The solenoid principle. A coil of wire is connected to some sort of electric current.
To determine the value of Stabilizing resistor Rs = Vs/Is = If(Rct +2Rl)/Is Where, Rs = resistance value of the stabilizing resistor Vs = voltage at which the relay will operate Is = current flowing through the stabilizing resitor and the relay If = maximum secondary fault current magnitude Rct = internal resistance of the current transformer Rl = resistance of attached wire leads
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.
Yes. As the length of a lever increases, the force needed to operate it (at the end of the lever) is increased.
In the U.S. 120 volts. <<>> Using the equation E = I x R, Volts = Amps x Resistance = 110 volts.