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.
No power is dissipated by a load composed exclusively of either capacitive or inductive reactance.
Increase the voltage across the resistor by 41.4% .
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.
real power (as opposed to imaginary power, which is not dissipated)
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. 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.
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.
.205 watts or 205 mw
I = 2A R = 1000Ω Power Dissipated P = I2R = (2A)2(1000Ω) = 4000W Voltage across resistor V = IR = (2A)(1000Ω) = 2000V
P = I^2 x R] P = 0.2^2 x 100 P = 4 W
Power dissipated = I2R 0.022 x 1000 = 0.4 watts
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