A pure resistive load always has a power factor of one. This is because the current and voltage waveforms are in phase in an AC circuit.
A resistor doesn't have a power factor. However, if a circuit is pure resistance in nature the power factor will be one when a voltage is applied and a current flows in the circuit. The power factor is a measure of the relative phases of the current and voltage in a circuit.
Power Factor of an electrodynamometer can be improved by connecting a large resistor in series with the current coil.
The power generated in a resistor is converted into heat. and that can be power which is converted into heat is the product of the voltage across the resistor and, current passing through the resistor. or the product of square of the current and the resistance offered by the resistor.
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No, because the power dissipated in a resistor is proportional to the square of the current through the resistor but only directly proportional to the resistance of the resistor (I^2 * R) and the current through the lower value resistor will be higher than the current through the higher value resistor, the lower value resistor will usually dissipate more power.
When a resistor and an inductor are both connected to an AC supply, the current in the resistor is in phase with the voltage, while the current in the inductor is a quarter-cycle (90 degrees) behind. Supposing they both draw 1 amp on a 12-volt AC supply. The resistor will dissipate 12 watts, while the inductor will dissipate no power. Any power that enters the inductor comes back to the generator in a later part of the cycle. But the current drawn from the supply is 1.414 amps, so this would be a load with a power factor of 0.707.
It depends on the voltage applied across it. But the maximum current is limited by the power-rating of the resistor (power divided by the square of the voltage).
The heat generated by any particular resistor depends (at least electrically) solely on the power it dissipates. Power dissipation in a resistor is equal to current squared times resistance, and the current through the resistor is equal to the voltage across it divided by the resistance. If we take a 10 ohm resistor ('your resistor') and put it in a series circuit such that there is 10 volts across your resistor, the current through it will be 1 ampere (10/10=1). the power dissipated will be 10 watts (1^2 * 10=10). If we put your resistor in a parallel circuit that also puts 10 volts across it, then the current and power will be the same. Your resistor does not know or care where the voltage came from. From this point of view, once you get down to the voltage across the resistor, it does not matter what type of circuit it is in. On the other hand, for any given power supply voltage, then the type of circuit and the value of external components certainly does affect the terminal voltage and thus the current through as well as the power dissipated by the resistor. In a parallel circuit, the voltage across your resistor remains basically the same no matter what resistance you put in parallel with it (unless you overload the power supply or the power supply has high internal resistance). In this case, the voltage across the resistor is the same as the power supply, current is I=E/R, R being that resistor only, and power is P=I^2 * R. In a series circuit the current through the resistors is I=E/R, R being the total resistance (including the other resistor(s)). The power dissipation in your resistor will then be P=I^2 * R, I being the series current we just calculated, and R being your resistor only. Since the other resistors affect the current, and since the current is the same no matter where you measure in a series circuit, then the voltage across your resistor and thus the power dissipation will be affected. The voltage across your resistor will be E=I*R, I being the series current we just calculated, and R being your resistor only. So, while the calculation for power dissipated in a particular resistor does not change relative to what type of circuit it is in, the calculation to arrive at the voltage across the resistor and/or the current through it (which you will then need to calculate power) does. Keep in mind there are other mechanical parameters that influence the actual case temperature of the resistor. Physical size of the case, composition, and airflow velocity, if any, will alter the case-to-ambient thermal conductivity. Ambient temperature will also be a factor in the final temperature.
The current would be about 20 volts.
Depends on the current. Put a resistor in-line with the current, then measure the voltage across the resistor. V=RI. So, divide the measured voltage by resistor value. Be careful with the size of the resistor, as Power dissipated in a resistor is R*I^2 or V^2/2. So, a 1-Amp current into a 1 Ohm resistor will result in a 1Watt power dissipated in the resistor. If it's too small, it'll burn. Also, notice that if you do that, you haven't measured the current in the original circuit. You've measured the current when an extra resistor is installed in the original circuit, and that's different.
The current can't be calculated from the information given in the question.The power rating of a resistor is the maximum power it can dissipate before it overheatsand its resistance possibly changes permanently. The power rating is not the amount ofpower it always dissipates.So, all we really know about the resistor in the question is that its resistance is 21 ohms.And all we can say about the current through it is:Current through the resistor = (voltage between the ends of the resistor) divided by (21).
The power factor depends on the phase angle between the voltage and current on a conductor. The amplitude of the current has no effect on it.