Use the following formula:
P= R x I²
So, when we apply it: P= 300Ω x (0.02A)²
P= 0.12 W
So, power dissipated in the resistance, in this case, is 0.12W
Use the formula: P=IR (power = current x resistance).
No. Ohm's law relates voltage, current and resistance: V=IR. ("I" is the symbol for current.)
The power used, assuming Unity Power Factor (resistive load), is the product of resistance and the square of the current -- or 1210 Watts.
The relation is:P = I2RWhere:I is the current (for example, in amperes)R is the resistance (for example, in ohms)P is the power (energy per second) converted from electrical energy to heat. If the current is in amperes and the resistance in ohms, then power is in watts (equal to joules/second).
Ohms.
ohms is a measure of resistance(R) in a circuit. Watts is a measure of the power(P), in this case lets assume it is the power used by the resistive element (lamp, heater etc). Power(watts)=Current(Amps)x Current(amps) x Resistance(ohms) or Resistance (ohms)=Power(W)/(current x current)
When you increase voltage (V) then, to get the same total power (W), the current (I) must be decreased. This result comes from the Power Law: Power = voltage x current Ohms Law does not deal with power at all, it deals only with the relationship between voltage, resistance and current: Voltage = resistance x current
10.2 kilo ohms is the resistance necessary for 1 volt to induce a current of 98.04 micro amperes. Ohm's law: voltage equals current times resistance.
9 amperes.
Power = 1 A and resistance = 1100 Ohms.
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
Ohms do not relate to power per se. Ohms do however contribute how much power a circuit can deliver. In a given circuit the lower the resistance (measured in ohms) the higher the current & higher the power. This is assuming the voltage remains constant.
The apparent power (VA) is a quantity which applies to alternating current with a reactive component. It does not apply to DC and, with AC, the apparent power and real power are the same if there is only resistance present. With AC, other than just resistance there can be capacitors or inductors (coils) present. Perfect examples of these types of components do not dissipate power but do conduct current if AC voltage is applied. In your example, the resistive component is 30 ohms while the reactive impedance is 40 ohms meaning the total impedance is 50 ohms (Pythagoras). You can't just add reactance and impedance. The current through the circuit will therefore be 2.4 amps (I=V/R OR 120/50). The real or true power will be 172.8 watts ( I2 x R OR 2.42 x 30) while the apparent power will be 288 VA (2.42 x 50).
Ohms law states that E=I * R, or voltage equals current times resistance. Therefore current equals voltage divided by resistance. 120v divided by 16 ohms equals 7.5 amps.
The formula you are looking for is R = E/I
Use the formula: P=IR (power = current x resistance).
10.2 kilo ohms is the resistance necessary for 1 volt to induce a current of 98.04 micro amperes. Ohm's law: voltage equals current times resistance.