Comments (3) on “Calculate the resistance of a 7 W LED (actual power consumption) bulb operating with a 62.0A current.”
Answerthe energy required is 1.60 × 10⁻¹⁸ j, to three significant figures.explanationwe define electric potential as the ratio electric potential energy to the charge, [tex]v = \dfrac{e_p}{q}[/tex]then the following gives us the potential difference, [tex]\delta v = v_b - v_a[/tex] [tex]\begin{aligned} \delta v & = \dfrac{e_p_b}{q} - \dfrac{e_p_a}{q} \\ & = \dfrac{\delta e_p}{q} \end{aligned}[/tex]the formula [tex]\delta v = \delta e_p / q[/tex] may be on your formula sheet.the change in electric potential energy would be the energy that would needed to move it from point a to point b. solving for that, we get [tex]\delta e_p = q\delta v[/tex]using q = 8.00 × 10⁻¹⁹ c and δv = 2.00 v, we get [tex]\begin{aligned} \delta e_p & = (8.00\times10^{-19} \text{ c})(2.00\text{ v}) \\ & = 1.60\times 10^{-18}\text{ j} \end{aligned}[/tex]the energy required is 1.60 × 10⁻¹⁸ j, to three significant figures.
Answerthe energy required is 1.60 × 10⁻¹⁸ j, to three significant figures.explanationwe define electric potential as the ratio electric potential energy to the charge, [tex]v = \dfrac{e_p}{q}[/tex]then the following gives us the potential difference, [tex]\delta v = v_b - v_a[/tex] [tex]\begin{aligned} \delta v & = \dfrac{e_p_b}{q} - \dfrac{e_p_a}{q} \\ & = \dfrac{\delta e_p}{q} \end{aligned}[/tex]the formula [tex]\delta v = \delta e_p / q[/tex] may be on your formula sheet.the change in electric potential energy would be the energy that would needed to move it from point a to point b. solving for that, we get [tex]\delta e_p = q\delta v[/tex]using q = 8.00 × 10⁻¹⁹ c and δv = 2.00 v, we get [tex]\begin{aligned} \delta e_p & = (8.00\times10^{-19} \text{ c})(2.00\text{ v}) \\ & = 1.60\times 10^{-18}\text{ j} \end{aligned}[/tex]the energy required is 1.60 × 10⁻¹⁸ j, to three significant figures.
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Your answer is 32 hope i