PROBLEM:

A typical 12-volts battery is rated according to "ampere-hrs". A 70-Ampere-hour battery for example, at a discharge rate of 3.5-Ampere has a life of 20 hrs.

Questions:
a.) assuming the voltage remains constant, obtain the energy and power delivered.

b.) Repeat for a discharge rate of 7.0 Ampere
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Please provide solution, so that i can figure out to arrive at the solution, Thanks in advance!

God Bless...

V=voltage, 12v

I=current, 3.5 amp.
T=time of discharge in seconds = 20 hours=20*3600 seconds

Energy delivered, = VIT joules

Same idea for other discharge rates.

To solve this problem, we need to understand the concepts of energy and power.

a.) To obtain the energy delivered by the battery, we can use the formula:

Energy (in watt-hours) = Voltage * Ampere-hours

In this case, the voltage is 12 volts and the Ampere-hours is 70. So, the energy delivered by the battery is:

Energy = 12 volts * 70 Ampere-hours = 840 watt-hours

To determine the power delivered by the battery, we can use the formula:

Power (in watts) = Energy / Time

The given discharge rate is 3.5 Amperes, and it has a life of 20 hours. Therefore, the power delivered by the battery is:

Power = 840 watt-hours / 20 hours = 42 watts

So, the battery delivers 840 watt-hours of energy and a power of 42 watts when the discharge rate is 3.5 Amperes.

b.) Now let's repeat the calculations for a discharge rate of 7.0 Amperes.

Using the same formula as before, the energy delivered by the battery is:

Energy = 12 volts * 70 Ampere-hours = 840 watt-hours

Next, for the power delivered:

Power = 840 watt-hours / 20 hours = 42 watts

The battery still delivers the same amount of energy, 840 watt-hours, but the power is the same as before, 42 watts.

In summary,

a.) At a discharge rate of 3.5 Amperes, the battery delivers 840 watt-hours of energy and a power of 42 watts.

b.) At a discharge rate of 7.0 Amperes, the battery still delivers 840 watt-hours of energy and a power of 42 watts.

I hope this explanation helps you understand how to arrive at the solution. If you have any further questions, feel free to ask!