If the battery is charged at an average constant voltage of 13 V, and discharged at an average

If the battery is charged at an average constant voltage of...

Key Concepts

Voltaic Efficiency

Voltaic efficiency is a concept that measures the ratio of the voltage during discharge to the voltage during charge of an electrochemical cell or battery. It provides an indicator of how much of the input energy stored during charging is effectively delivered during discharging, accounting for inherent losses in the system.

Battery Charging and Discharging

The process of battery charging involves applying an external voltage to store energy, while discharging is the release of that stored energy through conversion into electrical energy. Understanding these processes is crucial because differences between the charging and discharging voltages indicate energy losses due to factors such as internal resistance and overpotentials.

Transcript

00:01 So in this problem, we're trying to determine the efficiency of a rechargeable battery.

00:07 So what that's going to be is we're going to be comparing the energy delivered to the load by the rechargeable battery, divided by the energy delivered to the battery by the power supply.

00:19 So we're going to start by looking at the scenario where the battery is charging, and we want to find the current that's going through the circuit as it's charging.

00:28 So we know that the power supply charging the battery is 14 .7 volts.

00:33 So we can go ahead and take that emf, and we can equate it to the emf of the battery, 13 .2.

00:43 And we also have to add in the potential difference across this internal resistance in the battery.

00:49 So we have the charging current times 0 .85 oms.

00:57 So now that we have this expression, we can go.

01:00 Ahead and solve for the charging current which will give us a value of 1 .76 amps.

01:09 And now that we have the charging current we can go ahead and find the energy that's delivered by the power supply.

01:16 And we're going to go ahead and do that by using the expression power times time and we can rewrite the power as the voltage or rather the emf of the power supply.

01:36 Times the charging current, times the time which it's charging.

01:43 And we're given the time in hours, so what i'm going to do is just add this conversion factor where we have 3 ,600 seconds over one hour so that our units are correct.

01:57 We'll find that the energy delivered by the power supply is going to be 168 kilojoules.

02:06 So we now have one part of, the efficiency that we're looking for.

02:13 So what we can do now is we can go ahead find the energy stored in the battery.

02:19 So in order to find that, we're going to go ahead use the same equation that we used here.

02:25 This time we're using the emf of the battery, times the charging current times the time at which it's charging.

02:33 And again, i'll add this conversion factor to seconds.

02:37 So when we go ahead and plug in the values that we know, what we're going to find is that the energy stored in the battery is 151 kilojoules.

02:50 And since we know that the energy that's stored in the battery is going to be equal to the energy that's released by the battery, what we can do is we can set up an equation in order to find the discharging current.

03:03 And we'll do that by using the same equation for energy rate here.

03:07 So we'll take 151 kilojoules as the total energy that's being released during discharge...

Please give Ace some feedback