Power dissipated by the resistor = I^2 * R or V^2 / R, where R = its resistance value, I = the current in the resistor, and V = the voltage drop across the two terminals of the resistor. You need to measure or find the information of either I (using an ammeter) or V (a voltmeter).
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.
P = I^2 x R] P = 0.2^2 x 100 P = 4 W
In order to determine this, it will be necessary to find which resistor 'maxes out' at the lowest voltage. This can be found using the equation Vi=sqrt (Pi*Ri) for each resistor, where Pi is the power rating of resistor i and Ri is the value of resistor i. Once this is found, the power dissipation of each other resistor can be found using the equation Pi=(Vl^2)/Ri, where Vl is the voltage that maxes out the resistor which maxes out at the lowest voltage, and Ri is the resistance of each resistor. The equivalent power rating would then be the sum of the power dissipated across each resistor.
The power dissipated by a circuit with a voltage of 12V and a current of 3A is 36W. Watts is Volts times Amps.
Power dissipated is always Volts times Amps. W= V*I because of ohm's law, V=I*R, you can substitute either the voltage or amperage with the other value; W= V^2/R or W= I^2*R.
Increase the voltage across the resistor by 41.4% .
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.
real power (as opposed to imaginary power, which is not dissipated)
.205 watts or 205 mw
I = 2A R = 1000Ω Power Dissipated P = I2R = (2A)2(1000Ω) = 4000W Voltage across resistor V = IR = (2A)(1000Ω) = 2000V
Power dissipated = I2R 0.022 x 1000 = 0.4 watts
P = I^2 x R] P = 0.2^2 x 100 P = 4 W
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
In order to determine this, it will be necessary to find which resistor 'maxes out' at the lowest voltage. This can be found using the equation Vi=sqrt (Pi*Ri) for each resistor, where Pi is the power rating of resistor i and Ri is the value of resistor i. Once this is found, the power dissipation of each other resistor can be found using the equation Pi=(Vl^2)/Ri, where Vl is the voltage that maxes out the resistor which maxes out at the lowest voltage, and Ri is the resistance of each resistor. The equivalent power rating would then be the sum of the power dissipated across each resistor.
You may find it helpful to use Ohm's law and the definition of electrical power.
There is insufficient information in the question to answer it. You need to provide either the voltage across the resistor, or the power dissipated by the resistor. please restate the question.
66W 230/800= 0,2875. 230= 66w