P = (E2)/R = 81/9 = 9 watts
2½ ie 2 to the power of one-half. The power "one-half" applied to any value signifies that value's square root. Do not confuse with the power "-1" which indicates the reciprocal of the given value.
5/2 = 2.5 Newtons per square meter = 2.5 Pascals
Could be Pulse Repetition FrequencyActually it is a bit more mundane, it is PReFerred, as in the Preferred number. It replaces an entire list of previous prefix's to military numbers such as C for Connector, R for Resistor, W for Wire, etc.An example:MIL-PRF-27/356C succeeds MIL-T-27/356B and is the specification for:TRANSFORMERS AND INDUCTORS (AUDIO, POWER AND HIGH-POWER PULSE), INDUCTORS, POWER, TF5S04ZZ
Clyde made the error of not telling us what he wanted simplified.
(2.5 * 3)3 = 7.53 = 421.875 or 2.5 * 33 = 2.5 * 27 = 67.5 Note: The third power can be applied either before multiplying the two numbers or after. Hence I have given both. Question must be rephrased to mention what exactly is required. In that case, the incorrect answer can be removed.
Increase the voltage across the resistor by 41.4% .
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
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
Power dissipated in a resistance = E2/R = (100)2/100 = 100 watts.
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)
I = E / RIf the voltage across the resistor is 90 volts, and the resistance of the resistoris 9 ohms, then the current through the resistor is90/9 = 10 Amperes.Don't try this at home!The power dissipated by the resistor is E2/R = (90)2/9 = 900 watts. That's comparable to the power (heat) dissipated by a small toaster. A common composition resistor will get hot and possibly explode if it's asked to dissipate that kind of power.