Power = I2 R = (0.02)2 x (1,000) = 0.4 watt
Power dissipated = I2R 0.022 x 1000 = 0.4 watts
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.
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).
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
The power dissipated across a resistor, or any device for that matter, is watts, or voltage times current. If you don't know one of voltage or current, you can calculate it from Ohm's law: voltage equals resistance times current. So; if you know voltage and current, power is voltage times current; if you know voltage and resistance, watts is voltage squared divided by resistance; and if you know current and resistance, watts is current squared times resistance.
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
Voltage times current. You obtain current from the division of voltage and resistance, so: I[A] = U[V] / R[ohm] and P[W] = U[V] * I[A] it follows, that P[W] = U[V] * (U[V] / R[ohm]) = U[V] ^ 2 * R[ohm] So, voltage squared divided by resistance will give you the power that will be dissipated in a resistor. Whether the resistor will take that abuse is up to its power dissipation rating, however.
The power dissipated by a resistance 'R' carrying a current 'I' is [ I2R ]. The power is dissipated as heat, and you can see from [ I2R ] that for a given current, it's directly proportional to 'R'.
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.
I = E / RIf the voltage across the resistor is 90 volts, and the resistance of the resistoris 9 ohms, then the current through the resistor is90/9 = 10 Amperes.Don't try this at home!The power dissipated by the resistor is E2/R = (90)2/9 = 900 watts. That's comparable to the power (heat) dissipated by a small toaster. A common composition resistor will get hot and possibly explode if it's asked to dissipate that kind of power.
.205 watts or 205 mw