Calculating Energy Delivered to a Battery using Ohm's Law

  • Thread starter Thread starter tigerguy
  • Start date Start date
  • Tags Tags
    Law Ohm's law
Click For Summary
SUMMARY

The discussion centers on calculating the energy delivered to a battery using Ohm's Law. A battery charger delivers a current of 7.5 A for 3.5 hours at a constant voltage of 12 V. The correct formula for energy is E = P · t = U · I · t, where U is voltage, I is current, and t is time in seconds. The energy delivered to the battery is calculated as E = 12 V · 7.5 A · (3.5 hours × 3600 seconds/hour) = 151,200 joules.

PREREQUISITES
  • Understanding of Ohm's Law
  • Basic knowledge of electrical units (voltage, current, resistance)
  • Familiarity with energy calculations in physics
  • Ability to convert time units (hours to seconds)
NEXT STEPS
  • Study the application of Ohm's Law in electrical circuits
  • Learn about energy calculations using E = P · t
  • Explore the relationship between current, voltage, and resistance
  • Investigate practical applications of battery charging and energy delivery
USEFUL FOR

Electrical engineers, physics students, and anyone involved in battery technology or energy calculations will benefit from this discussion.

tigerguy
Messages
32
Reaction score
0
A battery charger is connected to a dead battery and delivers a current of 7.5 A for 3.5 hours, keeping the voltage across the battery terminals at 12 V in the process. How much energy is delivered to the battery?

I'm not exactly sure what I am doing wrong. I understand that we have to use Ohm's Law in this question. So doing that, I took the given current, and converted it into the total number of coloumbs used over the 3.5 hours. I then calculated the resistance, and then I multiplied the total coloumbs by the resistance, and got joules (151,200). That's wrong, so I'm not sure where I went wrong.

I'd appreciate any help. Thanks.
 
Physics news on Phys.org
tigerguy said:
A battery charger is connected to a dead battery and delivers a current of 7.5 A for 3.5 hours, keeping the voltage across the battery terminals at 12 V in the process. How much energy is delivered to the battery?

I'm not exactly sure what I am doing wrong. I understand that we have to use Ohm's Law in this question. So doing that, I took the given current, and converted it into the total number of coloumbs used over the 3.5 hours. I then calculated the resistance, and then I multiplied the total coloumbs by the resistance, and got joules (151,200). That's wrong, so I'm not sure where I went wrong.

I'd appreciate any help. Thanks.

use E = P \cdot t = U \cdot I \cdot t with t in seconds.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
4
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 7 ·
Replies
7
Views
7K
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K