Power=current squared times resistance
You may be mixing up 'impedance' with 'apparent power'.Impedance is the vector sum of an a.c. load's resistance and reactance, and is expressed in ohms. Apparent power is the vector sum of true power and reactive power, and is expressed in volt amperes.Apparent power is also the product of the square of the load current and the impedance of the load.
an ohmmeter contains its own power source that it uses to test the circuit. if power is still on in the circuit, at the least the meter will give incorrect readings, at the worst the meter may be damaged rendering it useless.
You have it backwards, the resistance controls the current not the current controls the resistance. I = E/R . Your question should read, "If the voltage is constant and the resistance in the circuit is increased what happens to the current?" Say the voltage is 120 volts and the resistance is 30 ohms, I = 120/30 = 4 amps. Now we double the resistance to 60 ohms, then I = 120/60 = 2 amps. So now you can see if you increase the resistance the current drops.
For single phase the formula is HP = I x E x %eff. x pf/746. For three phase the formula is HP = I x E x 1.73 x %eff. x pf/746. Where I = amps, E = volts, %eff. = efficiency of the motor and pf is the power factor of the motor.
The electrical resistance of a light bulb increases when it is turned on As a resistor, the tungsten light bulb has a positive resistance coefficient. This means that the electrical resistance goes up when the filament becomes hot. For example, a 100 watt light bulb operated at 120 volts - it does not matter if it is AC or DC for this calculation - will have a resistance of 144 ohms when hot and draw .833 ampere. When cold the filament typically has a resistance of only 10 ohms which increases as the filament heats up.
You may be mixing up 'impedance' with 'apparent power'.Impedance is the vector sum of an a.c. load's resistance and reactance, and is expressed in ohms. Apparent power is the vector sum of true power and reactive power, and is expressed in volt amperes.Apparent power is also the product of the square of the load current and the impedance of the load.
You may be mixing up 'impedance' with 'apparent power'.Impedance is the vector sum of an a.c. load's resistance and reactance, and is expressed in ohms. Apparent power is the vector sum of true power and reactive power, and is expressed in volt amperes.Apparent power is also the product of the square of the load current and the impedance of the load.
Length is not a value in itself. It is an attribute of objects and, in the context of an object, it may have a value. That value can be expressed as a power of 10.Length is not a value in itself. It is an attribute of objects and, in the context of an object, it may have a value. That value can be expressed as a power of 10.Length is not a value in itself. It is an attribute of objects and, in the context of an object, it may have a value. That value can be expressed as a power of 10.Length is not a value in itself. It is an attribute of objects and, in the context of an object, it may have a value. That value can be expressed as a power of 10.
Efficiency = Output/Input. This is usually expressed as a percentage but may be given in the form of a ratio.Another AnswerEfficiency is output power divided by input power, normally expressed as a percentage.
Establish the federal reserve system
check your timing
In electronic circuits, resistance is used to empede the flow of current. You may want a specific voltage at a given location, so by using resistors, you can obtain that voltage. You may want to limit current, which a resistor can do for you. But resistance is inherent in everything. Copper wire has resistance, your toaster is really nothing more than a resistor. Resistance is generally considered the load in electrical systems. If it were not for resistance, everything would be considered a short circuit. So you see resistance, although many times not desired, is required.
The most important of the expressed presidential powers in the United States is probably the power to veto. A president who passes laws just to agree with Congress is not going to achieve the goals they set forth when running for office.
They are usually rated by their power factor. With 100 volts at 1 amp the load uses 100 watts maximum, in general it would be 100 watts times the power-factor.For example a small induction motor might have a PF of 0.7.AnswerIn practise, there are very few purely-inductive loads. Most are resistive-inductive, and rated according to their apparent power expressed in volt amperes. A purely-inductive load would be rated according to its reactive power, expressed in reactive volt amperes. No load is ever 'rated' according to its power factor.
The abbreviation for "five times a day" is "5 times a day" or can also be expressed as "5xd." In medical contexts, it may be referred to as "q.i.d." (quater in die), which means four times a day, but for strictly five times, "5xd" is more appropriate.
I'm not sure there is a term other than "power"; V*A, or the vector sum of real and reactive power is equivalent to the power calculated by multiplying the voltage times the current, ignoring phase shift.AnswerThe product of voltage and current in an a.c. circuit is called 'apparent power', expressed in volt amperes, in order to distinguish it from 'true power' (in watts) and 'reactive power' (in reactive volt amperes).
Yes, the efficiency increases. Yes, it is linear. Power lost in a current-carrying conductor is: P = I^2 * R So, if you halve the resistance, you halve the power loss. Note though that the current (I) term is squared. So if you can decrease the current by increasing the transmission voltage, the increase in efficiency is not linear, but exponential! Halve the current (and double the voltage to get the same power), and you reduce losses by four times! This is why utilities use such high voltages for transmission. Superconductors are no different, you are still talking about a reduction in resistance, superconductors just achieve a much lower resistance than a standard conductor. The question is whether the cost of superconductors and their cooling systems (currently very high) outweigh the modest gain in transmission efficiency.