A car battery with a 12 V emf and an internal resistance of 0.037 Ω is being charged with a current of 45 A. Note that in this process, the battery is being charged.
(a) What is the potential difference (in V) across its terminals?
(b) At what rate is thermal energy being dissipated in the battery (in W)?
(c) At what rate is electric energy being converted to chemical energy (in W)?
(d) What is Pr in this case?

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aristocles aristocles · Answer 1 · 2019-11-17T05:20:57+01:00

Answer:

Part a)

[tex]V = 13.665 V[/tex]

Part b)

[tex]P = 74.9 Watt[/tex]

Part c)

[tex]P = 540 watt[/tex]

Part d)

[tex]P_r = 0.878[/tex]

Explanation:

Part a)

As we know that the during the charging process of the battery the terminal voltage of the cell is given as

[tex]V = E + iR[/tex]

[tex]V = 12 + (0.037)(45)[/tex]

[tex]V = 13.665 V[/tex]

Part b)

Thermal energy dissipated in the battery is due to its internal resistance

so it is given as

[tex]P = i^2 R[/tex]

here we have

[tex]P = 45^2 (0.037)[/tex]

[tex]P = 74.9 Watt[/tex]

Part c)

rate of energy conversion in the in the battery is given as

[tex]P = E. i[/tex]

[tex]P = 12(45)[/tex]

[tex]P = 540 watt[/tex]

Part d)

percentage of the power conversion is given as

[tex]Pr = \frac{P_{out}}{P_{total}}[/tex]

[tex]Pr = \frac{540}{540 + 74.9}[/tex]

[tex]P_r = 0.878[/tex]