Question

A car battery with a 12 V emf and an internal resistance of 0.11 Ω is being charged with a current of 56 A. What are (a) the potential difference across the terminals, (b) the rate Pr of energy dissipation inside the battery and (c) the rate Pemf of energy conversion to chemical form? When the battery is used to supply 56 A to the starter motor, what are (d) V and (e) Pr

Answer 1

Part a)

V = 18.16 V

Part b)

`P_r = 345 Watt`

Part c)

P = 672 Watt

Part d)

V = 5.84 V

Part e)

`P_r = 345 Watt`

Step-by-step explanation:

Part a)

When battery is in charging mode

then the potential difference at the terminal of the cell is more than its EMF and it is given as

`\Delta V = E + i r`

here we have

`E = 12 V`

`i = 56 A`

`r = 0.11`

now we have

`\Delta V = 12 + (0.11)(56) = 18.16 V`

Part b)

Rate of energy dissipation inside the battery is the energy across internal resistance

so it is given as

`P_r = i^2 r`

`P_r = 56^2 (0.11)`

`P_r = 345 W`

Part c)

Rate of energy conversion into EMF is given as

`P_(emf) = i E`

`P_(emf) = (56)(12)`

`P_(emf) = 672 Watt`

Now battery is giving current to other circuit so now it is discharging

now we have

Part d)

`V = E - i r`

`V = 12 - (56)(0.11)`

`V = 12 - 6.16 = 5.84 V`

Part e)

now the rate of energy dissipation is given as

`P_r = i^2 r`

`P_r = 56^2 (0.11)`

`P_r = 345 W`