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Due to energy usage and/or the reduction in conductance (increase in resistance) in a given load or resistor, some electrical energy is lost through that component. As such, a proportional drop in current and voltage occurs.

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11y ago

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Related Questions

What is the power dissipated by a reactive load?

No power is dissipated by a load composed exclusively of either capacitive or inductive reactance.


What process will double the power dissipated by a resistor?

Increase the voltage across the resistor by 41.4% .


2 resistors unequal value are in parallel. Would the power dissipated by the resistor with larger ohmic value be greater than the power dissipated by the resistor of lesser value?

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.


Ac power that is converted to heat in a resistor is known as?

real power (as opposed to imaginary power, which is not dissipated)


How Determine the power dissipated by the resistor?

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).


A series circuit has 100mA flowing through a 1.5kohm load What is the power dissipated by the load?

A series circuit has 100mA flowing through a 1.5kohm load. The power dissipated by the load is equivalent to 15 Watt. This is based on the formula, power is equals to square current times load.


The no-load output voltage of a DC power supply is measured at 15V When a 600 ohms load is connected to the output the output drops to 13.7V Calculate the internal resistance of the power supply?

This question can be answered using voltage dividers. Assume the power supply consists of a voltage source and a resistor. With no load, all of the voltage source's voltage is dissipated by the internal resistor of 15V. When there is a load, there are two resistors in series. To calculate the internal resistance:1. I=V/R. You know the 600ohm resistor dissipated 13.7V. So that would mean a current of 13.7/600=22.8mA2. If the 600ohm resistor dropped 13.7, kirchoff's voltage law would tell us the internal resistor dropped 15-13.7=1.3V.3. R=V/I, Use the current to calculate the internal resistance. 1.3/22.8mA = 56.9ohmsCommentFurther to the above answer, a voltage-source's voltage is not 'dissipated by the internal resistance when on no load'. On no load, there is no current passing through the internal resistance, so no 'voltage dissipation' can takes plac -i.e. the non-load voltage is 15 V.


What is the formula for calculating the power dissipated in a resistor, given the current flowing through it and the resistance value, known as the i2r power?

The formula for calculating the power dissipated in a resistor, known as the i2r power, is P I2 R, where P is the power in watts, I is the current in amperes, and R is the resistance in ohms.


What power is dissipated by an 8.2 k resistor if a current of 0.005 amps pass's through the resistor?

.205 watts or 205 mw


What is the power dissipated by a resistor with a current of 2 amps and a resistance of 1000 ohms?

I = 2A R = 1000Ω Power Dissipated P = I2R = (2A)2(1000Ω) = 4000W Voltage across resistor V = IR = (2A)(1000Ω) = 2000V


What is the power dissipated a resistor with a current of 0.02 amps and a resistance of 1000 ohms?

P = I^2 x R] P = 0.2^2 x 100 P = 4 W


What is the power dissipated by a resistor with a current of 0.02 A adn a resistance of 1000 ohms?

Power dissipated = I2R 0.022 x 1000 = 0.4 watts