A resistor is a device that converts electrical energy entirely into heat. We can say then that the resistor dissipates the electrical energy it receives from the circuit. Thus, the electrical power consumed by a resistor is dissipated.
As we know, this power is given by:
P = Ui (I)
U – is the potential difference (ddp)
i – is the electric current intensity
P – is the power dissipated
By Ohm’s law:
U = Ri ( II)
Where R is the electrical resistance of the resistor.
Substituting (II) into (I), we have:
P = (R . I) . i → P = R . i 2
The electrical energy transformed into thermal energy, at the end of a time interval ∆t, is given by:
This formula translates Joule’s law, which can be stated as follows:
“The electrical energy dissipated in a resistor, in a given time interval ∆t, is directly proportional to the square of the current intensity that flows through it.”
Being , the electrical power dissipated can also be given by:
We can then conclude that, when the ddp is constant, the electrical power dissipated in a resistor is inversely proportional to its electrical resistance.
In 0.5 kg of water contained in a container, a resistor of electrical resistance 2 Ω is immersed for 7 min.
Data: specific heat of water 1 cal/g °C and 1 cal = 4.2 J.
If an electric current of 5 A flows through the resistor, calculate the rise in temperature of the water, assuming there is no change of state.
The resistor transforms the electrical energy entirely into heat, heating the water.
Thus, we have:
E el = Q
We know that:
E el = P.∆t and Q = m . ç . ∆T
P.∆t = m . ç . ∆T
Ri 2 . ∆t = m. ç . ∆T
The data provided by the problem are:
R = 2 Ω
i = 5 A
∆t = 7min = 420 s
m = 0.5 kg = 500 g
c = 1 cal/g ºC = 4.2 J/g ºC
Replacing the given in the equation:
R . i 2.∆t = m . ç . ∆T
2 . 5 2 . 420 = 500 . 4.2. ∆T
∆T = 2 . 25 . 420
500 . 4.2
∆T = 21000
∆T = 10 ºC
Therefore, we can conclude that the temperature rises by 10 ºC.