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∙ 14y agoPower dissipation of a resistor or any load is the amount of power (in watts) that is converted to heat, light, or other form of energy. In a resistor, power dissipation is defined by Ohm's law P = I^2 * R Power dissipated equals current through the resistor squared times the resistance in ohms. Since the power is converted to heat, a resistor has a maximum dissipation rating set by the manufacturer, above which the resistor will be damaged.
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∙ 15y agoWiki User
∙ 15y agoPower dissipation is measured in watts. Watts is Amps times Volts. Measure the voltage across the transistor, measure the current flow through it, and you can calculate watts very easily. (Of course, you need to make sure your measuring devices do not introduce error beyond what you can tolerate in your measurements.) In an AC circuit with reactance, its a bit more complex (pun intended) as you need to look at phase angle of the current, differentiating between watts and vars, but the equation is the same.
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∙ 14y agoGiven voltage and resistance, power is voltage squared divided by resistance.
Ohm's law: Current is voltage divided by resistance
Power law: Power is voltage times current, or voltage squared divided by resistance
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∙ 14y agoP = VRMS * IRMS. If using DC voltage, the power will be the current times the voltage.
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∙ 14y agoResistors dissipate energy as heat.
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.
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.
Voltage can be calculated using Ohm's Law:Voltage = Current (A) x Resistance (Ω)Voltage = 4A x 3Ω = 12 VoltsTherefore, the battery is a 12 Volts.The power dissipated is Voltage x CurrentPower = 4A x 12V = 48 Watts
The voltage across the resistor is whatever voltage is applied. The only maximum here would be a voltage that would damage the resistor. If you think this might happen, you'll have to look up such a voltage from the data sheets.
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).
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.
Like Ohm's Law, the formula for calculating power is a simple product of two quantities. It is given by the formula P = VI, where V is the voltage in volts and I is the current in amperes (or simply amps). So, if you know the value of any two of the quantities, you can easily calculate the third with simple arithmetic. For example, if the current flowing through a resistor is two amps and the voltage drop across that resistor is five volts, the power dissipated by the resistor is, P = VI = 5 volts * 2 amps = 10 watts. If you are given the power and the voltage, you can easily find the current. For example, if you are told that the voltage drop across a resistor is five volts and is dissipating 10 watts, the current through the resistor is 10 watts/5 volts = 2 amps.
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. If a voltage is applied across a resistor, a current flows through it.
Voltage can be calculated using Ohm's Law:Voltage = Current (A) x Resistance (Ω)Voltage = 4A x 3Ω = 12 VoltsTherefore, the battery is a 12 Volts.The power dissipated is Voltage x CurrentPower = 4A x 12V = 48 Watts
The voltage across the resistor is whatever voltage is applied. The only maximum here would be a voltage that would damage the resistor. If you think this might happen, you'll have to look up such a voltage from the data sheets.
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.
I = 2A R = 1000Ω Power Dissipated P = I2R = (2A)2(1000Ω) = 4000W Voltage across resistor V = IR = (2A)(1000Ω) = 2000V
The applied voltage is 53+28 = 81V.
The power dissipated by a circuit with a voltage of 12V and a current of 3A is 36W. Watts is Volts times Amps.