Power Dissipation in Resistance
When current flows in a resistance, heat is produced because friction between the moving free electrons and the atoms obstructs the path of electron flow. The heat is evidence that power is used in producing current. This is how a fuse opens, as heat resulting from excessive current melts the metal link in the fuse.
The power is generated by the source of applied voltage and consumed in the resistance in the form of heat. As much power as the resistance dissipates in heat must be supplied by the voltage source; otherwise, it cannot maintain the potential difference required to produce the current.
Any one of the three formulas can be used to calculate the power dissipated in a resistance. The one to be used is just a matter of convenience, depending on which factors are known.
In the following diagram, the power dissipated with 2 A through the resistance and 6 V across it is 2 x 6 = 12 W. Or, calculating in terms of just the current and resistance, we get 22 times 3, which equals 12 W. Using the voltage and resistance, the power can be calculated as 62 or 36, divided by 3, which also equals 12 W.
We have introduced a new schematic symbol here too. The schematic symbol of a battery is shown at the left. Note the small bar at the top is the negative terminal. The direction of current flow is shown correctly, from negative to positive. No matter which equation is used, 12 W of power is dissipated, in the form of heat. The battery must generate this amount of power continuously in order to maintain the potential difference of 6 V that produces the 2 A current against the opposition of 3 ohms.
In some applications, the electrical power dissipation is desirable because the component must produce heat in order to do its job. For instance, a 600 W toaster must dissipate this amount of power to produce the necessary amount of heat. Similarly, a 300 W light bulb must dissipate this power to make the filament white hot so that it will have the incandescent glow that furnishes the light. In other applications, however, the heat may be just an undesirable by-product of the need to provide current through the resistance in a circuit. In any case, though, whenever there is current in a resistance, it dissipates power equal to I2R.
The term I2R is used many times to describe unwanted resistive power losses in a circuit.
You will hear of the expression I2R losses as we go through this course
When current flows in a resistance, heat is produced because friction between the moving free electrons and the atoms obstructs the path of electron flow. The heat is evidence that power is used in producing current. This is how a fuse opens, as heat resulting from excessive current melts the metal link in the fuse.
The power is generated by the source of applied voltage and consumed in the resistance in the form of heat. As much power as the resistance dissipates in heat must be supplied by the voltage source; otherwise, it cannot maintain the potential difference required to produce the current.
Any one of the three formulas can be used to calculate the power dissipated in a resistance. The one to be used is just a matter of convenience, depending on which factors are known.
In the following diagram, the power dissipated with 2 A through the resistance and 6 V across it is 2 x 6 = 12 W. Or, calculating in terms of just the current and resistance, we get 22 times 3, which equals 12 W. Using the voltage and resistance, the power can be calculated as 62 or 36, divided by 3, which also equals 12 W.
We have introduced a new schematic symbol here too. The schematic symbol of a battery is shown at the left. Note the small bar at the top is the negative terminal. The direction of current flow is shown correctly, from negative to positive. No matter which equation is used, 12 W of power is dissipated, in the form of heat. The battery must generate this amount of power continuously in order to maintain the potential difference of 6 V that produces the 2 A current against the opposition of 3 ohms.
In some applications, the electrical power dissipation is desirable because the component must produce heat in order to do its job. For instance, a 600 W toaster must dissipate this amount of power to produce the necessary amount of heat. Similarly, a 300 W light bulb must dissipate this power to make the filament white hot so that it will have the incandescent glow that furnishes the light. In other applications, however, the heat may be just an undesirable by-product of the need to provide current through the resistance in a circuit. In any case, though, whenever there is current in a resistance, it dissipates power equal to I2R.
The term I2R is used many times to describe unwanted resistive power losses in a circuit.
You will hear of the expression I2R losses as we go through this course
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