Batteries are recharged by connecting them to a power supply (i.e., another batt

Question

Batteries are recharged by connecting them to a power supply (i.e., another battery) of greater emf in such a way that the current flows into the positive terminal of the battery being recharged. This reverse current through the battery replenishes its chemicals. The current is kept fairly low so as not to overheat the battery being recharged by dissipating energy in its internal resistance.

Suppose the real battery of the figure is rechargeable. What emf power supply should be used for a 0.85 A recharging current?

If this power supply charges the battery for 10 minutes, how much energy goes into the battery?

How much is dissipated as thermal energy in the internal resistance?

Explanation / Answer

Current i = 0.85 A

Internal resistance r = 1 ohm

Emf of the battery being charged E = 1.5 volt

From Ohm's law voltage across internal resistance V = ir

V = 0.85x1 = 0.85 volt

From Kirchoff's second law (Voltage law),

E + V = Emf of the power supply doing the charging

1.5 volt + 0.85 volt = E '

So,emf power supply should be used for a 0.85 A recharging current E ' = 2.35 volt

(b).Power delivered to battery P = E ' i

= 2.35 x0.85

= 1.9975 watt

power supply charges the battery for 10 minutes then energy goes into the battery

U = power x time

= 1.9975 watt x 10 min

= 1.9975 watt x 10x60 s

= 1198.5 J

(c). Power dissipated in internal resistance P ' = V i

P ' = 0.85 x 0.85

= 0.7225 watt

dissipated as thermal energy in the internal resistance U ' = P ' t = 0.7225 wattx 600 s

U ' = 433.5 J

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