A student places a potential difference $V$ of $(4.0 \pm 0.1)\,\text{V}$ across a resistor with resistance $R$ of $(10.0 \pm 0.3)\,\Omega$ for a duration $t$ of $(50 \pm 1)\,\text{s}$. The student then finds the energy $E$ dissipated by using the equation shown below: $$E = \frac{V^2 t}{R} = \frac{4.0^2 \times 50}{10.0} = 80\,\text{J}.$$ What is the absolute uncertainty in the energy value calculated?
- A$1.5\,\text{J}$
- B$3\,\text{J}$
- C$6\,\text{J}$
- D$8\,\text{J}$