Power 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.
Power 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.
Given 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
P = VRMS * IRMS. If using DC voltage, the power will be the current times the voltage.
Resistors 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.
Increase the voltage across the resistor by 41.4% .
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).
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.
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.
I = 2A R = 1000Ω Power Dissipated P = I2R = (2A)2(1000Ω) = 4000W Voltage across resistor V = IR = (2A)(1000Ω) = 2000V
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 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.
A resistor's resistance is measured in ohms. The higher the resistance the less current will flow with a constant voltage applied across the resistor. In terms of Ohm's Law Voltage = Current x Resistance.