A certain 13 V dry-cell battery, completely discharged, requires a current of 400 mA for 4.1 hr to completely recharge. What is the energy storage capacity of the battery, assuming the voltage does not depend on its charge status?
2007-09-04 04:31:35 · 2 answers · asked by ihaveaquestion 2 in Science & Mathematics ➔ Engineering
Power = voltage x current = enegy used/time
So Power to recharge the battery = 13 V x 0.4 Amps = 5.2 Watts
But 1 Watt = 1 Joule/second and Power = Enegry/time
It takes 4.1 hrs =14,760 seoncds to recharge so
Energy = 5.2 x 14,760 = 76,752 Joules
2007-09-04 04:44:27 · answer #1 · answered by nyphdinmd 7 · 1⤊ 0⤋
Here's another way to look at it:
400 mA * 4.1 hr = 1.64 Amp-hr (the usual way to state the capacity of a battery).
But, this is the equivalent to the answer above.
1.64 Amp-hr * 13 V = 21.32 Watt-hr
21.32 Watt-hr * 3600 sec/hr = 76752 Watt-second = 76752 Joule
.
2007-09-04 12:14:44 · answer #2 · answered by tlbs101 7 · 1⤊ 0⤋