A 12 ohm resistor with 6 volts across it will dissipate 3 watts of power.
Current = voltage divided by resistance = 6 / 12 = 0.5 amperes.
Power = voltage times current = 6 * 0.5 = 3 watts.
Volts = Amps (i) X Resistance (or Impedence in AC)
Here I = 6 volts/12 ohms = 0.5 Amps
Power= Amps (I) X Volts
Power = 0.5 X 6 Volts
Power = 3 Watts
You first have to know the voltage that is going to be submitted to. Then the power
that will dissipate will be given by: P = V2/R
It will depend on the voltage u apply..... It can be calculated as.... V/12 ; V is the voltege that u apply... Regards; Sujay; http://sujay00.blogspot.com
power = voltage x current = 24 volts x 2 amperes = 48 watts
power = current2 x resistance = (2 amperes)2 x 12 ohms = 48 watts
P = E2 / R = 3,600 / 620 = 5.8065 watts(rounded)
48 watts
3 watts
You have two resistors, each with resistance of 12Ω, and a 12-volt battery. 1). The resistors are in series across the battery. ..... A. voltage across each resistor ..... B. current through each resistor ..... C. power dissipated by each resistor ..... D. total power delivered by the battery 2). The resistors are in parallel across the battery. ..... A. voltage across each resistor ..... B. current through each resistor ..... C. power dissipated by each resistor ..... D. total power delivered by the battery ============================================ 1). ... A. 6 volts ... B. 0.5 Amp ... C. 3 watts ... D. 6 watts 2). ... A. 12 volts ... B. 1 Amp ... C. 12 watts ... D. 24 watts
0.069444444444444444444444444444444444 ohms. P/E^2=R. P = power in watts. E = electricity in volts. R = resistance in ohms.
there might be ways to get the power rating by measuring the size of the resistor. but as the physical size of the resistor increases, its power rating also increases..
Well, if the question says "use P = VI", just do it! You have the values of voltage (V) and current (I) right there.
Power is defined as the rate at which energy is used,generated, moved, converted, or dissipated.
Increase the voltage across the resistor by 41.4% .
P = (E2)/R = 81/9 = 9 watts
The power dissipated by a 10 ohm resistor with 800v across it is 64 kw.Ohm's law: current is voltage divided by resistancePower law: power is voltage times current, so power is voltage squared divided by resistanceDon't even think about trying this. 64 kw is a lot of power. The resistor will probably explode, or catch fire. At best, the 80 amps required will trip your circuit breaker, if you are lucky.
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).
I = 2A R = 1000Ω Power Dissipated P = I2R = (2A)2(1000Ω) = 4000W Voltage across resistor V = IR = (2A)(1000Ω) = 2000V
Power dissipated in a resistance = E2/R = (100)2/100 = 100 watts.
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
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.
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.
You just stated that the voltage across the resistor is 15 volts, so that's your answer ! If the resistor is connected to a 15-V battery or to the output of a 15-V power supply, then a meter across the resistor is also across the power supply, and reads 15 volts. The current through the resistor is (V/R) = (15/2700) = 5.56 mA. The power dissipated by the resistor (and delivered by the battery) is (V2/R) = (225/2700) = 0.083 watt.
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).
real power (as opposed to imaginary power, which is not dissipated)