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.

http://session.masteringphysics.com/problemAsset/1200626/4/P23-56.jpg

Suppose the real battery of the figure is rechargeable. What emf power supply should be used for a 0.95 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?

thanks!

Explanation / Answer

a.

What will the voltage across the 1 ohm internal resistance be if 0.95 A is flowing through it?

Using Ohm's Law v = i * r

v = 0.95 * 1
v = 0.95 volts

Using Kirchhoff's voltage law, [emf of battery being charged] + [voltage across internal resistance] = [emf of the power supply doing the charging]

1.5 v + 0.95 v = 2.45 volts <<<<<<<<<<<<<<<<<<

b.

Power delivered to battery = v * i
= 2.1 * 0.95
= 1.995 W
Energy = power * time [in seconds]
10 minutes = 600 s
Energy = 1.995 W * 600 s
= 1197 J <<<<<<<<<<<<<<<<<<<<

Power dissipated in internal resistance = v * i
= 0.95 V * 0.95 A
= 0.9025 W

Again in 600 s, that's
=.9025 W * 600 s = 541.5 J <<<<<<<<<<<<<<<<<<<<

[So that leaves 1197 - 541.5 = 655.5 J actually stored in the battery ]

Related Questions

Navigate

• Browse (All)
• Browse B
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.