Question

The charge flowing through resistor r=2ohm with time as q=3-4t^2. The heat produces in 2 second is about ?

Answer 1

Answer:

the amount of heat produced for 2s of time is 341.3J.

Explanation:

Given the resistance of the resistor, $R=2\Omega$

The charge is given by, $q=3-4t^2$

Electric current is defined as the rate of flow of charge.

This can be written as,

$I=\dfrac{dq}{dt}$

Substituting the charge and differentiating with respect to time,

$I=\dfrac{d(3-4t^2)}{dt}=-8t$

According Joule's law of heating, the amount of heat produced in a current carrying conductor in a time t is given by

$\dfrac{dH}{dt} =I^{2} R\implies dH=I^{2} Rdt$

Integrating on both sides and substituting the respective values

$\int dH = \int\limits^2_0 (-8t)^{2} \times 2dt$

$\implies H=128 \int\limits^2_0 t^{2} dt$

          $=128\bigg[ \dfrac{t^{3}}{3}\bigg]\limits^2_0 =128\times \dfrac{8}{3}=341.3J$

Therefore, the amount of heat produced for 2s of time is 341.3J.

So, the correct answer is option 2.

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