Answered

An unknown resistor is connected between the terminals of a 12.0 V battery. Energy is dissipated in the resistor at the rate of 0.630 W. The same resistor is then connected between the terminals of a 10.5 V battery. At what rate is energy now dissipated?

Answer :

Answer:

The energy is dissipated in the resistor at the rate of 0.482 W

Explanation:

Given :

Potential difference resistor [tex]V_{1} =[/tex] 12 V

Energy dissipated in the resistor [tex]P_{1} =[/tex] 0.630 W

New potential difference [tex]V_{2} =[/tex] 10.5 V

We know that energy dissipated per unit time through resistor is given by,

⇒    [tex]P = VI = \frac{V^{2} }{R}[/tex]

Rate of energy dissipated is proportional to the square of the applied potential.

So for new potential we can write,

     [tex]\frac{P_{2} }{P_{1} } = \frac{V_{2}^{2} }{V_{1} ^{2} }[/tex]

    [tex]P_{2} = \frac{0.63 \times 110.25}{144}[/tex]

    [tex]P_{2} = 0.482[/tex] W

So new rate of energy dissipation is 0.482 W

Other Questions