Increase the voltage across the resistor by 41.4% .
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
real power (as opposed to imaginary power, which is not dissipated)
Power dissipated = I2R 0.022 x 1000 = 0.4 watts
You may find it helpful to use Ohm's law and the definition of electrical power.
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.
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
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).
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.
.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
AS:total power = P1 + P2 + P3 + .......so,total power = 5(50mW)= 0.25Wtotal power dissipated by the five resistors is 0.25W.
You may find it helpful to use Ohm's law and the definition of electrical power.
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.