The fundamental equation for the power of any load is the product of the voltage across the load and the current through it: P = U I.
Since voltage is the product of current and resistance (U = I R), we can substitute for voltage in the original equation:
P = U I = (IR) I = I2R
The formula you are looking for is W = E(squared)/R, W = Voltage (squared) divided by Resistance.
There is insufficient information in the question to properly answer it. You need to also know the voltage or, alternatively, the power in order to find the resistance given the current. Please restate the question.
Power dissipation of a resistor or any load is the amount of power (in watts) that is converted to heat, light, or other form of energy. In a resistor, power dissipation is defined by Ohm's law P = I^2 * R Power dissipated equals current through the resistor squared times the resistance in ohms. Since the power is converted to heat, a resistor has a maximum dissipation rating set by the manufacturer, above which the resistor will be damaged.
Power=current squared times resistance
Okay. Resistance by ohms law is given by R = V/I But Power P = V * I Dividing R/P = 1/ I 2 Or R = P / I squared For a constant power, resistance is inversely proportional to I squared and not simply proportional to.
For example, you can write statements based on:* Ohm's Law: V = IR (voltage = current x resistance) * Power dissipation: P = I squared R
The formula you are looking for is W = E(squared)/R, W = Voltage (squared) divided by Resistance.
P = I2R (power = current squared times resistance). Therefore, if the current doubles, the amount of dissipated electrical energy will increase by a factor of 4.P = I2R (power = current squared times resistance). Therefore, if the current doubles, the amount of dissipated electrical energy will increase by a factor of 4.P = I2R (power = current squared times resistance). Therefore, if the current doubles, the amount of dissipated electrical energy will increase by a factor of 4.P = I2R (power = current squared times resistance). Therefore, if the current doubles, the amount of dissipated electrical energy will increase by a factor of 4.
power in watts = voltage in volts x current in amps. or power in watts = current in amps x (resistance in ohms) squared i think what you meant was power in watts =(current in amps)squared x resistance in ohms
Copper losses are purely voltage-drop losses (I squared R) caused by the resistance of the windings, as opposed to hysteresis losses and eddy current losses (so-called iron losses), which are magnetic in nature. They are called copper losses whether the winding conductors are made of copper or not, by the way.
a squared + b squared = c squared
Energy dissipated/used by a load =(voltage across the load) x (current through it) x (time)(voltage squared) x (time) divided by (load impedance/resistance)(current squared) x (load impedance/resistance) x (time)(power) x (time)
Side length squared is the formula for finding the area of a square.
x2+y2
a squared + b squared=c squared
(watts equal) voltage times current (e x I ) 120 x 20 resistance times current squared (r x I squared) 6 ohms x20 squared voltage squared divided by resistance (E squared divided by resistance) 120squared divided by 6 check OHMS LAW,
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.