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)
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).
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.
A typical resistor will burn out when it dissipates power in excess of double its power dissipation rating for an extended period of time. The power dissipated by a resistor is equal to I2R or E2/R, where E = the voltage across the resistor I = the current through the resistor R = the resistance of the resistor
Who can tell? The power rating of a resistor simply tells us the maximum power that resistor is capable of handling; it doesn't tell us anything about the actual power being produced for any given current. So, to find out the voltage drop across that resistor, you will need to find out its resistance, and multiply this value by the current you specify.
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 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.
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.
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 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.
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.
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).
A typical resistor will burn out when it dissipates power in excess of double its power dissipation rating for an extended period of time. The power dissipated by a resistor is equal to I2R or E2/R, where E = the voltage across the resistor I = the current through the resistor R = the resistance of the resistor
It depends on the purpose for installing the resistor. If the intent is to decrease current flow, the resistor must be connected in series with the load. If the purpose is to increase current flow, the resistor must be connected in parallel with the load. To connect a resistor in series, connect the resistor to one side of the power source, in line with the load. This will decrease circuit current flow. To connect a resistor in parallel, connect the resistor between the positive and negative sides of the power source, which will effectively connect the resistor across the load . This will increase current flow through the circuit. However, before connecting a component in parallel, make sure the increase in current flow will not exceed the current rating of the circuit or fuses/breakers will blow.
The power in a resistor (in watts) is simply the product of the current (in amperes) times the voltage (in volts).The power in a resistor (in watts) is simply the product of the current (in amperes) times the voltage (in volts).The power in a resistor (in watts) is simply the product of the current (in amperes) times the voltage (in volts).The power in a resistor (in watts) is simply the product of the current (in amperes) times the voltage (in volts).
Four times as much.